Question 513 Marks
Shabana's age 10 years hence, will be twice juhi's present age. 6 years back shabana's age was $\frac{5}{3}$ times Juhi's at that time find their present ages.
Answer
View full question & answer→26 years, 18 years
Questions · Page 2 of 3
3 Marks Question
Question 523 Marks
$64 m-45 n=289 ; 45 m-64 n=365$
Answer
View full question & answer→$m=1, n=-5$
Question 533 Marks
$\frac{1}{3} x+\frac{1}{4} y=4 ; \frac{5}{6} x-\frac{1}{8} y=4$
Answer
View full question & answer→$x=6, y=8$
Question 543 Marks
$\frac{1}{3} x+5 y=13 ; 2 x+\frac{1}{2} y=19$
Answer
View full question & answer→$x=9, y=2$
Question 553 Marks
$47 x+31 y=63 ; 31 x+47 y=15$
Answer
View full question & answer→$x=2, y=-1$
Question 563 Marks
Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Answer
View full question & answer→Let the speed of $Joseph = x km / h$
Let the speed of Hamid be $=y km / h$
When approaching each other, combined speed $=(x+y) km / h$
Time taken to meet $=\frac{30}{x+y}=\frac{1}{3}(20 mins )$
$\therefore x + y =90 \ldots I$
When moving away from each other, combined speed $=(x-y) km / h$
Time taken for Hamid to catch up $=\frac{30}{x-y}=3$
$\therefore x - y =10 \ldots II$
Equating I and II,
$\begin{aligned}
& x+y=90 \\
& x-y=10 \\
& 2 x=100 \\
& x=\frac{100}{2}=50
\end{aligned}$
Substituting $x=50$ in eq. I
$50+y=90$
$\begin{aligned}
& y=90-50 \\
& y=40
\end{aligned}$
Hamid's speed $50 km / hr$.
Joseph's speed $40 km / hr$.
Let the speed of Hamid be $=y km / h$
When approaching each other, combined speed $=(x+y) km / h$
Time taken to meet $=\frac{30}{x+y}=\frac{1}{3}(20 mins )$
$\therefore x + y =90 \ldots I$
When moving away from each other, combined speed $=(x-y) km / h$
Time taken for Hamid to catch up $=\frac{30}{x-y}=3$
$\therefore x - y =10 \ldots II$
Equating I and II,
$\begin{aligned}
& x+y=90 \\
& x-y=10 \\
& 2 x=100 \\
& x=\frac{100}{2}=50
\end{aligned}$
Substituting $x=50$ in eq. I
$50+y=90$
$\begin{aligned}
& y=90-50 \\
& y=40
\end{aligned}$
Hamid's speed $50 km / hr$.
Joseph's speed $40 km / hr$.
Question 573 Marks
In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Answer
View full question & answer→Ratio of skilled and unskilled worker's salary $=5: 3$
Let it be $5 x$ and $3 x$
Total of one day's salary $=₹ 720$
$\begin{aligned}
& \text { So, } 5 x+3 x=720 \\
& 8 x=720 \\
& x=\frac{720}{8} \\
& x=90
\end{aligned}$
Skilled worker's wages $=5 x =5 \times 90=₹ 450$.
unskilled worker's wages $3 x =3 \times 90=₹ 270$
Let it be $5 x$ and $3 x$
Total of one day's salary $=₹ 720$
$\begin{aligned}
& \text { So, } 5 x+3 x=720 \\
& 8 x=720 \\
& x=\frac{720}{8} \\
& x=90
\end{aligned}$
Skilled worker's wages $=5 x =5 \times 90=₹ 450$.
unskilled worker's wages $3 x =3 \times 90=₹ 270$
Question 583 Marks
View full question & answer→Question 593 Marks
Answer
View full question & answer→Let $\frac{1}{3 x+4 y}=m$ and $\frac{1}{2 x-3 y}=n$
$\begin{aligned}
& \frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \ldots (1) \\
& 5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots (2)
\end{aligned}$
Equating Eq. I and II
$\begin{aligned}
& 10 m+4 n=5 \\
& 10 m-4 n=-3 \\
& 20 m=2
\end{aligned}$
$\begin{aligned}
& m =\frac{2}{20} \\
& m =\frac{1}{10}
\end{aligned}$
Substituting $m =\frac{1}{10}$ in Eq. I
$\begin{aligned}
& 10 \times \frac{1}{10}+4 n=5 \\
& 1+4 n=5 \\
& 4 n=5-1
\end{aligned}$
4n=4
$\begin{aligned}
& n =\frac{4}{4} \\
& n =1 \\
& \quad \frac{1}{3 x +4 y }= m \Rightarrow \frac{1}{3 x +4 y }=\frac{1}{10} \Rightarrow 3 x +4 y =10 \ldots (3) \\
& \therefore \frac{1}{2 x -3 y }= n \Rightarrow \frac{1}{2 x -3 y }=1 \Rightarrow 2 x -3 y =1 \ldots (4)
\end{aligned}$
Multiply Eq. III by 3 and Eq. IV by 4 and Equate
$\begin{aligned}
& 9 x+12 y=30 \\
& 8 x-12 y=4 \\
& 17 x=34 \\
& x=\frac{34}{17} \\
& x=2
\end{aligned}$
Substituting $x=2$ in Eq. $V$
$\begin{aligned}
& 9 \times 2+12 y=30 \\
& 18+12 y=30 \\
& 12 y=30-18
\end{aligned}$
$\begin{aligned}
& 12 y=12 \\
& y=\frac{12}{12} \\
& y=1
\end{aligned}$
Hence, $(x, y)=(2,1)$
$\begin{aligned}
& \frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \ldots (1) \\
& 5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots (2)
\end{aligned}$
Equating Eq. I and II
$\begin{aligned}
& 10 m+4 n=5 \\
& 10 m-4 n=-3 \\
& 20 m=2
\end{aligned}$
$\begin{aligned}
& m =\frac{2}{20} \\
& m =\frac{1}{10}
\end{aligned}$
Substituting $m =\frac{1}{10}$ in Eq. I
$\begin{aligned}
& 10 \times \frac{1}{10}+4 n=5 \\
& 1+4 n=5 \\
& 4 n=5-1
\end{aligned}$
4n=4
$\begin{aligned}
& n =\frac{4}{4} \\
& n =1 \\
& \quad \frac{1}{3 x +4 y }= m \Rightarrow \frac{1}{3 x +4 y }=\frac{1}{10} \Rightarrow 3 x +4 y =10 \ldots (3) \\
& \therefore \frac{1}{2 x -3 y }= n \Rightarrow \frac{1}{2 x -3 y }=1 \Rightarrow 2 x -3 y =1 \ldots (4)
\end{aligned}$
Multiply Eq. III by 3 and Eq. IV by 4 and Equate
$\begin{aligned}
& 9 x+12 y=30 \\
& 8 x-12 y=4 \\
& 17 x=34 \\
& x=\frac{34}{17} \\
& x=2
\end{aligned}$
Substituting $x=2$ in Eq. $V$
$\begin{aligned}
& 9 \times 2+12 y=30 \\
& 18+12 y=30 \\
& 12 y=30-18
\end{aligned}$
$\begin{aligned}
& 12 y=12 \\
& y=\frac{12}{12} \\
& y=1
\end{aligned}$
Hence, $(x, y)=(2,1)$
Question 603 Marks
Solve the following simultaneous equations.
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
Answer
$\begin{aligned} & \frac{7 x-2 y}{x y}=5 \Rightarrow \frac{7 x}{x y}-\frac{2 y}{x y}=5 \Rightarrow \frac{7}{y}-\frac{2}{x}=5 \ldots \text { (I) } \\ & \frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \ldots \text { (II) } \\ & \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\ & 7 n-2 m=5 \ldots \text { (III) } \\ & 8 n+7 m=15 \ldots \text { (IV) } \\ & \text { Multiply Eq. } 1 \text { by } 7 \text { and Eq.II by } 2 \\ & 49 n-14 m=35 \ldots \text { (V) } \\ & 16 n+14 m=30 \ldots \text { (VI) } \\ & 65 n=65 \\ & n=\frac{65}{65} \\ & n=1 \text { Substituting value in Eq.VI } \\ & 16 \times 1+14 m=30 \\ & 14 m=30-16 \\ & 14 m=14\end{aligned}$
$\begin{aligned}
& m =\frac{14}{14} \\
& m =1 \\
& \therefore \frac{1}{ x }= m \Rightarrow \frac{1}{ x }=1 \Rightarrow x =1 \\
& \quad \frac{1}{ y }= n \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1
\end{aligned}$
View full question & answer→$\begin{aligned} & \frac{7 x-2 y}{x y}=5 \Rightarrow \frac{7 x}{x y}-\frac{2 y}{x y}=5 \Rightarrow \frac{7}{y}-\frac{2}{x}=5 \ldots \text { (I) } \\ & \frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \ldots \text { (II) } \\ & \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\ & 7 n-2 m=5 \ldots \text { (III) } \\ & 8 n+7 m=15 \ldots \text { (IV) } \\ & \text { Multiply Eq. } 1 \text { by } 7 \text { and Eq.II by } 2 \\ & 49 n-14 m=35 \ldots \text { (V) } \\ & 16 n+14 m=30 \ldots \text { (VI) } \\ & 65 n=65 \\ & n=\frac{65}{65} \\ & n=1 \text { Substituting value in Eq.VI } \\ & 16 \times 1+14 m=30 \\ & 14 m=30-16 \\ & 14 m=14\end{aligned}$
$\begin{aligned}
& m =\frac{14}{14} \\
& m =1 \\
& \therefore \frac{1}{ x }= m \Rightarrow \frac{1}{ x }=1 \Rightarrow x =1 \\
& \quad \frac{1}{ y }= n \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1
\end{aligned}$
Question 613 Marks
Solve the following simultaneous equations.
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
Answer
$\begin{aligned}
& \frac{148}{x}+\frac{231}{y}=\frac{527}{x y} \Rightarrow \frac{148 y+231 x}{x y}=\frac{527}{x y} \Rightarrow 231 x+148 y=527 \ldots(1) \\
& \frac{231}{x}+\frac{148}{y}=\frac{610}{x y} \Rightarrow \frac{231 y+148 x}{x y}=\frac{610}{x y} \Rightarrow 148 x+231 y=610 \ldots(2)
\end{aligned}$
Adding Eq. I and II
$\begin{aligned}
& 379 x+379 y=1137 \\
& x+y=3 \ldots \text { (III) }
\end{aligned}$
Subtracting Eq. I and II
$\begin{aligned}
& 83 x-83 y=-83 \\
& x-y=-1 \ldots \text { (IV) }
\end{aligned}$
Equating I and II
$\begin{aligned}
& x+y=3 \\
& x-y=-1 \\
& 2 x=2 \\
& x=\frac{2}{2} \\
& x=1
\end{aligned}$
Substituting $x =1$ in Eq. I
$\begin{aligned}
& 1+y=3 \\
& y=3-1 \\
& y=2
\end{aligned}$
Hence, $(x, y)=(1,2)$
View full question & answer→$\begin{aligned}
& \frac{148}{x}+\frac{231}{y}=\frac{527}{x y} \Rightarrow \frac{148 y+231 x}{x y}=\frac{527}{x y} \Rightarrow 231 x+148 y=527 \ldots(1) \\
& \frac{231}{x}+\frac{148}{y}=\frac{610}{x y} \Rightarrow \frac{231 y+148 x}{x y}=\frac{610}{x y} \Rightarrow 148 x+231 y=610 \ldots(2)
\end{aligned}$
Adding Eq. I and II
$\begin{aligned}
& 379 x+379 y=1137 \\
& x+y=3 \ldots \text { (III) }
\end{aligned}$
Subtracting Eq. I and II
$\begin{aligned}
& 83 x-83 y=-83 \\
& x-y=-1 \ldots \text { (IV) }
\end{aligned}$
Equating I and II
$\begin{aligned}
& x+y=3 \\
& x-y=-1 \\
& 2 x=2 \\
& x=\frac{2}{2} \\
& x=1
\end{aligned}$
Substituting $x =1$ in Eq. I
$\begin{aligned}
& 1+y=3 \\
& y=3-1 \\
& y=2
\end{aligned}$
Hence, $(x, y)=(1,2)$
Question 623 Marks
Solve the following simultaneous equations.
$\frac{7}{2 x +1}+\frac{13}{ y +2}=27 ; \frac{13}{2 x +1}+\frac{7}{ y +2}=33$
$\frac{7}{2 x +1}+\frac{13}{ y +2}=27 ; \frac{13}{2 x +1}+\frac{7}{ y +2}=33$
Answer
$\begin{aligned}
& \text { Let } \frac{1}{2 x+1}=m \text { and } \frac{1}{y+2}=n \\
& 7 m+13 n=27 \ldots \text { (I) } \\
& 13 m+7 n=33 \ldots \text { (II) }
\end{aligned}$Adding Eq. I and II
$20 m +20 n =60 \Rightarrow m + n =3 \ldots(III)$
Subtract Eq. I and II
$-6 m+6 n=-6 \Rightarrow-m+n=-1 \ldots(IV)$
Equating Eq. III and IV
$\begin{aligned}
& m+n=3 \\
& -m+n=-1 \\
& \hline 2 n=2 \\
& n=1
\end{aligned}$
Substituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore \frac{1}{2 x +1}= m \Rightarrow \frac{1}{2 x +1}=2 \Rightarrow 2(2 x +1)=1 \Rightarrow 4 x +2=1 \Rightarrow 4 x =1-2 \\
& \Rightarrow 4 x =-1 \Rightarrow x =-\frac{1}{4} \\
& \therefore \frac{1}{ y +2}= n \Rightarrow \frac{1}{ y +2}=1 \Rightarrow y +2=1 \Rightarrow y =1-2 \Rightarrow y =-1
\end{aligned}$
Hence, $(x, y)=\left(-\frac{1}{4},-1\right)$
View full question & answer→$\begin{aligned}
& \text { Let } \frac{1}{2 x+1}=m \text { and } \frac{1}{y+2}=n \\
& 7 m+13 n=27 \ldots \text { (I) } \\
& 13 m+7 n=33 \ldots \text { (II) }
\end{aligned}$Adding Eq. I and II
$20 m +20 n =60 \Rightarrow m + n =3 \ldots(III)$
Subtract Eq. I and II
$-6 m+6 n=-6 \Rightarrow-m+n=-1 \ldots(IV)$
Equating Eq. III and IV
$\begin{aligned}
& m+n=3 \\
& -m+n=-1 \\
& \hline 2 n=2 \\
& n=1
\end{aligned}$
Substituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore \frac{1}{2 x +1}= m \Rightarrow \frac{1}{2 x +1}=2 \Rightarrow 2(2 x +1)=1 \Rightarrow 4 x +2=1 \Rightarrow 4 x =1-2 \\
& \Rightarrow 4 x =-1 \Rightarrow x =-\frac{1}{4} \\
& \therefore \frac{1}{ y +2}= n \Rightarrow \frac{1}{ y +2}=1 \Rightarrow y +2=1 \Rightarrow y =1-2 \Rightarrow y =-1
\end{aligned}$
Hence, $(x, y)=\left(-\frac{1}{4},-1\right)$
Question 633 Marks
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
Answer
Let $\frac{1}{x}=m$ and $\frac{1}{y}=n$
$2 m+\frac{2}{3} n=\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n}=1 \Rightarrow 12 m+4 n=1 \ldots(1)$
$3 m+2 n=0 \ldots(2)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots(3)$
Subtract Eq.III from Eq. I
$\begin{aligned}
& 12 m+4 n=1 \\
& -6 m-4 n=0 \\
& \hline 6 m=1 \\
& m=\frac{1}{6}
\end{aligned}$
Substitute $m=1 / 6$ in Eq. I
$\begin{aligned}
& 12 \times \frac{1}{6}+4 n=1 \\
& 2+4 n=1 \\
& 4 n=1-2
\end{aligned}$
$\begin{aligned}
& 4 n =-1 \\
& n =-\frac{1}{4} \\
& \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 \\
& \therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4
\end{aligned}$
Hence, $(x, y)=(6,-4)$
View full question & answer→Let $\frac{1}{x}=m$ and $\frac{1}{y}=n$
$2 m+\frac{2}{3} n=\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n}=1 \Rightarrow 12 m+4 n=1 \ldots(1)$
$3 m+2 n=0 \ldots(2)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots(3)$
Subtract Eq.III from Eq. I
$\begin{aligned}
& 12 m+4 n=1 \\
& -6 m-4 n=0 \\
& \hline 6 m=1 \\
& m=\frac{1}{6}
\end{aligned}$
Substitute $m=1 / 6$ in Eq. I
$\begin{aligned}
& 12 \times \frac{1}{6}+4 n=1 \\
& 2+4 n=1 \\
& 4 n=1-2
\end{aligned}$
$\begin{aligned}
& 4 n =-1 \\
& n =-\frac{1}{4} \\
& \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 \\
& \therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4
\end{aligned}$
Hence, $(x, y)=(6,-4)$
Question 643 Marks
Solve the following equations by Cramer’s method.
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
Answer
View full question & answer→Let,
$\begin{aligned}
& \frac{x+y-8}{2}=\frac{x+2 y-14}{3} \\
& \Rightarrow 3 x+3 y-24=2 x+4 y-28 \\
& \Rightarrow x-y=-4 \ldots \text { (1) }
\end{aligned}$
Also,
$\begin{aligned}
& \text { Let } \frac{x+2 y-14}{3}=\frac{3 x-y}{4} \\
& \Rightarrow 4 x+8 y-56=9 x-3 y \\
& \Rightarrow 5 x-11 y=-56 \ldots \text { (2) }
\end{aligned}$
Hence the two equations are:
$\begin{aligned}
& x-y=-4 \ldots(1) \\
& 5 x-11 y=-56
\end{aligned}$
Now,
$\begin{aligned}
& D=\left|\begin{array}{cc}
1 & -1 \\
5 & -11
\end{array}\right| \\
& \Rightarrow D=(-11-(-5))=-6
\end{aligned}$
Also, $D _{ x }=\left|\begin{array}{cc}-4 & -1 \\ -56 & -11\end{array}\right|$
$D_x=44-56=-12$
And,
$\begin{aligned}
& D_y=\left|\begin{array}{cc}
1 & -4 \\
5 & -56
\end{array}\right| \\
& \Rightarrow D_y=-56+20=-36
\end{aligned}$
Now,
$x=\frac{D_x}{D}=\frac{-12}{-6}=2$
And,
$y=\frac{D_y}{D}=\frac{-36}{-6}=6$
Hence, $(2,6)$ is the solution
$\begin{aligned}
& \frac{x+y-8}{2}=\frac{x+2 y-14}{3} \\
& \Rightarrow 3 x+3 y-24=2 x+4 y-28 \\
& \Rightarrow x-y=-4 \ldots \text { (1) }
\end{aligned}$
Also,
$\begin{aligned}
& \text { Let } \frac{x+2 y-14}{3}=\frac{3 x-y}{4} \\
& \Rightarrow 4 x+8 y-56=9 x-3 y \\
& \Rightarrow 5 x-11 y=-56 \ldots \text { (2) }
\end{aligned}$
Hence the two equations are:
$\begin{aligned}
& x-y=-4 \ldots(1) \\
& 5 x-11 y=-56
\end{aligned}$
Now,
$\begin{aligned}
& D=\left|\begin{array}{cc}
1 & -1 \\
5 & -11
\end{array}\right| \\
& \Rightarrow D=(-11-(-5))=-6
\end{aligned}$
Also, $D _{ x }=\left|\begin{array}{cc}-4 & -1 \\ -56 & -11\end{array}\right|$
$D_x=44-56=-12$
And,
$\begin{aligned}
& D_y=\left|\begin{array}{cc}
1 & -4 \\
5 & -56
\end{array}\right| \\
& \Rightarrow D_y=-56+20=-36
\end{aligned}$
Now,
$x=\frac{D_x}{D}=\frac{-12}{-6}=2$
And,
$y=\frac{D_y}{D}=\frac{-36}{-6}=6$
Hence, $(2,6)$ is the solution
Question 653 Marks
Solve the following equations by Cramer’s method.
7x + 3y = 15; 12y – 5x = 39
7x + 3y = 15; 12y – 5x = 39
Answer
$\begin{aligned} & D =\left[\begin{array}{cc}7 & 3 \\ -5 & 12\end{array}\right]=(7 \times 12)-(3 \times-5)=84+15=99 \\ & D _{ x }=\left[\begin{array}{cc}15 & 3 \\ 39 & 12\end{array}\right]=(15 \times 12)-(3 \times 39)=180-117=63 \\ & D _{ y }=\left[\begin{array}{cc}7 & 15 \\ -5 & 39\end{array}\right]=(7 \times 39)-(15 \times-5)=273+75=348 \\ & x =\frac{ D _{ x }}{ D }=\frac{63}{99}=\frac{7}{11} y =\frac{ D _{ y }}{ D }=\frac{348}{99}=\frac{116}{33} \\ & \therefore( x , y )=\left(\frac{7}{11}, \frac{116}{33}\right)\end{aligned}$
View full question & answer→$\begin{aligned} & D =\left[\begin{array}{cc}7 & 3 \\ -5 & 12\end{array}\right]=(7 \times 12)-(3 \times-5)=84+15=99 \\ & D _{ x }=\left[\begin{array}{cc}15 & 3 \\ 39 & 12\end{array}\right]=(15 \times 12)-(3 \times 39)=180-117=63 \\ & D _{ y }=\left[\begin{array}{cc}7 & 15 \\ -5 & 39\end{array}\right]=(7 \times 39)-(15 \times-5)=273+75=348 \\ & x =\frac{ D _{ x }}{ D }=\frac{63}{99}=\frac{7}{11} y =\frac{ D _{ y }}{ D }=\frac{348}{99}=\frac{116}{33} \\ & \therefore( x , y )=\left(\frac{7}{11}, \frac{116}{33}\right)\end{aligned}$
Question 663 Marks
Solve the following equations by Cramer’s method.
$3 x-2 y=\frac{5}{2} ; \frac{1}{3} x+3 y=-\frac{4}{3}$
$3 x-2 y=\frac{5}{2} ; \frac{1}{3} x+3 y=-\frac{4}{3}$
Answer
$\begin{aligned} & 3 x -2 y =\frac{5}{2} \Rightarrow 6 x -4 y =5 \\ & \frac{1}{3} x +3 y =-\frac{4}{3} \Rightarrow \frac{ x +9 y }{3}=-\frac{4}{3} \Rightarrow x +9 y =-4 \\ & D =\left[\begin{array}{cc}6 & -4 \\ 1 & 9\end{array}\right]=(6 \times 9)-(-4 \times 1)=54+4=58 \\ & D _{ x }=\left[\begin{array}{cc}5 & -4 \\ -4 & 9\end{array}\right]=(5 \times 9)-(-4 \times-4)=45-16=29 \\ & D _{ y }=\left[\begin{array}{cc}6 & 5 \\ 1 & -4\end{array}\right]=(6 \times-4)-(5 \times 1)=-24-5=-29 \\ & x=\frac{D_x}{D}=\frac{1}{2}, y=\frac{D_y}{D}=\frac{(-29)}{58}=\frac{(-1)}{2} \\ & \therefore( x , y )=(1 / 2,-1 / 2)\end{aligned}$
View full question & answer→$\begin{aligned} & 3 x -2 y =\frac{5}{2} \Rightarrow 6 x -4 y =5 \\ & \frac{1}{3} x +3 y =-\frac{4}{3} \Rightarrow \frac{ x +9 y }{3}=-\frac{4}{3} \Rightarrow x +9 y =-4 \\ & D =\left[\begin{array}{cc}6 & -4 \\ 1 & 9\end{array}\right]=(6 \times 9)-(-4 \times 1)=54+4=58 \\ & D _{ x }=\left[\begin{array}{cc}5 & -4 \\ -4 & 9\end{array}\right]=(5 \times 9)-(-4 \times-4)=45-16=29 \\ & D _{ y }=\left[\begin{array}{cc}6 & 5 \\ 1 & -4\end{array}\right]=(6 \times-4)-(5 \times 1)=-24-5=-29 \\ & x=\frac{D_x}{D}=\frac{1}{2}, y=\frac{D_y}{D}=\frac{(-29)}{58}=\frac{(-1)}{2} \\ & \therefore( x , y )=(1 / 2,-1 / 2)\end{aligned}$
Question 673 Marks
Solve the following equations by Cramer’s method.
4m – 2n = –4; 4m + 3n = 16
4m – 2n = –4; 4m + 3n = 16
Answer
$\begin{aligned} & D =\left[\begin{array}{cc}4 & -2 \\ 4 & 3\end{array}\right]=(4 \times 3)-(-2 \times 4)=12+8=20 \\ & D_x=\left[\begin{array}{cc}-4 & -2 \\ 16 & 3\end{array}\right]=(-4 \times 3)-(-2 \times 16)=-12+32=20 \\ & D_y=\left[\begin{array}{cc}4 & -4 \\ 4 & 16\end{array}\right]=(4 \times 16)-(-4 \times 4)=64+16=80 \\ & x=\frac{D_x}{D}=\frac{20}{20}=1 y=\frac{D_y}{D}=\frac{80}{20}=4 \\ & \therefore(x, y)=(1,4)\end{aligned}$
View full question & answer→$\begin{aligned} & D =\left[\begin{array}{cc}4 & -2 \\ 4 & 3\end{array}\right]=(4 \times 3)-(-2 \times 4)=12+8=20 \\ & D_x=\left[\begin{array}{cc}-4 & -2 \\ 16 & 3\end{array}\right]=(-4 \times 3)-(-2 \times 16)=-12+32=20 \\ & D_y=\left[\begin{array}{cc}4 & -4 \\ 4 & 16\end{array}\right]=(4 \times 16)-(-4 \times 4)=64+16=80 \\ & x=\frac{D_x}{D}=\frac{20}{20}=1 y=\frac{D_y}{D}=\frac{80}{20}=4 \\ & \therefore(x, y)=(1,4)\end{aligned}$
Question 683 Marks
Solve the following equations by Cramer’s method.
6x – 3y = –10; 3x + 5y – 8 = 0
6x – 3y = –10; 3x + 5y – 8 = 0
Answer
$\begin{aligned} & 6 x-3 y=-10 \\ & 3 x+5 y=8 \\ & D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39 \\ & D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 \\ & D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \times 3)=48+30=78 \\ & x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} y=\frac{D_y}{D}=\frac{78}{39}=2 \\ & \therefore(x, y)=\left(-\frac{2}{3}, 2\right)\end{aligned}$
View full question & answer→$\begin{aligned} & 6 x-3 y=-10 \\ & 3 x+5 y=8 \\ & D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39 \\ & D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 \\ & D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \times 3)=48+30=78 \\ & x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} y=\frac{D_y}{D}=\frac{78}{39}=2 \\ & \therefore(x, y)=\left(-\frac{2}{3}, 2\right)\end{aligned}$
Question 693 Marks
Find the values of each of the following determinants.
(1) $\left|\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right|$
(3) $\left|\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right|$
(1) $\left|\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right|$
(3) $\left|\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right|$
Answer
View full question & answer→(1) $D =\left[\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right]=(4 \times 7)-(3 \times 2)=28-6=22$
(2) $D=\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]=(5 \times 1)-2 \times-3=5-6=-1$
(3) $D =\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
(2) $D=\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]=(5 \times 1)-2 \times-3=5-6=-1$
(3) $D =\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
Question 703 Marks
Solve the following simultaneous equation graphically.
3x + y = 10; x – y = 2
3x + y = 10; x – y = 2
Answer
View full question & answer→
Question 713 Marks
Solve the following simultaneous equation graphically.
3x – y – 2 = 0; 2x + y = 8
3x – y – 2 = 0; 2x + y = 8
Answer
View full question & answer→The given simultaneous equations are
$3 x-y-2=0$
$\therefore \quad y=3 x-2$
$2 x+y=8$
$\therefore \quad y=8-2 x$
The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
$3 x-y-2=0$
$\therefore \quad y=3 x-2$
| x | 0 | 1 | 3 | -1 |
| y | -2 | 1 | 7 | -5 |
| (x, y) | (0, -2) | (1, 1) | (3, 7) | (-1, -5) |
$2 x+y=8$
$\therefore \quad y=8-2 x$
| x | 0 | 4 | 1 | 3 |
| y | 8 | 0 | 6 | 2 |
| (x, y) | (0, 8) | (4, 0) | (1, 6) | (3, 2) |
The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
Question 723 Marks
Solve the following simultaneous equation graphically.
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
Answer
View full question & answer→The given simultaneous equations are
$5 x-6 y+30=0$
$\therefore \quad 6 y=5 x+30$
$\therefore \quad y=\frac{5 x+30}{6}$
$5 x+4 y-20=0$
$\therefore \quad 4 y=20-5 x$
$\therefore \quad y=\frac{20-5 x}{4}$
The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.
$5 x-6 y+30=0$
$\therefore \quad 6 y=5 x+30$
$\therefore \quad y=\frac{5 x+30}{6}$
| x | 0 | -6 | 6 | 3 |
| y | 5 | 0 | 10 | 7.5 |
| (x, y) | (0, 5) | (-6, 0) | (6, 10) | (3, 7.5) |
$5 x+4 y-20=0$
$\therefore \quad 4 y=20-5 x$
$\therefore \quad y=\frac{20-5 x}{4}$
| x | 0 | 4 | -4 | 2 |
| y | 5 | 0 | 10 | 2.5 |
| (x, y) | (0, 5) | (4, 0) | (-4, 10) | (2, 2.5) |
The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.
Question 733 Marks
Solve the following simultaneous equation graphically.
x – 3y = 1; 3x – 2y + 4 = 0
x – 3y = 1; 3x – 2y + 4 = 0
Answer
View full question & answer→The given simultaneous equations are
$x-3 y=1$
$\therefore \quad 3 y=x-1$
$\therefore \quad y=\frac{x-1}{3}$
$3 x-2 y+4=0$
$\therefore \quad 2 y=3 x+4$
$\therefore \quad y=\frac{3 x+4}{2}$
The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.
$x-3 y=1$
$\therefore \quad 3 y=x-1$
$\therefore \quad y=\frac{x-1}{3}$
| x | 4 | -2 | -5 | 1 |
| y | 1 | -1 | -2 | 0 |
| (x, y) | (4, 1) | (-2, -1) | (-5, -2) | (1, 0) |
$\therefore \quad 2 y=3 x+4$
$\therefore \quad y=\frac{3 x+4}{2}$
| x | 0 | -2 | 2 | -4 |
| y | 2 | -1 | 5 | -4 |
| (x, y) | (0, 2) | (-2, -1) | (2, 5) | (-4, -4) |
The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.
Question 743 Marks
Solve the following simultaneous equation graphically.
2x + 3y = 12; x – y = 1
2x + 3y = 12; x – y = 1
Answer
View full question & answer→The given simultaneous equations are
$2 x+3 y=12$
$\therefore \quad 3 y=12-2 x$
$\therefore \quad y=\frac{12-2 x}{3}$
$x-y=1$
$\therefore \quad y=x-1$
The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.
$2 x+3 y=12$
$\therefore \quad 3 y=12-2 x$
$\therefore \quad y=\frac{12-2 x}{3}$
| x | 0 | 6 | 3 | -3 |
| y | 4 | 0 | 2 | 6 |
| (x, y) | (0, 4) | (6, 0) | (3, 2) | (-3, 6) |
$x-y=1$
$\therefore \quad y=x-1$
| x | 0 | 2 | 4 | 5 |
| y | -1 | 1 | 3 | 4 |
| (x, y) | (0, -1) | (2, 1) | (4, 3) | (-4, 4) |
The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.
Question 753 Marks
Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Answer
View full question & answer→Let Manish's present age be $x$
Let Savita's present age be y
According to $1^{\text {st }}$ situation,
$x+y=31 \dots (1)$
According to second situation,
$\begin{aligned}
& x-3=4(y-3) \\
& x-3=4 y-12 \\
& x-4 y=-12+3 \\
& x-4 y=-9 \ldots (2)
\end{aligned}$
Subtracting Eq. II from I
$\begin{aligned}
& x+y=31 \\
& -x+4 y=9 \\
& 5 y=40 \\
& y=\frac{40}{5} \\
& y=8
\end{aligned}$
Substitute $y=8$ in eq. I
$\begin{aligned}
& x+8=31 \\
& x=31-8 \\
& x=23
\end{aligned}$
Manisha's age 23 years
Savita's age 8 years.
Let Savita's present age be y
According to $1^{\text {st }}$ situation,
$x+y=31 \dots (1)$
According to second situation,
$\begin{aligned}
& x-3=4(y-3) \\
& x-3=4 y-12 \\
& x-4 y=-12+3 \\
& x-4 y=-9 \ldots (2)
\end{aligned}$
Subtracting Eq. II from I
$\begin{aligned}
& x+y=31 \\
& -x+4 y=9 \\
& 5 y=40 \\
& y=\frac{40}{5} \\
& y=8
\end{aligned}$
Substitute $y=8$ in eq. I
$\begin{aligned}
& x+8=31 \\
& x=31-8 \\
& x=23
\end{aligned}$
Manisha's age 23 years
Savita's age 8 years.
Question 763 Marks
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
Question 773 Marks
Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Question 783 Marks
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Answer
View full question & answer→Let the numerator and denominator of the fraction be $x$ and $y$ respectively.
Fraction $=\frac{\bar{x}}{ y }$
Given,
Denominator $=2$ (Numerator $)+4$
$\begin{aligned}
& \Rightarrow y =2 x +4 \\
& \Rightarrow 2 x - y =(-4) \ldots I
\end{aligned}$
According to the given condition, we have
$\begin{aligned}
& y-6=12(x-6) \\
& \Rightarrow y-6=12 x-72 \\
& \Rightarrow 12 x-y=66 \ldots II
\end{aligned}$
Equating Eq. I and II,
$\begin{aligned}
& 2 x-y=-4 \\
& -12 x+y=-66 \\
& -10 x=-70 \\
& x=\frac{70}{10}
\end{aligned}$
$x=7$
Putting $x=7$ in equation $I$, we get
$\begin{aligned}
& \Rightarrow 2 \times 7-y=-4 \\
& \Rightarrow 14-y=-4 \\
& \Rightarrow y=14+4 \\
& \Rightarrow y=18
\end{aligned}$
Hence, required fraction $=\frac{7}{18}$
Fraction $=\frac{\bar{x}}{ y }$
Given,
Denominator $=2$ (Numerator $)+4$
$\begin{aligned}
& \Rightarrow y =2 x +4 \\
& \Rightarrow 2 x - y =(-4) \ldots I
\end{aligned}$
According to the given condition, we have
$\begin{aligned}
& y-6=12(x-6) \\
& \Rightarrow y-6=12 x-72 \\
& \Rightarrow 12 x-y=66 \ldots II
\end{aligned}$
Equating Eq. I and II,
$\begin{aligned}
& 2 x-y=-4 \\
& -12 x+y=-66 \\
& -10 x=-70 \\
& x=\frac{70}{10}
\end{aligned}$
$x=7$
Putting $x=7$ in equation $I$, we get
$\begin{aligned}
& \Rightarrow 2 \times 7-y=-4 \\
& \Rightarrow 14-y=-4 \\
& \Rightarrow y=14+4 \\
& \Rightarrow y=18
\end{aligned}$
Hence, required fraction $=\frac{7}{18}$
Question 793 Marks
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Answer
View full question & answer→Suppose father's age(in years) be $x$ and that son's age be $y$.
Then,
$\begin{aligned}
& x+2 y=70 \dots(1) \\
& 2 x+y=95 \dots(2)
\end{aligned}$
Multiply Eq.I by 2 and equate
$\begin{gathered}
2 x+4 y=140 \\
-2 x-y=-95 \\
\hline 3 y=45
\end{gathered}$
$\begin{aligned}
& y=\frac{45}{3} \\
& y=15
\end{aligned}$
Substituting $y=15$ in Eq.II
$\begin{aligned}
& 2 x+15=95 \\
& 2 x=95-15 \\
& 2 x=80 \\
& x=\frac{80}{2} \\
& x=40
\end{aligned}$
∴ Son’s age is 15 years, father’s age is 40 years.
Then,
$\begin{aligned}
& x+2 y=70 \dots(1) \\
& 2 x+y=95 \dots(2)
\end{aligned}$
Multiply Eq.I by 2 and equate
$\begin{gathered}
2 x+4 y=140 \\
-2 x-y=-95 \\
\hline 3 y=45
\end{gathered}$
$\begin{aligned}
& y=\frac{45}{3} \\
& y=15
\end{aligned}$
Substituting $y=15$ in Eq.II
$\begin{aligned}
& 2 x+15=95 \\
& 2 x=95-15 \\
& 2 x=80 \\
& x=\frac{80}{2} \\
& x=40
\end{aligned}$
∴ Son’s age is 15 years, father’s age is 40 years.
Question 803 Marks
Answer
$\begin{aligned} & \text { Length of rectangle } \Rightarrow 2 x+y+8=4 x-y \\
& \Rightarrow 2 x-4 x+y+y=-8 \\
& \Rightarrow-2 x+2 y=-8 \\
& \Rightarrow-x+y=-4 \ldots \ldots(I)
\end{aligned}$
$\begin{aligned}
& \text { Breadth of the rectangle }=2 y=x+4 \\
& \Rightarrow-x+2 y=4 \ldots . . \text { (II) }
\end{aligned}$
Equating Eq. I and II and change sign of Eq. II
$\begin{gathered}
-x+y=-4 \\
x-2 y=-4 \\
\hline -y=-8 \\
y=8
\end{gathered}$
Substituting $y=8$ in Eq.I
$\begin{aligned}
& -x+8=-4 \\
& -x=-4-8 \\
& -x=-12 \\
& x=12
\end{aligned}$
$\begin{aligned} & \text { Length }=2 \times 12+8+8=40 \\ & \text { Breadth }=2 \times 8=16 \\ & \text { Area }=\text { Length } \times \text { breadth }=40 \times 16=640 \text { sq. unit } \\ & \text { Perimeter }=2 \text { (Length }+ \text { Breadth) }=2(40+16)=2(56)=112 \text { unit. }\end{aligned}$
View full question & answer→$\begin{aligned} & \text { Length of rectangle } \Rightarrow 2 x+y+8=4 x-y \\
& \Rightarrow 2 x-4 x+y+y=-8 \\
& \Rightarrow-2 x+2 y=-8 \\
& \Rightarrow-x+y=-4 \ldots \ldots(I)
\end{aligned}$
$\begin{aligned}
& \text { Breadth of the rectangle }=2 y=x+4 \\
& \Rightarrow-x+2 y=4 \ldots . . \text { (II) }
\end{aligned}$
Equating Eq. I and II and change sign of Eq. II
$\begin{gathered}
-x+y=-4 \\
x-2 y=-4 \\
\hline -y=-8 \\
y=8
\end{gathered}$
Substituting $y=8$ in Eq.I
$\begin{aligned}
& -x+8=-4 \\
& -x=-4-8 \\
& -x=-12 \\
& x=12
\end{aligned}$
$\begin{aligned} & \text { Length }=2 \times 12+8+8=40 \\ & \text { Breadth }=2 \times 8=16 \\ & \text { Area }=\text { Length } \times \text { breadth }=40 \times 16=640 \text { sq. unit } \\ & \text { Perimeter }=2 \text { (Length }+ \text { Breadth) }=2(40+16)=2(56)=112 \text { unit. }\end{aligned}$
Question 813 Marks
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Answer
View full question & answer→Let the greater no. be $x$ and smaller no. be $x-3$
As per given situation,
$\begin{aligned}
& 2(x-3)+3(x)=19 \\
& \Rightarrow 2 x-6+3 x=19 \\
& \Rightarrow 5 x-6=19 \\
& \Rightarrow 5 x=19+6 \\
& \Rightarrow 5 x=25 \\
& \Rightarrow x=\frac{25}{5}=5
\end{aligned}$
$\therefore$ smaller no is $x-3 \Rightarrow 5-3=2$
Hence, The numbers are 5 and 2.
As per given situation,
$\begin{aligned}
& 2(x-3)+3(x)=19 \\
& \Rightarrow 2 x-6+3 x=19 \\
& \Rightarrow 5 x-6=19 \\
& \Rightarrow 5 x=19+6 \\
& \Rightarrow 5 x=25 \\
& \Rightarrow x=\frac{25}{5}=5
\end{aligned}$
$\therefore$ smaller no is $x-3 \Rightarrow 5-3=2$
Hence, The numbers are 5 and 2.
Question 823 Marks
Solve the following simultaneous equation.
$\begin{aligned} & \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\ & \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}\end{aligned}$
$\begin{aligned} & \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\ & \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}\end{aligned}$
Answer
$\begin{aligned}& \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\
& \frac{1}{[2(3 x+y)]}-\frac{1}{[2(3 x-y)]}=\frac{1}{8} \\
& \text { Let } \frac{1}{3 x+y}=m \text { and } \frac{1}{3 x-y}=n \\
& m+n=\frac{3}{4} \Rightarrow 4(m+n)=3 \Rightarrow 4 m+4 n=3 \ldots \text { (I) } \\
& \frac{1}{2} m-\frac{1}{2} n=\frac{1}{8} \Rightarrow 8(m-n)=1 \times 2 \Rightarrow 8 m-8 n=2 \ldots (II)
\end{aligned}$
Multiply Eq. I by 2
$8 m+8 n=6$ $\dots$ (a)
$8 m-8 n=2$ $\dots$ (b)Add (a) and (b) to get,
$8 m+8 n+8 m-8 n=8$
$\Rightarrow 16 m =8$
$m =\frac{8}{16}$
$m =\frac{1}{2}$
Substituting $m =\frac{1}{2}$ in Eq. II
$8 \times \frac{1}{2}-8 n=2$
⇒ 4 - 8n = 2
⇒ - 8n = 2 - 4
⇒ - 8n = -2
⇒ 8n = 2
$\begin{aligned}
\Rightarrow & n=\frac{2}{8} \\
\Rightarrow & n=\frac{1}{4} \\
& \therefore m=\frac{1}{2(3 x+y)} \\
& \frac{1}{2(3 x+y)}=\frac{1}{2} \\
\Rightarrow & 2=2(3 x+y) \\
\Rightarrow & 2=6 x+2 y \ldots \ldots . \text { III } \\
& n=\frac{1}{2(3 x-y)} \\
\Rightarrow & \frac{1}{2(3 x-y)}=\frac{1}{4} \\
\Rightarrow & 4=2(3 x-y) \\
\Rightarrow & 4=6 x-2 y \ldots . I V
\end{aligned}$
Add Eq. III and IV
$\begin{aligned}
& 6 x+2 y=2 \\
& \frac{6 x-2 y=4}{12 x=6} \\
& x=\frac{6}{12} \\
& x=\frac{1}{2}
\end{aligned}$
Substituting $x =\frac{1}{2}$ in Eq. III
$\begin{aligned}
& 6 \times \frac{1}{2}+2 y=2 \\
& \Rightarrow 3+2 y=2 \\
& \Rightarrow 2 y=2-3 \\
& \Rightarrow 2 y=-1 \\
& y=-\frac{1}{2}
\end{aligned}$
Hence $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
View full question & answer→$\begin{aligned}& \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\
& \frac{1}{[2(3 x+y)]}-\frac{1}{[2(3 x-y)]}=\frac{1}{8} \\
& \text { Let } \frac{1}{3 x+y}=m \text { and } \frac{1}{3 x-y}=n \\
& m+n=\frac{3}{4} \Rightarrow 4(m+n)=3 \Rightarrow 4 m+4 n=3 \ldots \text { (I) } \\
& \frac{1}{2} m-\frac{1}{2} n=\frac{1}{8} \Rightarrow 8(m-n)=1 \times 2 \Rightarrow 8 m-8 n=2 \ldots (II)
\end{aligned}$
Multiply Eq. I by 2
$8 m+8 n=6$ $\dots$ (a)
$8 m-8 n=2$ $\dots$ (b)Add (a) and (b) to get,
$8 m+8 n+8 m-8 n=8$
$\Rightarrow 16 m =8$
$m =\frac{8}{16}$
$m =\frac{1}{2}$
Substituting $m =\frac{1}{2}$ in Eq. II
$8 \times \frac{1}{2}-8 n=2$
⇒ 4 - 8n = 2
⇒ - 8n = 2 - 4
⇒ - 8n = -2
⇒ 8n = 2
$\begin{aligned}
\Rightarrow & n=\frac{2}{8} \\
\Rightarrow & n=\frac{1}{4} \\
& \therefore m=\frac{1}{2(3 x+y)} \\
& \frac{1}{2(3 x+y)}=\frac{1}{2} \\
\Rightarrow & 2=2(3 x+y) \\
\Rightarrow & 2=6 x+2 y \ldots \ldots . \text { III } \\
& n=\frac{1}{2(3 x-y)} \\
\Rightarrow & \frac{1}{2(3 x-y)}=\frac{1}{4} \\
\Rightarrow & 4=2(3 x-y) \\
\Rightarrow & 4=6 x-2 y \ldots . I V
\end{aligned}$
Add Eq. III and IV
$\begin{aligned}
& 6 x+2 y=2 \\
& \frac{6 x-2 y=4}{12 x=6} \\
& x=\frac{6}{12} \\
& x=\frac{1}{2}
\end{aligned}$
Substituting $x =\frac{1}{2}$ in Eq. III
$\begin{aligned}
& 6 \times \frac{1}{2}+2 y=2 \\
& \Rightarrow 3+2 y=2 \\
& \Rightarrow 2 y=2-3 \\
& \Rightarrow 2 y=-1 \\
& y=-\frac{1}{2}
\end{aligned}$
Hence $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
Question 833 Marks
Solve the following simultaneous equation.
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
Answer
$\begin{aligned}& \frac{27}{x-2}+\frac{31}{y+3}=85 \\
& \frac{31}{x-2}+\frac{27}{y+3}=89
\end{aligned}$
Let $\frac{1}{x-2}=m$ and $\frac{1}{y+3}=n$
$27 m+31 n=85 \dots (1)$
$31 m+27 n=89 \dots (2)$
Adding both equations
$58 m+58 n=174$
Dividing both sides by 58
$m + n =3 \dots (3)$
Subtracting Eq. I and II
$\begin{aligned}
& 27 m+31 n=85 \\
& -31 m-27 n=-89 \\
& -4 m+4 n=-4
\end{aligned}$
Dividing both sides by 4
$-m+n=-1 \dots (4)$
Equating Eq. III and IV
$\begin{gathered}
m+n=3 \\
-m+n=-1 \\
\hline 2 n=2
\end{gathered}$
$\begin{aligned}
& n =\frac{2}{2} \\
& n =1
\end{aligned}$
Subsituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore m =\frac{1}{ x -2} \\
& \Rightarrow \frac{1}{ x -2}=2 \Rightarrow 2( x -2)=1 \\
& \Rightarrow 2 x -4=1 \Rightarrow 2 x =4+1 \\
& \Rightarrow 2 x =5 \Rightarrow x =\frac{5}{2} \\
& \therefore n =\frac{1}{ y +3} \\
& \Rightarrow \frac{1}{ y +3}=1 \\
& \Rightarrow y +3=1 \\
& \Rightarrow y =1-3 \\
& \Rightarrow y =-2 \\
& y =2
\end{aligned}$
Hence $(x, y)=\left(\frac{5}{2},-2\right)$
View full question & answer→$\begin{aligned}& \frac{27}{x-2}+\frac{31}{y+3}=85 \\
& \frac{31}{x-2}+\frac{27}{y+3}=89
\end{aligned}$
Let $\frac{1}{x-2}=m$ and $\frac{1}{y+3}=n$
$27 m+31 n=85 \dots (1)$
$31 m+27 n=89 \dots (2)$
Adding both equations
$58 m+58 n=174$
Dividing both sides by 58
$m + n =3 \dots (3)$
Subtracting Eq. I and II
$\begin{aligned}
& 27 m+31 n=85 \\
& -31 m-27 n=-89 \\
& -4 m+4 n=-4
\end{aligned}$
Dividing both sides by 4
$-m+n=-1 \dots (4)$
Equating Eq. III and IV
$\begin{gathered}
m+n=3 \\
-m+n=-1 \\
\hline 2 n=2
\end{gathered}$
$\begin{aligned}
& n =\frac{2}{2} \\
& n =1
\end{aligned}$
Subsituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore m =\frac{1}{ x -2} \\
& \Rightarrow \frac{1}{ x -2}=2 \Rightarrow 2( x -2)=1 \\
& \Rightarrow 2 x -4=1 \Rightarrow 2 x =4+1 \\
& \Rightarrow 2 x =5 \Rightarrow x =\frac{5}{2} \\
& \therefore n =\frac{1}{ y +3} \\
& \Rightarrow \frac{1}{ y +3}=1 \\
& \Rightarrow y +3=1 \\
& \Rightarrow y =1-3 \\
& \Rightarrow y =-2 \\
& y =2
\end{aligned}$
Hence $(x, y)=\left(\frac{5}{2},-2\right)$
Question 843 Marks
Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
Answer
$\begin{aligned} & \frac{10}{x+y}+\frac{2}{x-y}=4 \\
& \frac{15}{x+y}-\frac{5}{x-y}=-2 \\
& \text { Let } \frac{1}{x+y}=m \text { and } \frac{1}{x-y}=n \\
& 10 m+2 n=4 \ldots \text { (I) } \\
& 15 m-5 n=-2 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 5 and Eq.II by 2
$\begin{gathered}
50 m+10 n=20 \\
\frac{30 m-10 n=-4}{80 m=16}
\end{gathered}$
$\begin{aligned}
m & =\frac{16}{80} \\
m & =\frac{1}{5}
\end{aligned}$
$\begin{aligned}
& \text { Substituting } m=\frac{1}{5} \text { in Eq. I } \\
& 10 \times \frac{1}{5}+2 n=4 \\
& 2+2 n=4 \\
& 2 n=4-2 \\
& 2 n=2 \\
& n=\frac{2}{2} \\
& n=1
\end{aligned}$
$\begin{aligned}
\therefore & m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5 \dots (3)\\
& n =\frac{1}{ x - y } \Rightarrow \frac{1}{x-y}=1 \Rightarrow x - y =1 \dots (4)
\end{aligned}$
Now, equating Eq. III and IV
$\begin{gathered}
x+y=5 \\
x-y=1 \\
\hline 2 x=6 \\
x=\frac{6}{2} \\
x=3
\end{gathered}$
Subsituting value of $x=3$ in Eq. III
$\begin{aligned}
& 3+y=5 \\
& y=5-3 \\
& y=2
\end{aligned}$
Hence $(x, y)=(3,2)$
View full question & answer→$\begin{aligned} & \frac{10}{x+y}+\frac{2}{x-y}=4 \\
& \frac{15}{x+y}-\frac{5}{x-y}=-2 \\
& \text { Let } \frac{1}{x+y}=m \text { and } \frac{1}{x-y}=n \\
& 10 m+2 n=4 \ldots \text { (I) } \\
& 15 m-5 n=-2 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 5 and Eq.II by 2
$\begin{gathered}
50 m+10 n=20 \\
\frac{30 m-10 n=-4}{80 m=16}
\end{gathered}$
$\begin{aligned}
m & =\frac{16}{80} \\
m & =\frac{1}{5}
\end{aligned}$
$\begin{aligned}
& \text { Substituting } m=\frac{1}{5} \text { in Eq. I } \\
& 10 \times \frac{1}{5}+2 n=4 \\
& 2+2 n=4 \\
& 2 n=4-2 \\
& 2 n=2 \\
& n=\frac{2}{2} \\
& n=1
\end{aligned}$
$\begin{aligned}
\therefore & m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5 \dots (3)\\
& n =\frac{1}{ x - y } \Rightarrow \frac{1}{x-y}=1 \Rightarrow x - y =1 \dots (4)
\end{aligned}$
Now, equating Eq. III and IV
$\begin{gathered}
x+y=5 \\
x-y=1 \\
\hline 2 x=6 \\
x=\frac{6}{2} \\
x=3
\end{gathered}$
Subsituting value of $x=3$ in Eq. III
$\begin{aligned}
& 3+y=5 \\
& y=5-3 \\
& y=2
\end{aligned}$
Hence $(x, y)=(3,2)$
Question 853 Marks
Solve the following simultaneous equation.
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
Answer
$\begin{aligned} & \frac{2}{x}-\frac{3}{y}=15 \\
& \frac{8}{x}+\frac{5}{y}=77 \\
& \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\
& 2 m-3 n=15 \ldots \text { (I) } \\
& 8 m+5 n=77 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 4
$8 m-12 n=60 \ldots \text { (III) }$
Equating Eq. II and III. Change the signs of Eq. III
$\begin{gathered}
8 m+5 n=77 \\
-8 m+12 n=-60 \\
\hline 17 n=17
\end{gathered}$
$\begin{aligned}
& n =\frac{17}{17} \\
& n =1
\end{aligned}$
Substituting $n =1$ in Eq. II
$\begin{aligned}
& 8 m+5 \times 1=77 \\
& 8 m+5=77 \\
& 8 m=77-5 \\
& 8 m=72 \\
& m=\frac{72}{8} \\
& m=9
\end{aligned}$
$\begin{aligned} & \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{ x }=9 \Rightarrow x =\frac{1}{9} \\ & \therefore n =\frac{1}{ y } \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1 \\ & \text { Hence }( x , y )=\left(\frac{1}{9}, 1\right)\end{aligned}$
View full question & answer→$\begin{aligned} & \frac{2}{x}-\frac{3}{y}=15 \\
& \frac{8}{x}+\frac{5}{y}=77 \\
& \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\
& 2 m-3 n=15 \ldots \text { (I) } \\
& 8 m+5 n=77 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 4
$8 m-12 n=60 \ldots \text { (III) }$
Equating Eq. II and III. Change the signs of Eq. III
$\begin{gathered}
8 m+5 n=77 \\
-8 m+12 n=-60 \\
\hline 17 n=17
\end{gathered}$
$\begin{aligned}
& n =\frac{17}{17} \\
& n =1
\end{aligned}$
Substituting $n =1$ in Eq. II
$\begin{aligned}
& 8 m+5 \times 1=77 \\
& 8 m+5=77 \\
& 8 m=77-5 \\
& 8 m=72 \\
& m=\frac{72}{8} \\
& m=9
\end{aligned}$
$\begin{aligned} & \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{ x }=9 \Rightarrow x =\frac{1}{9} \\ & \therefore n =\frac{1}{ y } \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1 \\ & \text { Hence }( x , y )=\left(\frac{1}{9}, 1\right)\end{aligned}$
Question 863 Marks
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
Answer
$\begin{aligned} & 2 x+3 y=2 \\ & x-\frac{y}{2}=\frac{1}{2} \Rightarrow 2 x-y=1 \\ & D=\left[\begin{array}{cc}2 & 3 \\ 2 & -1\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 \\ & D_x=\left[\begin{array}{cc}2 & 3 \\ 1 & -1\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 \\ & D_y=\left[\begin{array}{ll}2 & 2 \\ 2 & 1\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) \\ & x=\frac{D_x}{D}=\frac{-5}{-8}=\frac{5}{8} y=\frac{D_y}{D}=\frac{-2}{-8}=\frac{1}{4} \\ & \therefore(x, y)=\left(\frac{5}{8}, \frac{1}{4}\right) \text { is solution. }\end{aligned}$
View full question & answer→$\begin{aligned} & 2 x+3 y=2 \\ & x-\frac{y}{2}=\frac{1}{2} \Rightarrow 2 x-y=1 \\ & D=\left[\begin{array}{cc}2 & 3 \\ 2 & -1\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 \\ & D_x=\left[\begin{array}{cc}2 & 3 \\ 1 & -1\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 \\ & D_y=\left[\begin{array}{ll}2 & 2 \\ 2 & 1\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) \\ & x=\frac{D_x}{D}=\frac{-5}{-8}=\frac{5}{8} y=\frac{D_y}{D}=\frac{-2}{-8}=\frac{1}{4} \\ & \therefore(x, y)=\left(\frac{5}{8}, \frac{1}{4}\right) \text { is solution. }\end{aligned}$
Question 873 Marks
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54; 3m + 2n = 28
4m + 6n = 54; 3m + 2n = 28
Answer
$\begin{aligned} & 4 m+6 n=54 \\
& 3 m+2 n=28 \\
& D=\left[\begin{array}{ll}
4 & 6 \\
3 & 2
\end{array}\right]=(4 \times 2)-(6 \times 3)=8-18=10 \\
& D_x=\left[\begin{array}{ll}
54 & 6 \\
28 & 2
\end{array}\right]=(54 \times 2)-(6 \times 28)=108-168=60 \\
& D_y=\left[\begin{array}{ll}
4 & 54 \\
3 & 28
\end{array}\right]=(4 \times 28)-(54 \times 3)=112-162=50 \\
& x=\frac{D_x}{D}=\frac{60}{10}=6 y=\frac{D_y}{D}=\frac{50}{10}=5
\end{aligned}$
$\therefore(x, y)=(6,5)$ is solution.
View full question & answer→$\begin{aligned} & 4 m+6 n=54 \\
& 3 m+2 n=28 \\
& D=\left[\begin{array}{ll}
4 & 6 \\
3 & 2
\end{array}\right]=(4 \times 2)-(6 \times 3)=8-18=10 \\
& D_x=\left[\begin{array}{ll}
54 & 6 \\
28 & 2
\end{array}\right]=(54 \times 2)-(6 \times 28)=108-168=60 \\
& D_y=\left[\begin{array}{ll}
4 & 54 \\
3 & 28
\end{array}\right]=(4 \times 28)-(54 \times 3)=112-162=50 \\
& x=\frac{D_x}{D}=\frac{60}{10}=6 y=\frac{D_y}{D}=\frac{50}{10}=5
\end{aligned}$
$\therefore(x, y)=(6,5)$ is solution.
Question 883 Marks
Solve the following simultaneous equations using Cramer’s rule.
6x – 4y = –12; 8x – 3y = –2
6x – 4y = –12; 8x – 3y = –2
Answer
$\begin{aligned} & 6 x-4 y=-12 \\
& 8 x-3 y=-2 \\
& D=\left[\begin{array}{ll}
6 & -4 \\
8 & -3
\end{array}\right]=(6 \times-3)-(-4 \times 8)=-18+32=14 \\
& D_x=\left[\begin{array}{cc}
-12 & -4 \\
-2 & -3
\end{array}\right]=(-12 \times-3)-(-4 \times-2)=36-8=28 \\
& D_y=\left[\begin{array}{cc}
6 & -12 \\
8 & -2
\end{array}\right]=(6 \times-2)-12 \times 8=12+96=108 \\
& x=\frac{D_x}{D}=\frac{28}{14}=2 y=\frac{D_y}{D}=\frac{108}{14}=6
\end{aligned}$
$\therefore(x, y)=(2,6)$ is solution.
View full question & answer→$\begin{aligned} & 6 x-4 y=-12 \\
& 8 x-3 y=-2 \\
& D=\left[\begin{array}{ll}
6 & -4 \\
8 & -3
\end{array}\right]=(6 \times-3)-(-4 \times 8)=-18+32=14 \\
& D_x=\left[\begin{array}{cc}
-12 & -4 \\
-2 & -3
\end{array}\right]=(-12 \times-3)-(-4 \times-2)=36-8=28 \\
& D_y=\left[\begin{array}{cc}
6 & -12 \\
8 & -2
\end{array}\right]=(6 \times-2)-12 \times 8=12+96=108 \\
& x=\frac{D_x}{D}=\frac{28}{14}=2 y=\frac{D_y}{D}=\frac{108}{14}=6
\end{aligned}$
$\therefore(x, y)=(2,6)$ is solution.
Question 893 Marks
Solve the following simultaneous equations using Cramer’s rule.
x + 2y = –1; 2x – 3y = 12
x + 2y = –1; 2x – 3y = 12
Answer
$\begin{aligned} & x+2 y=-1 \\ & 2 x-3 y=12 \\ & D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 \\ & D_x=\left[\begin{array}{cc}-1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21 \\ & D_y=\left[\begin{array}{cc}1 & -1 \\ 2 & 12\end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14 \\ & x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2 \\ & \therefore(x, y)=(3,-2) \text { is solution. }\end{aligned}$
View full question & answer→$\begin{aligned} & x+2 y=-1 \\ & 2 x-3 y=12 \\ & D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 \\ & D_x=\left[\begin{array}{cc}-1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21 \\ & D_y=\left[\begin{array}{cc}1 & -1 \\ 2 & 12\end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14 \\ & x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2 \\ & \therefore(x, y)=(3,-2) \text { is solution. }\end{aligned}$
Question 903 Marks
Solve the following simultaneous equations using Cramer’s rule.
4x + 3y – 4 = 0; 6x = 8 – 5y
4x + 3y – 4 = 0; 6x = 8 – 5y
Answer
$\begin{aligned}
& 4 x+3 y=4 \\
& 6 x+5 y=8 \\
& D=\left[\begin{array}{ll}
4 & 3 \\
6 & 5
\end{array}\right]=(4 \times 5)-(3 \times 6)=20-18=2 \\
& D_x=\left[\begin{array}{ll}
4 & 3 \\
8 & 5
\end{array}\right]=(4 \times 5)-(3 \times 8)=20-24=-4 \\
& D_y=\left[\begin{array}{ll}
4 & 4 \\
6 & 8
\end{array}\right]=(4 \times 8)-(4 \times 6)=32-24=8 \\
& x=\frac{D_x}{D}=-\frac{4}{2}=-2 y=\frac{D_y}{D}=\frac{8}{2}=4
\end{aligned}$
$\therefore( x , y )=(-2,4)$ is the solution.
View full question & answer→$\begin{aligned}
& 4 x+3 y=4 \\
& 6 x+5 y=8 \\
& D=\left[\begin{array}{ll}
4 & 3 \\
6 & 5
\end{array}\right]=(4 \times 5)-(3 \times 6)=20-18=2 \\
& D_x=\left[\begin{array}{ll}
4 & 3 \\
8 & 5
\end{array}\right]=(4 \times 5)-(3 \times 8)=20-24=-4 \\
& D_y=\left[\begin{array}{ll}
4 & 4 \\
6 & 8
\end{array}\right]=(4 \times 8)-(4 \times 6)=32-24=8 \\
& x=\frac{D_x}{D}=-\frac{4}{2}=-2 y=\frac{D_y}{D}=\frac{8}{2}=4
\end{aligned}$
$\therefore( x , y )=(-2,4)$ is the solution.
Question 913 Marks
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10; 4x + 3y = 5
3x – 4y = 10; 4x + 3y = 5
Answer
$\begin{aligned}
& 3 x-4 y=10 \\
& 4 x+3 y=5 \\
& D=\left|\begin{array}{cc}
3 & -4 \\
4 & 3
\end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25 \\
& D_x=\left[\begin{array}{cc}
10 & -4 \\
5 & 3
\end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50 \\
& D_y=\left[\begin{array}{cc}
3 & 10 \\
4 & 5
\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25 \\
& x=\frac{D_x}{D}=\frac{50}{25}=2 \\
& x=\frac{D_y}{D}=-\frac{25}{25}=-1
\end{aligned}$
$\therefore(x, y)=(2,-1)$ is the solution
View full question & answer→$\begin{aligned}
& 3 x-4 y=10 \\
& 4 x+3 y=5 \\
& D=\left|\begin{array}{cc}
3 & -4 \\
4 & 3
\end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25 \\
& D_x=\left[\begin{array}{cc}
10 & -4 \\
5 & 3
\end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50 \\
& D_y=\left[\begin{array}{cc}
3 & 10 \\
4 & 5
\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25 \\
& x=\frac{D_x}{D}=\frac{50}{25}=2 \\
& x=\frac{D_y}{D}=-\frac{25}{25}=-1
\end{aligned}$
$\therefore(x, y)=(2,-1)$ is the solution
Question 923 Marks
Find the values of following determinants.
(1) $\left|\begin{array}{cc}-1 & 7 \\ 2 & 4\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & 3 \\ -7 & 0\end{array}\right|$
(3) $\left|\begin{array}{ll}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{array}\right|$
(1) $\left|\begin{array}{cc}-1 & 7 \\ 2 & 4\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & 3 \\ -7 & 0\end{array}\right|$
(3) $\left|\begin{array}{ll}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{array}\right|$
Answer
View full question & answer→we know, determinant of a $2 \times 2$ matrix
$\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|$
(1) $(-1 \times 4)-(7 \times 2)=-4-14=-18$
(2) $(5 \times 0)-(3 \times-7)=0-(-21)=21$
$\begin{aligned}
& \text { (3) } \frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6} \\
& =-\frac{8}{6} \\
& =-\frac{4}{3}
\end{aligned}$
$\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|$
(1) $(-1 \times 4)-(7 \times 2)=-4-14=-18$
(2) $(5 \times 0)-(3 \times-7)=0-(-21)=21$
$\begin{aligned}
& \text { (3) } \frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6} \\
& =-\frac{8}{6} \\
& =-\frac{4}{3}
\end{aligned}$
Question 933 Marks
Solve the following simultaneous equation graphically.
2x – 3y = 4 ; 3y – x = 4
2x – 3y = 4 ; 3y – x = 4
Answer
View full question & answer→The given simultaneous equations are
$2 x-3 y=4$
$\therefore \quad 3 y=2 x-4$
$\therefore \quad y=\frac{2 x-4}{3}$
$3 y-x=4$
$\therefore \quad 3 y=x+4$
$\therefore \quad y=\frac{x+4}{3}$
The two lines intersect at point (8, 4).
∴ x = 8 and y = 4 is the solution of the simultaneous equations 2x – 3y = 4 and 3y – x = 4.
$2 x-3 y=4$
$\therefore \quad 3 y=2 x-4$
$\therefore \quad y=\frac{2 x-4}{3}$
| x | 2 | -1 | 5 | 8 |
| y | 0 | -2 | 2 | 4 |
| (x. y) | (2, 0) | (-1, -2) | (-5, 2) | (8, 4) |
$\therefore \quad 3 y=x+4$
$\therefore \quad y=\frac{x+4}{3}$
| x | 2 | -4 | 5 | -1 |
| y | 2 | 0 | 3 | 1 |
| (x. y) | (2, 2) | (-4, -0) | (5, 3) | (-1, 1) |
The two lines intersect at point (8, 4).
∴ x = 8 and y = 4 is the solution of the simultaneous equations 2x – 3y = 4 and 3y – x = 4.
Question 943 Marks
Solve the following simultaneous equation graphically.
3x – 4y = –7; 5x – 2y = 0
3x – 4y = –7; 5x – 2y = 0
Answer
The given simultaneous equations are
$3 x-4 y=-7$
$\therefore \quad 4 y=3 x+7$
$\therefore \quad y=\frac{3 x+7}{4}$
$5 x-2 y=0$
$\therefore \quad 2 y=5 x$
$\therefore \quad y=\frac{5}{2} x$
The two lines intersect at point (1, 2.5).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
View full question & answer→The given simultaneous equations are
$3 x-4 y=-7$
$\therefore \quad 4 y=3 x+7$
$\therefore \quad y=\frac{3 x+7}{4}$
| x | -1 | -5 | 3 | 5 |
| y | 1 | -2 | 7 | -5. 5 |
| (x, y) | (-1, 1) | (5, -2) | (3, 4) | (5, 5.5) |
$\therefore \quad 2 y=5 x$
$\therefore \quad y=\frac{5}{2} x$
| x | 0 | 2 | -2 | 1 |
| y | 0 | 5 | -5 | 2 |
| (x, y) | (0, 0) | (2, 5) | (-2 -5) | (11, 02) |
The two lines intersect at point (1, 2.5).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
Question 953 Marks
Solve the following simultaneous equation graphically.
3x – y = 2; 2x – y = 3
3x – y = 2; 2x – y = 3
Answer
View full question & answer→The given simultaneous equations are
$3 x-y=2$
$\therefore \quad y=3 x-2$
$2 x-y=3$
$\therefore \quad y=2 x-3$
The two lines intersect at point (1, 2.5).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
$3 x-y=2$
$\therefore \quad y=3 x-2$
| x | 0 | 1 | -1 | 2 |
| y | -2 | 1 | -5 | 4 |
| (x, y) | (0, -2) | (1, 1) | (-1, -5) | (2, 4) |
$\therefore \quad y=2 x-3$
| x | 0 | 1 | 2 | 3 |
| y | -3 | -1 | 1 | 3 |
| (x, y) | (0, -3) | (1, -1) | (2, 1) | (3, 3) |
The two lines intersect at point (1, 2.5).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
Question 963 Marks
Solve the following simultaneous equation graphically.
x + y = 0; 2x – y = 9
x + y = 0; 2x – y = 9
Answer
View full question & answer→The given simultaneous equations are
$x+y=0$
$\therefore \quad y=-x$
$2 x-y=9$
$\therefore \quad y=2 x-9$
The two lines intersect at point (3, -3).
∴ x = 3 and y = -3 is the solution of the simultaneous equations x + y = 0 and 2x – y = 9.
$x+y=0$
$\therefore \quad y=-x$
| x | 0 | 2 | -2 | -1 |
| y | 0 | -2 | 2 | 1 |
| (x, y) | (0, 0) | (2, -2) | (-2, 2) | (-1, 1) |
$2 x-y=9$
$\therefore \quad y=2 x-9$
| x | 0 | 2 | 5 | 4 |
| y | -9 | -5 | 1 | -1 |
| (x, y) | (0, -9) | (2, -5) | (5, 1) | (4, -1) |
The two lines intersect at point (3, -3).
∴ x = 3 and y = -3 is the solution of the simultaneous equations x + y = 0 and 2x – y = 9.
Question 973 Marks
Solve the following simultaneous equation graphically.
x + y = 5; x – y = 3
x + y = 5; x – y = 3
Answer
View full question & answer→The given simultaneous equations are
$x+y=5$
$\therefore \quad y=5-x$
$x-y=3$
$\therefore \quad y=x-3$
The two lines intersect at point (4, 1). .
∴ x = 4 and y = 1 is the solution of the simultaneous equations x+y = 5 and x – y = 3.
$x+y=5$
$\therefore \quad y=5-x$
| x | 0 | 5 | 2 | 3 |
| y | 5 | 0 | 3 | 2 |
| (x, y) | (0, 5) | (5, 0) | (2, 3) | (3, 2) |
$\therefore \quad y=x-3$
| x | 0 | 3 | 2 | 1 |
| y | -3 | 0 | -1 | -2 |
| (x, y) | (0, -3) | (3, 0) | (2, -1) | (1, -2) |
The two lines intersect at point (4, 1). .
∴ x = 4 and y = 1 is the solution of the simultaneous equations x+y = 5 and x – y = 3.
Question 983 Marks
Solve the following simultaneous equation graphically.
x + y = 6; x – y = 4
x + y = 6; x – y = 4
Answer
View full question & answer→The given simultaneous equations are
x + y = 6
∴ y = 6 – x
x – y = 4
∴ y = x – 4
The two lines intersect at point (5, 1).
∴ x = 5 and y = 1 is the solution of the simultaneous equations x + y = 6 and x – y = 4.
x + y = 6
∴ y = 6 – x
| x | 0 | 6 | 3 | 4 |
| y | 6 | 0 | 3 | 2 |
| (x, y) | (0, 6) | (6, 0) | (3, 3) | (4, 2) |
∴ y = x – 4
| x | 0 | 4 | 2 | 5 |
| y | -4 | 0 | -2 | 1 |
| (x, y) | (0, -4) | (4, 0) | (2, -2) | (5, 1) |
The two lines intersect at point (5, 1).
∴ x = 5 and y = 1 is the solution of the simultaneous equations x + y = 6 and x – y = 4.
Question 993 Marks
Answer
View full question & answer→i. x + y = 3
ii. x – y = 4
| x | 3 | -2 | 0 |
| y | 0 | 5 | 3 |
| (x, y) | (3, 0) | (-2, 5) | (0, 3) |
ii. x – y = 4
| x | 4 | -1 | 0 |
| y | 0 | -5 | -4 |
| (x, y) | (4, 0) | (-1, 5) | (0, -4) |
Question 1003 Marks
A certain amount is equally distributed among certain number of students. Each would get ₹ 2 less if 10 students were more and each would get ₹ 6 more if 15 students were less. Find the number of students and the amount distributed.