Question 1:  Solve the pair of simultaneous equations using substitution method.

(i) $2x+3y = 12$ and $3x+2y=18$

The given system of equation is:

$2x+3y = 12$ … … … … … (i)

$3x+2y=18$ … … … … … (ii)

From equation (i), we get

$x =$ $\frac{1}{2}$ $(12-3y) \Rightarrow x = 6 -$ $\frac{3}{2}$ $y$

Substituting $x = 6 -$ $\frac{3}{2}$ $y$ in equation (ii), we get

$3 ( 6 -$ $\frac{3}{2}$ $y) +2y=18$

$\Rightarrow 18 -$ $\frac{9}{2}$ $y = 18$

$\frac{9}{2}$ $y = 0 \Rightarrow y = 0$

Putting $y = 0$ in $x = 6 -$ $\frac{3}{2}$ $y$ we get $x = 6$

Hence the solution of the given system of equations is $x =6, y = 0$

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(ii) $\frac{b}{a}$ $x +$ $\frac{a}{b}$ $y = a^2 + b^2$ and $x+y = 2ab$

The given system of equation is:

$\frac{b}{a}$ $x +$ $\frac{a}{b}$ $y = a^2 + b^2$ … … … … … (i)

$x+y = 2ab$… … … … … (ii)

From equation (ii), we get

$x = (2ab-y)$

Substituting $x = (2ab-y)$ in equation (i), we get

$\frac{b}{a}$ $(2ab-y) +$ $\frac{a}{b}$ $y = a^2 + b^2$

$\Rightarrow 2b^2 -$ $\frac{b}{a}$ $y + +$ $\frac{a}{b}$ $y = a^2 + b^2$

$\Rightarrow$ $\frac{a^2-b^2}{ab}$ $y = a^2 + b^2$

$\Rightarrow y = ab$

Putting $y = ab$ in $x = (2ab-y)$ $y$ we get $x = ab$

Hence the solution of the given system of equations is $x =ab, y = ab$

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(iii) $\frac{x}{6}$ $= y-6$ and $\frac{3x}{4}$ $= 1+y$

The given system of equation is:

$\frac{x}{6}$ $= y-6$… … … … … (i)

$\frac{3x}{4}$ $= 1+y$ … … … … … (ii)

From equation (i), we get

$y =$ $\frac{x}{6}$ $+ 6$

Substituting $y =$ $\frac{x}{6}$ $+ 6$ in equation (ii), we get

$\frac{3}{4}$ $x = 1+$ $\frac{x}{6}$ $+ 6$

$\Rightarrow \Big($ $\frac{3}{4}$ $-$ $\frac{1}{6}$ $\Big) x = 7$

$\Rightarrow x = 12$

Putting $x = 12$ in $y =$ $\frac{12}{6}$ $+ 6$ we get $y = 8$

Hence the solution of the given system of equations is $x =12, y = 8$

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(iv) $\frac{x}{2}$ $+$ $\frac{2y}{3}$ $= -1$ and $x -$ $\frac{y}{3}$ $= 3$

The given system of equation is:

$\frac{x}{2}$ $+$ $\frac{2y}{3}$ $= -1$ … … … … … (i)

$x -$ $\frac{y}{3}$ $= 3$… … … … … (ii)

From equation (ii), we get

$x =$ $\frac{y}{3}$ $+3$

Substituting $x =$ $\frac{y}{3}$ $+3$ in equation (i), we get

$\frac{1}{2}\Big($ $\frac{y}{3}$ $+ 3 \Big) +$ $\frac{2y}{3}$ $= -1$

$\Rightarrow$ $\frac{y}{6}$ $+$ $\frac{3}{2}$ $+$ $\frac{2y}{3}$ $= -1$

$\Rightarrow y \Big($ $\frac{1}{6}$ $+$ $\frac{2}{3}$ $\Big) = -1 -$ $\frac{3}{2}$

$\Rightarrow y$ $\frac{5}{6}$ $= -$ $\frac{5}{2}$

$\Rightarrow y = -3$

Putting $y = -3$ we get $x = 3 +$ $\frac{-3}{3}$ $= 2$

Hence the solution of the given system of equations is $x =2, y = -3$

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(v) $9-(x-4)= y + 7$ and $2(x+y) = 4-3y$

The given system of equation is:

$9-(x-4)= y + 7 \Rightarrow x+y = 6$ … … … … … (i)

$2(x+y) = 4-3y \Rightarrow 2x+5y =4$ … … … … … (ii)

From equation (i), we get

$x = 6-y$

Substituting $x = 6-y$  in equation (ii), we get

$2(6-y)+5y = 4$

$\Rightarrow 12 - 2y + 5y = 4$

$\Rightarrow y = -$ $\frac{8}{3}$

Putting $y = -$ $\frac{8}{3}$ in $x = 6 - ($ $\frac{- 8}{3}$ $)$  we get $x =$ $\frac{26}{3}$

Hence the solution of the given system of equations is $x =$ $\frac{26}{3}$$y = -$ $\frac{8}{3}$

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Question 2: Solve the system of equation using the method of elimination.

(i)   $\frac{2}{x}$ $+$ $\frac{2}{3y}$ $=$ $\frac{1}{6}$   and   $\frac{3}{x}$ $+$ $\frac{2}{y}$ $=0$

Taking $\frac{1}{x}$ $= u$ and $\frac{1}{y}$ $= v$, the given system of equations become

$12u + 4v = 2$ … … … … … (i)

$3u+2v= 0$  … … … … … (ii)

Now multiplying equation (ii) by $4$ and subtracting it from equation (i) we get

$12u+4v=1$

$\underline{ (-) \hspace{0.5cm} 12u+8v=0}$

$-4v=1$

$\Rightarrow v = -$ $\frac{1}{4}$

Substituting $v = -$ $\frac{1}{4}$ in (ii) we get

$3u+ 2 (-$ $\frac{1}{4}$ $) = 0 \Rightarrow u =$ $\frac{1}{6}$

Therefore $x =$ $\frac{1}{u}$ $= 6$ and $y =$ $\frac{1}{v}$ $= -4$

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(ii)   $8v-3u=5uv$   and   $6v-5u=-2uv$

If we put $v=0$ in either of the equations, we get $u = 0$. Similarly if we put $u=0$ in either of the equations, we get $v = 0$. Therefore $u=0$ and $v=0$ is one of the solutions that satisfies the two equation. However, for $u \neq 0$ and $v \neq 0$, we follow the following approach.

Dividing each of the given equations by uv we get:

$\frac{8}{u}$ $-$ $\frac{3}{v}$ $=$ $5$ … … … … … (i)

$\frac{6}{u}$ $-$ $\frac{5}{v}$ $= -2$ … … … … … (ii)

Taking $\frac{1}{u}$ $= x$ and $\frac{1}{v}$ $= y$, the given system of equations become

$8x - 3y = 5$ … … … … … (iii)

$6x - 5y= -2$  … … … … … (iv)

Now multiplying equation( (i) by $3$ and equation (ii) by $4$  we get

$24x - 9y = 15$  … … … … … (v)

$24x - 20y= -8$  … … … … … (vi)

Subtracting (vi) from (v) we get

$24x - 9y = 15$

$\underline{ (-) \hspace{0.5cm} 24x - 20y= -8}$

$11y = 23$

$\Rightarrow y =$ $\frac{23}{11}$

Substituting $y =$ $\frac{23}{11}$ in (iii) we get

$8x - 3($ $\frac{23}{11}$ $) = 5 \Rightarrow x =$ $\frac{31}{22}$

Therefore $u =$ $\frac{1}{x}$ $=$ $\frac{22}{31}$ and $v =$ $\frac{1}{y}$ $=$ $\frac{11}{23}$

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(iii)   $\frac{1}{2(2x+3y)}$ $+$ $\frac{12}{7(3x-2y)}$ $=$ $\frac{1}{2}$   and   $\frac{7}{2x+3y}$ $+$ $\frac{4}{3x-2y}$ $= 2$

Taking $\frac{1}{2x+3y}$ $= u$ and $\frac{1}{3x-2y}$ $= v$, the given system of equations become

$\frac{1}{2}$ $u +$ $\frac{12}{7}$ $v =$ $\frac{1}{2}$ $\Rightarrow 7u + 24v = 7$ … … … … … (i)

$7u + 4v= 2$  … … … … … (ii)

Subtracting (ii) from (1) we get

$7u + 24v = 7$

$\underline{ (-) \hspace{0.5cm} 7u + 4v= 2}$

$20v = 5$

$\Rightarrow v =$ $\frac{1}{4}$

Substituting $v =$ $\frac{1}{4}$ in (i) we get

$7u + 24($ $\frac{1}{4}$ $) = 7 \Rightarrow u =$ $\frac{1}{7}$

Therefore

$\frac{1}{2x+3y}$ $= u$ $\Rightarrow 2x+3y = 7$  … … … … … (iii)

$\frac{1}{3x-2y}$ $= v$ $\Rightarrow 3x-2y = 4$  … … … … … (iv)

Now multiplying equation( (iv) by $2$ and equation (iii) by $3$  we get

$6x+9y=21$  … … … … … (v)

$6x-4y=8$  … … … … … (vi)

Subtracting (vi) from (v) we get

$6x+9y=21$

$\underline{ (-) \hspace{0.5cm} 6x-4y=8}$

$13y = 13$

$\Rightarrow y = 1$

Substituting $y =1$ in (v) we get

$6x+9(1)=21 \Rightarrow x = 2$

Therefore $x =2, y = 1$ is the solution for the given system of equations.

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(iv)   $0.4x+0.3y=1.7$   and   $0.7x-0.2y=0.8$

The given system of equation is:

$0.4x+0.3y=1.7$ … … … … … (i)

$0.7x-0.2y=0.8$ … … … … … (ii)

Now multiplying equation( (ii) by $30$ and equation (i) by $20$  we get

$8x+6y=34$  … … … … … (iii)

$21x-6y=24$  … … … … … (iv)

Adding (iv) and (iii) we get

$8x+6y=34$

$\underline{ (+) \hspace{0.5cm} 21x-6y=24}$

$29x = 58$

$\Rightarrow x = 2$

Substituting $x =2$ in (iii) we get

$8(2)+6y=34 \Rightarrow y = 3$

Therefore $x =2, y = 3$ is the solution for the given system of equations.

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(v)   $\frac{x}{2}$ $+ y = 0.8$   and   $\frac{7}{x+\frac{y}{2}}$ $= 10$

The given system of equation is:

$\frac{x}{2}$ $+ y = 0.8 \Rightarrow x + 2y = 1.6$ … … … … … (i)

$\frac{7}{x+\frac{y}{2}}$ $= 10 \Rightarrow 2x+y = 1.4$ … … … … … (ii)

Now multiplying equation( (i) by $2$ and equation (ii) by $1$  we get

$2x+4y=3.2$  … … … … … (iii)

$2x+y=1.4$  … … … … … (iv)

Subtracting (iv) and (iii) we get

$2x+4y=3.2$

$\underline{ (-) \hspace{0.5cm} 2x+y=1.4}$

$3y = 1.8$

$\Rightarrow y = 0.6$

Substituting $y = 0.6$ in (i) we get

$x = 1.6 - 2(0.6) = 0.4$

Therefore $x =0.4, y = 0.6$ is the solution for the given system of equations.

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(vi)   $\frac{x}{3}$ $+$ $\frac{y}{4}$ $= 11$   and   $\frac{5x}{6}$ $-$ $\frac{y}{3}$ $= -7$

The given system of equation is:

$\frac{x}{3}$ $+$ $\frac{y}{4}$ $= 11 \Rightarrow 4x+3y=132$ … … … … … (i)

$\frac{5x}{6}$ $-$ $\frac{y}{3}$ $= -7 \Rightarrow 5x-2y=-42$ … … … … … (ii)

Now multiplying equation( (i) by $2$ and equation (ii) by $3$  we get

$8x+6y = 264$  … … … … … (iii)

$15x-6y = -126$  … … … … … (iv)

Adding (iv) and (iii) we get

$8x+6y = 264$

$\underline{ (+) \hspace{0.5cm} 15x-6y = -126}$

$23x = 138$

$\Rightarrow x = 6$

Substituting $x = 6$ in (i) we get

$3y = 132 -4(6) -= 132 - 24 = 108 \Rightarrow y = 36$

Therefore $x =, y = 36$ is the solution for the given system of equations.

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(vii)   $3x-$ $\frac{y+7}{11}$ $+ 2 = 10$   and   $2y +$ $\frac{x+11}{7}$ $= 10$

The given system of equation is:

$3x-$ $\frac{y+7}{11}$ $+ 2 = 10$ $\Rightarrow 33x-y=95$ … … … … … (i)

$2y +$ $\frac{x+11}{7}$ $= 10$ $\Rightarrow x+14y = 59$ … … … … … (ii)

Now multiplying equation( (i) by $14$ and equation (ii) by $1$  we get

$462x-14y=1330$  … … … … … (iii)

$x+14y=59$  … … … … … (iv)

Adding (iv) and (iii) we get

$462x-14y=1330$

$\underline{ (+) \hspace{0.5cm} x+14y=59}$

$463 x = 1389$

$\Rightarrow x = 3$

Substituting $x = 3$ in (i) we get

$y = 33(3) - 95 = 99 - 95 = 4$

Therefore $x =3, y = 4$ is the solution for the given system of equations.

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(viii)   $\frac{1}{2x}$ $-$ $\frac{1}{y}$ $= -1$   and   $\frac{1}{x}$ $+$ $\frac{1}{2y}$ $= 8$

Taking $\frac{1}{x}$ $= u$ and $\frac{1}{y}$ $= v$, the given system of equations become

$u - 2v = -2$ … … … … … (i)

$2u+v=16$  … … … … … (ii)

Now multiplying equation (i) by $2$ and subtracting (ii) from equation (i) we get

$2u-4v=-4$

$\underline{ (-) \hspace{0.5cm} 2u+v=16}$

$-5v=-20$

$\Rightarrow v = 4$

Substituting $v = 4$ in (ii) we get

$u = 2(4)-2 = 6$

Therefore $x =$ $\frac{1}{6}$ and $y =$ $\frac{1}{4}$

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(ix)   $\frac{6}{x+y}$ $=$ $\frac{7}{x-y}$ $+ 3$   and   $\frac{1}{2(x+y)}$ $=$ $\frac{1}{3(x-y)}$

Taking $\frac{1}{x+y}$ $= u$ and $\frac{1}{x-y}$ $= v$, the given system of equations become

$6u-7v=3$ … … … … … (i)

$3u -2v =0$  … … … … … (ii)

Now multiplying equation (ii) by $2$ and subtracting it from equation (i) we get

$6u-7v=3$

$\underline{ (-) \hspace{0.5cm} 3u -2v =0}$

$-3v= 3$

$\Rightarrow v = -1$

Substituting $v = -1$ in (ii) we get

$u =$ $\frac{2}{3}$ $(-1) = -$ $\frac{2}{3}$

Therefore $x+y = -$ $\frac{3}{2}$  … … … … … (iii)

$x-y = -1$ … … … … … (vi)

Adding (iii) and (iv) we get $2x = -$ $\frac{5}{2}$ $\Rightarrow x = -$ $\frac{5}{4}$

from (iv) $y = x+1 = -$ $\frac{5}{4}$ $+ 1 = -$ $\frac{1}{4}$

$\\$

(x)   $\frac{5}{x+y}$ $-$ $\frac{2}{x-y}$ $= -1$   and   $\frac{15}{x+y}$ $+$ $\frac{7}{x-y}$ $= 10$

Taking $\frac{1}{x+y}$ $= u$ and $\frac{1}{x-y}$ $= v$, the given system of equations become

$5u-2v=-1$ … … … … … (i)

$15u+7v=10$  … … … … … (ii)

Now multiplying equation (i) by $3$ and subtracting (ii) from equation (i) we get

$15u-6v=-3$

$\underline{ (-) \hspace{0.5cm} 15u+7v=10}$

$-13v=-13$

$\Rightarrow v = 1$

Substituting $v = 1$ in (ii) we get

$u =$ $\frac{1}{5}$ $(2-1) =$ $\frac{1}{5}$

Therefore $x+y = 5$  … … … … … (iii)

$x-y = 1$ … … … … … (vi)

Adding (iii) and (iv) we get $2x = 6$ $\Rightarrow x = 3$

from (iv) $y = x-1 = 3-1 = 2$

Therefore $x = 3, y = 2$

$\\$

(xi)   $\frac{1}{2(x+2y)}$ $+$ $\frac{5}{3(3x-2y)}$ $= -\frac{3}{2}$   and   $\frac{5}{4(x+2y)}$ $-$ $\frac{3}{5(3x-2y)}$ $=$ $\frac{61}{60}$

Taking $\frac{1}{x+2y}$ $= u$ and $\frac{1}{3x-2y}$ $= v$, the given system of equations become

$3u+10v=-9$ … … … … … (i)

$75u-36v=61$  … … … … … (ii)

Now multiplying equation (i) by $25$ and subtracting (ii) from equation (i) we get

$75u+250v=-225$

$\underline{ (-) \hspace{0.5cm} 75u-36v=61}$

$286v=-286$

$\Rightarrow v = -1$

Substituting $v = -1$ in (ii) we get

$3u = -9 -10(-1) = 1 \Rightarrow u =$ $\frac{1}{3}$

Therefore $x+2y = 3$  … … … … … (iii)

$3x-2y = -1$ … … … … … (vi)

Adding (iii) and (iv) we get $4x = 2$ $\Rightarrow x =$ $\frac{1}{2}$

from (iv) $y =$ $\frac{1}{2}$ $(3 -$ $\frac{1}{2}$ $) =$ $\frac{5}{4}$

Therefore $x =$ $\frac{1}{2}$ $, y =$ $\frac{5}{4}$

$\\$

(xii)   $\frac{2}{3x+2y}$ $+$ $\frac{3}{3x-2y}$ $= \frac{17}{5}$   and   $\frac{5}{3x+2y}$ $+$ $\frac{1}{3x-2y}$ $=$ $2$

Taking $\frac{1}{3x+2y}$ $= u$ and $\frac{1}{3x-2y}$ $= v$, the given system of equations become

$10u+15v=17$ … … … … … (i)

$5u+v=2$  … … … … … (ii)

Now multiplying equation (ii) by $2$ and subtracting (ii) from equation (i) we get

$10u+15v=17$

$\underline{ (-) \hspace{0.5cm} 10u+2v=4}$

$13v = 13$

$\Rightarrow v = 1$

Substituting $v = 1$ in (ii) we get

$5u = 2-1= 1 \Rightarrow u =$ $\frac{1}{5}$

Therefore $3x+2y = 5$  … … … … … (iii)

$3x-2y = 1$ … … … … … (vi)

Adding (iii) and (iv) we get $6x = 6$ $\Rightarrow x = 1$

from (iii) $2y = 5-3(1) = 2 \Rightarrow y = 1$

Therefore $x = 1, y = 1$

$\\$

(xiii)   $\frac{5}{x+1}$ $-$ $\frac{2}{y-1}$ $= \frac{1}{2}$   and   $\frac{10}{x+1}$ $+$ $\frac{2}{y-1}$ $=$ $\frac{5}{2}$

Taking $\frac{1}{x+1}$ $= u$ and $\frac{1}{y-1}$ $= v$, the given system of equations become

$10u-4v=1$ … … … … … (i)

$20u+4v=5$  … … … … … (ii)

Adding (i) and (ii)  we get

$10u-4v=1$

$\underline{ (+) \hspace{0.5cm} 20u+4v=5}$

$30u = 6$

$\Rightarrow u =$ $\frac{1}{5}$

Substituting $u = \frac{1}{5}$ in (i) we get

$4v = 10(\frac{1}{5})-1 = 2-1 = 1 \Rightarrow v =$ $\frac{1}{4}$

Therefore $x+1 = 5 \Rightarrow x = 4$

$y-1 = 4 \Rightarrow y = 5$

Therefore $x = 4, y = 5$

$\\$

(xiv)   $x-y+z=4$, $x-2y-2z=9$   and   $2x+y+3z=1$

Given system of equations is

$x-y+z=4$ … … … … … (i)

$x-2y-2z=9$ … … … … … (ii)

$2x+y+3z=1$ … … … … … (iii)

From (i) $z = 4 - x +y$

Substituting this in (ii) and (iii) we get

$3x-4y = 17$  … … … … … (iv)

$x-4y = 11$  … … … … … (v)

Subtracting (v) from (iv) we get

$3x-4y = 17$

$\underline{ (-) \hspace{0.5cm} x-4y = 11}$

$x= 3$

From (v) $4y = 3 - 11 = -8 \Rightarrow y = -2$

Therefore $z = 4 -3 + (-2) = -1$

Hence $x = 3, y = -2 \ and \ z = -1$

$\\$

(xv)   $x-y+z=4$, $x+y+z=2$   and   $2x+y-3z=0$

Given system of equations is

$x-y+z=4$ … … … … … (i)

$x+y+z=2$ … … … … … (ii)

$2x+y-3z=0$ … … … … … (iii)

From (i) $y = x+z-4$

Substituting this in (ii) and (iii) we get

$2x+2z=6$  … … … … … (iv)

$3x-2z=4$  … … … … … (v)

Adding (v) from (iv) we get

$2x+2z=6$

$\underline{ (+) \hspace{0.5cm} 3x-2z=4}$

$5x= 10$

Therefore $x = 2$

From (v) $2z = 6-2(2) = 4 \Rightarrow z = 1$

Therefore $y = 2+1-4 = -1$

Hence $x = 2, y = -1 \ and \ z = 1$

$\\$

(xvi)   $\frac{10}{x+y}$ $+$ $\frac{2}{x-y}$ $= 4$   and   $\frac{15}{x+y}$ $-$ $\frac{9}{x-y}$ $=$ $-2$

Taking $\frac{1}{x+y}$ $= u$ and $\frac{1}{x-y}$ $= v$, the given system of equations become

$10u+2v=4$ … … … … … (i)

$15u-9v=-2$  … … … … … (ii)

Multiplying (i) by $3$ and (ii) by $2$ we get

$30u+6v=12$ … … … … … (iii)

$30u-18v=-4$  … … … … … (iv)

subtracting (iv) from (iii)

$30u+6v=12$

$\underline{ (-) \hspace{0.5cm} 30u-18v=-4}$

$24v=16$

$\Rightarrow v =$ $\frac{2}{3}$

Substituting $v =$ $\frac{2}{3}$ in (ii) we get

$10u = 4 - 2 ($ $\frac{2}{3}$ $) =$ $\frac{8}{3}$ $\Rightarrow u =$ $\frac{4}{15}$

Therefore $x+y =$ $\frac{15}{4}$ … … … … … (iii)

$x-y =$ $\frac{3}{2}$ … … … … … (vi)

Adding (iii) and (iv) we get $2x =$ $\frac{21}{4}$ $\Rightarrow x =$ $\frac{21}{8}$

from (iv) $y =$ $\frac{21}{8}$ $-$ $\frac{3}{2}$ $= \frac{9}{8}$

Hence $x =$ $\frac{21}{8}$ and $y =$ $\frac{9}{8}$

$\\$

(xvii)   $\frac{1}{3x+y}$ $+$ $\frac{1}{3x-y}$ $=$ $\frac{3}{4}$   and   $\frac{1}{2(3x+y)}$ $-$ $\frac{1}{2(3x-y)}$ $= -$ $\frac{1}{8}$

Taking $\frac{1}{3x+y}$ $= u$ and $\frac{1}{3x-y}$ $= v$, the given system of equations become

$4u+4v=3$ … … … … … (i)

$4u-4v=-1$  … … … … … (ii)

$4u+4v=3$

$\underline{ (-) \hspace{0.5cm} 4u-4v=-1}$

$8u=2$

$\Rightarrow u =$ $\frac{1}{4}$

Substituting $u =$ $\frac{1}{4}$ in (ii) we get

$v =$ $\frac{3}{4}$ $-$ $\frac{1}{2}$ $=$ $\frac{1}{2}$

Therefore $3x+y = 4$  … … … … … (iii)

$3x-y = 2$ … … … … … (vi)

Adding (iii) and (iv) we get $6x = 6$ $\Rightarrow x = 1$

from (iv) $y = 3(1) - 2 = 1$

Exercise 7(a) continued …