[5 marks sum]

Question 515 Marks

Solve the following pairs of equations$:\ \frac{x}{3}+\frac{y}{4}=11,\frac{5 x}{6}-\frac{y}{3}=-7$

Answer

$ \frac{x}{3}+\frac{y}{4}=11$
$\Rightarrow 4 x +3 y =132 ....(i)$
$\frac{5 x}{6}-\frac{y}{3}=-7$
$\Rightarrow 5 x -2 y =-42 ....(ii) $
Multiplying eqn. $(i)$ by $2$ and eqn. $(ii)$ by $3 ,$ we get
$ 8 x+6 y=264 ....(iii)$
$15 x-6 y=-126 ....(iv) $
Adding eqns. $(iii)$ and $(iv),$ we get
$ 23 x =138$
$\Rightarrow x =6 $
Substituting the value of $x$ in eqn. $(i),$ we get
$ 4(6)+3 y=132$
$\Rightarrow 24+3 y=132$
$\Rightarrow 3 y=108$
$\Rightarrow y=36 $
Thus, the solution set is $(6,36)$.

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Question 525 Marks

Solve the following simultaneous equations $:\ 2(3u - v) = 5uv,2(u + 3v) = 5uv$

Answer

$2(3 u-v)=5 u v$
$2(u+3 v)=5 u v$
$2(3 u-v)=5 u v$
$\Rightarrow 6 u-2 v=5 u v$
$\Rightarrow \frac{6}{v}-\frac{2}{u}=5 \ldots .(1)$
$2(u+3 v)=5 u v$
$\Rightarrow 2 u+6 v=5 u v$
$\Rightarrow \frac{2}{u}+\frac{6}{v}=5 \ldots . (2) $
Multiplying equation $(1)$ by $3 ,$
we get, $\frac{18}{v}-\frac{6}{u}=15 .....(3) $
Adding $(2)$ and $(3),$
$ \frac{20}{v}=20$
$\Rightarrow v =1$
$\therefore \frac{6}{u}$
$=5-\frac{2}{1}$
$=3$
$\Rightarrow u =\frac{6}{3}$
$=2 $
Thus, the solution set is $(2,1)$.

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Question 535 Marks

Solve the following simultaneous equation $:8v - 3u = 5uv,6v - 5u = -2uv$

Answer

$8 v-3 u=5 u v$
$6 v-5 u=-2 u v $
Dividing both sides of each equation by $uv,$
we get, $\frac{8}{u}-\frac{3}{v}=5 . . . (1)$
$\frac{6}{u}-\frac{5}{v}=-2 . . (2) $
Multiplying $(1)$ by $3$ and $(2)$ by $4,$
we get, $ \frac{24}{u}-\frac{9}{v}=15 \ldots . (3)$
$\frac{24}{u}-\frac{20}{v}=-8 . .(4) $
Subtracting $(4)$ from $(3), $
we get, $\frac{11}{v}=23$
$\Rightarrow v =\frac{11}{23}$
$\therefore \frac{6}{u}-\frac{5}{11} \times 23=-2$
$\Rightarrow \frac{6}{u}-\frac{115}{11}=-2$
$\Rightarrow \frac{6}{u} $
$ =-2+\frac{115}{11}$
$=\frac{-22+115}{11}$
$=\frac{93}{11}$
$\Rightarrow u =\frac{6 \times 11}{93}$
$=\frac{22}{31} $
Thus, the solution set is $\left(\frac{22}{11}, \frac{11}{23}\right)$.

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Question 545 Marks

Solve the following simultaneous equations by the substitution method$:\ 7(y + 3) - 2(x + 2) = 14;4(y - 2) + 3(x - 3) = 2$

Answer

The given equations are
$ 7(y+3)-2(x+2)=14 ....(i)$
$4(y-2)+3(x-3)=2 ....(ii) $
Consider
$ 7(y+3)-2(x+2)=14$
$\Rightarrow 7 y+21-2 x-4=14$
$\Rightarrow-2 x+7 y=-3$
$\Rightarrow 2 x-7 y=3$
$\Rightarrow 2 x=7 y+3$
$\Rightarrow x=\frac{7 y+3}{2} ....(iii) $
Now, consider equation
$ 4(y-2)+3(x-3)=2$
$\Rightarrow 4 y-8+3 x-9=2$
$\Rightarrow 3 x+4 y=19$
$\Rightarrow 3\left(\frac{7 y+3}{2}\right)+4 y=19 \ldots .[$ From $(iii)]$
$\Rightarrow \frac{21 y+9+8 y}{2}=19$
$\Rightarrow 29 y+9=38$
$\Rightarrow 29 y=29$
$\Rightarrow y=1 $
Substituting value of $y$ in eqn. $(iii),$ we get
$ x=\frac{7(1)+3}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,1)$.

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Question 555 Marks

Solve the following simultaneous equations by the substitution method$:\ 3 - (x + 5) = y + 2;2(x + y) = 10 + 2y$

Answer

The given equations are
$3 - (x + 5) = y + 2 ....(i)$
$2(x + y) = 10 + 2y ....(ii)$
Consider
$3 - (x + 5) = y + 2$
$\Rightarrow 3 - x - 5 = y + 2$
$\Rightarrow -x - 2 = y + 2$
$\Rightarrow x + y = -4$
$\Rightarrow x = -4 - y ....(iii)$
Now, consider equation
$2(x + y) = 10 + 2y$
$\Rightarrow 2x + 2y = 10 + 2y$
$\Rightarrow 2x = 10$
$\Rightarrow x = 5$
Substitutiing the value of $x$ in eqn. $(ii),$ we get
$5 = -4 - y$
$\Rightarrow y = -4 - 5$
$= -9$
Thus, the solution set is $(5, -9).$

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Question 565 Marks

Solve the following simultaneous equations by the substitution method$:\ 0.4x + 0.3y = 1.7,0.7x - 0.2y = 0.8$

Answer

The given equations are
$ 0.4 x+0.3 y=1.7 ....(i)$
$0.7 x-0.2 y=0.8 ....(ii) $
Multiplying both the equations by $10 ,$ we get
$ 4 x+3 y=17 ....(iii)$
$7 x-2 y=8 ....(iv) $
Now, consider equation
$ 4 x+3 y=17$
$\Rightarrow 4 x=17-3 y$
$\Rightarrow x=\frac{17-3 y}{4} ....(v) $
Substituting the value of $x$ in eqn. $(iv),$ we get
$ 7\left(\frac{17-3 y}{4}\right)-2 y=8$
$\Rightarrow \frac{119-21 y}{4}-2 y=8$
$\Rightarrow \frac{119-21 y-8 y}{4}=8$
$\Rightarrow 119-29 y=32$
$\Rightarrow-29 y=32-119$
$\Rightarrow-29 y=-87$
$\Rightarrow y=\frac{-87}{-29}$
$=3 $
Putting the value of $y$ in eqn. $(v),$ we get
$ x=\frac{17-3(3)}{4}$
$=\frac{17-9}{4}$
$=\frac{8}{4}$
$=2 $
Thus, the solution set is $(2,3)$.

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Question 575 Marks

Solve the following simultaneous equations by the substitution method$:\ 0.5x + 0.7y = 0.74,0.3x + 0.5y = 0.5$

Answer

The given equations are
$ 0.5 x+0.7 y=0.74 ....(i)$
$0.3 x+0.5 y=0.5 ....(ii) $
Now, consider equation
$ 0.5 x+0.7 y=0.74$
$\Rightarrow 0.5 x=0.74-0.7 y$
$\Rightarrow x=\frac{0.74-0.7 y}{0.5} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 0.3\left(\frac{0.74-0.7 y}{0.5}\right)+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.21}{0.5}+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.2 y+0.25}{0.5}=0.5$
$\Rightarrow 0.222+0.04 y=0.25$
$\Rightarrow 0.04 y=0.028$
$\Rightarrow y=\frac{0.028}{0.04}$
$=\frac{28}{40}$
$=\frac{7}{10}$
$=0.7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{0.74-0.7(0.7)}{0.5}$
$=\frac{0.74-0.49}{0.5}$
$=\frac{0.25}{0.5}$
$=\frac{25}{50}$
$=\frac{1}{2}$
$=0.5 $
Thus, the solution set is $(0.5,0.7)$.

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Question 585 Marks

Solve the following simultaneous equations by the substitution method$:\ 13 + 2y = 9x,3y = 7x$

Answer

The given equations are
$ 13+2 y=9 x ....(i)$
$3 y=7 x ....(ii) $
Now, consider equation
$ 3 y=7 x$
$\Rightarrow y=\frac{7}{3} x ....(iii) $
Substituting the value of $y$ in eqn. $(i),$
we get $13+2\left(\frac{7}{3} x\right)=9 x$
$\Rightarrow 13+\frac{14}{3} x=13$
$\Rightarrow 9 x-\frac{14}{3} x=13$
$\Rightarrow \frac{27 x-14 x}{3}=13$
$\Rightarrow 13 x=39$
$\Rightarrow x=\frac{39}{13}$
$=3 $
Putting the value of $x$ in eqn. $(iii),$
we get $y=\frac{7}{3} \times 3$
$=7$
Thus, the solution set is $(3,7)$.

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Question 595 Marks

Solve the following simultaneous equations by the substitution method$:\ 7x - 3y = 31;9x - 5y = 41$

Answer

The given equations are
$ 7 x-3 y=31 ...(i)$
$9 x-5 y=41 ....(ii) $
Now, consider equation
$ 7 x-3 y=31$
$\Rightarrow 7 x=31+3 y$
$\Rightarrow x=\frac{31+3 y}{7} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$
we get $9\left(\frac{31+3 y}{7}\right)-5 y=41$
$\Rightarrow \frac{279+27 y}{7}-5 y=41$
$\Rightarrow \frac{279+27 y-35 y}{7}=41$
$\Rightarrow 279-8 y=287$
$\Rightarrow-8 y=8$
$\Rightarrow y=-1 $
Putting the value of $y$ in eqn. $(iii).$
we get $ x=\frac{31+3(-1)}{7}$
$=\frac{31-3^7}{7}$
$=\frac{28^7}{7}$
$=4 $
Thus, the solution set is $(4,-1)$.

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Question 605 Marks

Solve the following simultaneous equations by the substitution method$:\ 2x + 3y = 31,5x - 4 = 3y$

Answer

The given equations are
$ 2 x+3 y=31 ....(i)$
$5 x-4=3 y ....(ii) $
Now, consider equation
$ 2 x+3 y=31$
$\Rightarrow 2 x=31-3 y$
$\Rightarrow x=\frac{31-3 y}{2} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 5\left(\frac{31-3 y}{2}\right)-4=3 y$
$\Rightarrow \frac{155-15 y}{2}-4=3 y$
$\Rightarrow \frac{155-15 y-8}{2}=3 y$
$\Rightarrow 147-15 y=6 y$
$\Rightarrow 21 y=147$
$\Rightarrow y=\frac{147}{21}=7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{31-3(7)}{2}$
$=\frac{31-21}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,7)$.

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Question 615 Marks

Solve the following simultaneous equations by the substitution method$:\ 2x + y = 8;3y = 3 + 4x$

Answer

The given equations are
$ 2 x+y=8 ....(i)$
$3 y=3+4 x ....(ii) $
Now, consider equation $2 x+y=8$
$ \Rightarrow y =8 \text { - } 2 x ....(iii)$
Substituting the value of $y$ in eqn. $(ii),$ we get
$ 3(8-2 x)=3+4 x$
$\Rightarrow 24-6 x=3+4 x$
$\Rightarrow 6 x-4 x=3-24$
$\Rightarrow-10 x=-21$
$\Rightarrow x=\frac{21}{10} $
Puutting the value of $x$ in eqn. $(iii),$ we get
$ y=8-2\left(\frac{21}{10}\right)$
$=8-\frac{21}{5}$
$=\frac{40-21}{5}$
$=\frac{19}{5} $
Thus, the solution set is $\left(\frac{21}{10}, \frac{19}{5}\right)$.

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Question 625 Marks

If the following three equations hold simultaneously for $x$ and $y,$ find the value of $'m\ '.2x + 3y + 6 = 0,4x - 3y - 8 = 0,x + my - 1 = 0$

Answer

The given equations are
$ 2 x+3 y+6=0 ....(i)$
$4 x-3 y-8=0 ....(ii)$
$x+m y-1=0 ....(ii) $
Adding eqns. $(i)$ and $(ii),$
we get $6 x-2=0$
$\Rightarrow 6 x=2$
$\Rightarrow x=\frac{1}{3} $
Substituting the value of $x$ in eqn. $(i),$
we get $2 \times \frac{1}{3}+3 y+6=0$
$\Rightarrow 3 y$
$=-6-\frac{2}{3}$
$=\frac{18-2}{3}$
$=\frac{-20}{3}$
$\Rightarrow y=-\frac{20}{9} $
Substituting the value of $x$ and in eqn. $(iii),$
we get $\frac{1}{3}+ m \times\left(-\frac{20}{9}\right)-1=0$
$\Rightarrow-\frac{20}{9} m$
$=1-\frac{1}{3}$
$=\frac{2}{3}$
$\Rightarrow m$
$=-\frac{2}{3} \times \frac{9}{20}$
$=-\frac{3}{10}.$

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MATHEMATICS [5 marks sum]