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Maths 2 Marks Questions

P-1 Linear Equations in Two Variables · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 27 questions.

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Question 12 Marks
Solve the following simultaneous equation.
$49x – 57y = 172; 57x – 49y = 252$
Answer
$49 x-57 y=172 \ldots$
$57 x-49 y=252 \ldots$
Adding both the Equations
$49 x-57 y=172 $
$57 x-49 y=252 $
$----------$
$ 106 x-106 y=424$
Dividing both sides by 106
$x-y=4$
Subtract equation (I) and (II)
$49 x-57 y  =172 $
$-57 x+49 y  =-252 $
$----------$
$-8 y-8 y =-80$
Divide both sides by $(-8)$
$x+y=10$
Equating Eq. (III) and (IV)
$x-y=4$
$\frac{x+y=10}{2 x=14} $
$x=\frac{14}{2} $
$x=7$
Substituting $x=7$ in Eq. IV
$7+y=10 $
$y=10-7$
$y=3$
$\therefore$ solution is $(x, y)=(7,3)$
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Question 22 Marks
Solve the following simultaneous equation.
99x + 101 y = 499; 101x + 99y = 501
Answer
$99 x+101 y=499 \ldots(I) $
$101 x+99 y=501 \ldots(II)$
Adding both the Equations
$99 x+101 y =499 $
$101 x+99 y =501$
$--------$
$200 x+200 y  =1000$
Dividing both sides by 200
$x+y=5\dots(III)$
Subtract equation (I) and (II)
$\begin{array}{c}99 x+101 y=499 \\-101 x-99 y=-501 \\\hline-2 x+2 y=-2\end{array}$
Divide both sides by ( -2 )
$x-y=1 \ldots( IV )$
Equating Eq. (III) and (IV)
$x+y=5 $
${x-y=1}$
$----$
${2 x=6} $
$x=\frac{6}{2} $
$x=3$
Substituting $x=3$ in Eq. III
$3+y=5 $
$y=5-3$
$y=2$
$\therefore$ solution is $(x, y)=(3,2)$
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Question 32 Marks
Solve the following simultaneous equation.
$\frac{1}{3} x+y=\frac{10}{3} ; 2 x+\frac{1}{4} y=\frac{11}{4}$
Answer

$\frac{1}{3} x + y =\frac{10}{3} \Rightarrow \frac{ x +3 y }{3}=\frac{10}{3} \Rightarrow x +3 y =10 \dots(I)$
$2 x +\frac{1}{4} y =\frac{11}{4} \Rightarrow \frac{8 x + y }{4}=\frac{11}{4} \Rightarrow 8 x + y =11\dots(II)$
Multiplying Eq. II by 3
$24 x+3 y=33 \ldots \text { (III) }$
Equating Eq. I and III, change the signs of Eq. III
$x+3y=10$
$\underline{-24x-3y=-33}$
$-23x=-23$
$x=1$
Substituting x = 1 in Eq. I
$1+3 y=10 $
$3 y=10-1 $
$ 3 y=9$
$ y=\frac{9}{3} $
$ y=3$
∴ solution is (x,y) = (1, 3)
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Question 42 Marks
Solve the following simultaneous equation.
5x + 2y = –3; x + 5y = 4
Answer
$5 x+2 y=-3 \ldots(I)$
$x+5 y=4 \ldots( II )$
Multiply Eq. I by 5 and Eq. II by 2
$25 x+10 y=-15 \ldots(III) $
$2 x+10 y=8 \ldots(\text { IV) }$
Change sign of Eq.(IV)
$25 x+10 y=-15 $
$\underline{-2 x-10 y=-8}$
$23 x=-23$
$x=-\frac{23}{23} $
$x=-1$
Subsituting $x=-1$ in Eq.II
$-1+5 y=4 $
$5 y=4+1$
$5 y=5$
$y=\frac{5}{5}$
$y=1$
∴ solution is (x, y) = (–1, 1)
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Question 52 Marks
Solve the following simultaneous equation.
5m – 3n = 19; m – 6n = –7
Answer

$5 m-3 n =19 \ldots \text { (I) } $
$m -6 n =-7 \ldots \text { (II) } $
$\text { Multiply Eq. II by } 5 $
$5 m-30 n =-35 \ldots \text { (III) } $
$\text { equating (II) and (III), change the sign of Eq. (III) } $
$5 m-3 n =19 $
$-5 m+30 n =35 $
$\text { Adding both we get } $
$\Rightarrow 27 n =54 $
$\Rightarrow n =\frac{54}{27}$
⇒ n = 2
Substituting n = 2 in Eq 1
⇒ 5m - 3(2) = 19
⇒ 5m - 6 = 19
⇒ 5m = 25
⇒ m = 5
∴ Solution is (m , n) = (5, 2)
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Question 62 Marks
Solve the following simultaneous equation.
2x – 3y = 9; 2x + y = 13
Answer
$2x-3y=9\dots(I)$
$2x+y=13\dots(II)$
Change the sign of Eq. (II)
$2x-3y=9$
$-2x+y=13$
$------$
$-4y=4$
$y=\frac{4}{4}$
$y= 1$
Substituting y = 1 in Eq. (II)
$2x +1 =13$
2x = 13 - 1
2x = 12
x = 6
∴ solution is (x, y) = (1,6)
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Question 72 Marks
Solve the following simultaneous equation.
x + 7y = 10; 3x – 2y = 7
Answer
$x+7 y=10 \ldots \text { (I) }$
$3 x-2 y=7 \ldots( II )$
Multiply Eq. I by 2 and Eq. II by 7
$2 x+14 y=20$
$\underline{21 x-14 y=49}$
$23 x=69 $
$x=\frac{69}{23} $
$x=3$
Substituting $x=3$ in Eq. 1
$3+7 y=10$
$7 y=10-3$
$7y=7$
$y= \frac{7}{7}$
$y=1$
∴ Solution is (x , y) = (3, 1)
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Question 82 Marks
Solve the following simultaneous equation.
$3a + 5b = 26; a + 5b = 22$
Answer
$3 a+5 b=26 \ldots \text { (I) }$
$a+5 b=22 \ldots \text { (II) }$
Change the sign of Eq. (II)
$3 a+5 b=26 $
$-a-5 b=-22$
$2 a=4 $
$a=\frac{4}{2}$
$a=2$
Substituting a $=2$ in Eq. (II)
$2+5 b=22 $
$5 b=22-2 $
$5 b=20 $
$b=\frac{20}{5} $
$b=4$
$\therefore$ solution is $(\mathrm{a}, \mathrm{b})=(2,4)$
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Question 92 Marks
Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 ……. (i)
2x + 3y = 12 ……… (ii)
Answer
5x + 3y = 9 ……. (i)

2x + 3y = 12 ……… (ii)

Subtracting equation (ii) from (i), we get,
(5x + 3y ) - (2x + 3y) = 9 - 12
5x - 2x + 3y - 3y = -3
3x = -3
x = -1
Putting the value of x in equation (i),
5(-1) + 3y = 9
-5 + 3y = 9
3y = 14
y = 14/3

Let’s add equations (I) and (II).

Hence, x = -1 and y = 14/3 is the solution of the equation.


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Question 152 Marks
Solve the following simultaneous equation.
49x – 57y = 172; 57x – 49y = 252
Answer

$\begin{aligned}
& 49 x-57 y=172 \ldots \\
& 57 x-49 y=252 \ldots
\end{aligned}$
Adding both the Equations
$\begin{gathered}
49 x-57 y=172 \\
57 x-49 y=252 \\
\hline 106 x-106 y=424
\end{gathered}$
Dividing both sides by 106
$x-y=4$
Subtract equation (I) and (II)
$\begin{aligned}
49 x-57 y & =172 \\
-57 x+49 y & =-252 \\
\hline-8 y-8 y & =-80
\end{aligned}$
Divide both sides by $(-8)$
$x+y=10$
Equating Eq. (III) and (IV)
$\begin{aligned}
& x-y=4 \\
& \frac{x+y=10}{2 x}=14 \\
& x=\frac{14}{2} \\
& x=7
\end{aligned}$
Substituting $x=7$ in Eq. IV
$\begin{aligned}
& 7+y=10 \\
& y=10-7 \\
& y=3
\end{aligned}$
$\therefore$ solution is $( x , y )=(7,3)$
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Question 162 Marks
Solve the following simultaneous equation.
99x + 101 y = 499; 101x + 99y = 501
Answer

$\begin{aligned}
& 99 x+101 y=499 \ldots \\
& 101 x+99 y=501 \ldots
\end{aligned}$
Adding both the Equations
$\begin{aligned}
99 x+101 y & =499 \\
101 x+99 y & =501 \\
200 x+200 y & =1000
\end{aligned}$
Dividing both sides by 200
$x+y=5$
Subtract equation (I) and (II)
$\begin{gathered}
99 x+101 y=499 \\
-101 x-99 y=-501 \\
\hline-2 x+2 y=-2
\end{gathered}$
Divide both sides by $(-2)$
$x-y=1 .... (iv)$
Equating Eq. (III) and (IV)
$\begin{aligned}
& x+y=5 \\
& \frac{x-y=1}{2 x=6} \\
& x=\frac{6}{2} \\
& x=3
\end{aligned}$
Substituting $x=3$ in Eq. III
$\begin{aligned}
& 3+y=5 \\
& y=5-3 \\
& y=2
\end{aligned}$
$\therefore$ solution is $( x , y )=(3,2)$
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Question 172 Marks
Solve the following simultaneous equation.
$\frac{1}{3} x+y=\frac{10}{3} ; 2 x+\frac{1}{4} y=\frac{11}{4}$
Answer

$\begin{aligned}
& \frac{1}{3} x+y=\frac{10}{3} \Rightarrow \frac{x+3 y}{3}=\frac{10}{3} \Rightarrow x+3 y=10 \ldots (1)\\
& 2 x+\frac{1}{4} y=\frac{11}{4} \Rightarrow \frac{8 x+y}{4}=\frac{11}{4} \Rightarrow 8 x+y=11 \ldots (2)
\end{aligned}$
Multiplying Eq. II by 3
$24 x +3 y =33 ...(3)$
Equating Eq. I and III, change the signs of Eq. III
$\begin{aligned}
& x+3 y=10 \\
& -24 x-3 y=-33 \\
& -23 x=-23 \\
& x=1
\end{aligned}$
Substituting $x=1$ in Eq. I
$\begin{aligned}
& 1+3 y=10 \\
& 3 y=10-1
\end{aligned}$
$\begin{aligned}
& 3 y=9 \\
& y=\frac{9}{3} \\
& y=3
\end{aligned}$
$\therefore$ solution is $(x, y)=(1,3)$
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Question 182 Marks
Solve the following simultaneous equation.
5x + 2y = –3; x + 5y = 4
Answer

$\begin{aligned}
& 5 x+2 y=-3 \ldots(1) \\
& x+5 y=4 \ldots (2)
\end{aligned}$
Multiply Eq. I by 5 and Eq. II by 2
$\begin{aligned}
& 25 x+10 y=-15 \ldots \text { (III) } \\
& 2 x+10 y=8 \ldots \text { (IV) }
\end{aligned}$
Change sign of Eq.(IV)
$\begin{gathered}
25 x+10 y=-15 \\
-2 x-10 y=-8 \\
\hline 23 x=-23
\end{gathered}$
$\begin{aligned}
& x=-\frac{23}{23} \\
& x=-1
\end{aligned}$
Subsituting $x=-1$ in Eq.II
$\begin{aligned}
& -1+5 y=4 \\
& 5 y=4+1 \\
& 5 y=5 \\
& y=\frac{5}{5} \\
& y=1
\end{aligned}$
$\therefore$ solution is $( x , y )=(-1,1)$
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Question 192 Marks
Solve the following simultaneous equation.
5m – 3n = 19; m – 6n = –7
Answer

$\begin{aligned}
& 5 m-3 n=19 \ldots (1) \\
& m-6 n=-7 \ldots (2)
\end{aligned}$
Multiply Eq. II by 5
$5 m-30 n=-35 \ldots (3)$
equating (I) and (III), change the sign of Eq. (III)
$\begin{aligned}
& 5 m -3 n =19 \\
& -5 m +30 n =35\text { Adding both we get } \\
& \Rightarrow 27 n =54
\end{aligned}$
$\begin{aligned}
\Rightarrow n=\frac{54}{27} \\
\Rightarrow n=2
\end{aligned}$
Substituting n = 2 in Eq 1
⇒ 5m - 3(2) = 19
⇒ 5m - 6 = 19
⇒ 5m = 25
⇒ m = 5
∴ Solution is (m , n) = (5, 2)
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Question 202 Marks
Solve the following simultaneous equation.
2x – 3y = 9; 2x + y = 13
Answer

$\begin{aligned}
& 2 x-3 y=9 \dots (1) \\
& 2 x+y=13 \dots (2)
\end{aligned}$
Change the sign of Eq. (II)
$\begin{gathered}
2 x-3 y=9 \\
\frac{-2 x-y=-13}{-4 y=4} \\
y=\frac{4}{4} \\
y=1
\end{gathered}$
Substituting $y=1$ in Eq. (II)
$\begin{aligned}
& 2 x+1=13 \\
& 2 x=13-1 \\
& 2 x=12 \\
& x=6
\end{aligned}$
$\therefore$ solution is $(x, y)=(1,6)$
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Question 222 Marks
Solve the following simultaneous equation.
3a + 5b = 26; a + 5b = 22
Answer

$\begin{aligned}
& 3 a+5 b=26 \ldots(1) \\
& a+5 b=22 \ldots(2)
\end{aligned}$
Change the sign of Eq. (II)
$\begin{aligned}
& 3 a+5 b=26 \\
& -a-5 b=-22 \\
& 2 a=4 \\
& a=\frac{4}{2} \\
& a=2
\end{aligned}$
Substituting a = 2 in Eq. (II)
$\begin{aligned}
& 2+5 b=22 \\
& 5 b=22-2 \\
& 5 b=20 \\
& b=\frac{20}{5} \\
& b=4
\end{aligned}$
$\therefore$ solution is $(a, b)=(2,4)$
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Question 232 Marks
solve :
$\frac{4}{x}+\frac{5}{y}=7 ; \frac{3}{x}+\frac{4}{y}=5$
Answer

$\begin{aligned}
& \frac{4}{x}+\frac{5}{y}=7 ; \frac{3}{x}+\frac{4}{y}=5 \\
& 4\left(\frac{1}{x}\right)+5\left(\frac{1}{y}\right)=7 \ldots \text { (I) } \\
& 3\left(\frac{1}{x}\right)+4\left(\frac{1}{y}\right)=5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x}\right)$ by $m$ and $\left(\frac{1}{y}\right)$ by $n$ in equations (I) and (II), we get
$\begin{array}{lll}
4 m+5 n=7 \ldots & \text { (III) } \\
3 m+4 n=5 \ldots & \text { (IV) }
\end{array}$
On solving these equations we get
$m=3, n=-1$
Now, $m=\frac{1}{x} \quad \therefore 3=\frac{1}{x} \quad \therefore x=\frac{1}{3}$
$n=\frac{1}{y} \quad \therefore-1=\frac{1}{y} \quad \therefore y=-1$
$\therefore$ Solution of given simultaneous equations is $(x, y)=\left(\frac{1}{3},-1\right)$
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Question 242 Marks
Solve the following simultaneous equations using Cramer’s Rule.
5x + 3y = -11 ; 2x + 4y = -10
Answer

Given equations
$\begin{aligned}
& 5 x+3 y=-11 \\
& 2 x+4 y=-10 \\
& D =\left|\begin{array}{ll}
5 & 3 \\
2 & 4
\end{array}\right| \quad=(5 \times 4)-(2 \times 3)=20-6=14 \\
& D _x=\left|\begin{array}{ll}
-11 & 3 \\
-10 & 4
\end{array}\right|=(-11) \times 4-(-10) \times 3=-44-(-30) \\
& =-44+30=-14 \\
& D_y=\left|\begin{array}{ll}
5 & -11 \\
2 & -10
\end{array}\right|=5 \times(-10)-2 \times(-11)=-50-(-22) \\
& =-50+22=-28 \\
\end{aligned}$
$x=\frac{ D _x}{ D }=\frac{-14}{14}=-1$
and $y=\frac{ D _y}{ D }=\frac{-28}{14}=-2$
$\therefore(x, y)=(-1,-2)$ is the solution.
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