# Solve the following pair of linear equations by the substitution method.(i) $x + y = 14, x – y = 4$(ii) $s – t = 3, \frac{s}{3} + \frac{t}{2} = 6$(iii) $3x – y = 3, 9x – 3y = 9$(iv) $0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3$(v) $\sqrt{2} x+\sqrt{3} y=0, \sqrt{3} x-\sqrt{8} y=0$(vi) $\frac{3 x}{2}-\frac{5 y}{3}=-2, \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$.

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To do:

We have to solve the given pair of equations by the substitution method.

Solution:

(i) $x+y=14$.........(i)

This implies,

$y = 14-x$

Putting the value of $y$ in equation $x – y = 4$, we get,

$x – (14 – x) = 4$

$x – 14 + x = 4$

$2x = 4 + 14$

$2x = 18$

$x = 9$

Putting $x = 9$ in equation (i), we get,

$9 + y = 14$

$y = 14 – 9$

$y = 5$

The values of $x$ and $y$ are $9$ and $5$ respectively.

(ii) $s-t=3$........(i)

This implies,

$s = 3+t$

Putting the value of $s$ in equation $\frac{s}{3} + \frac{t}{2} = 6$, we get,

$\frac{3+t}{3} + \frac{t}{2} = 6$

$\frac{2(3+t)+3(t)}{6} = 6$

$6+2t+3t = 6(6)$

$5t = 36-6$

$t = \frac{30}{5}$

$t=6$

Putting $t = 6$ in equation (i), we get,

$s-6 = 3$

$s = 3 + 6$

$s = 9$

The values of $s$ and $t$ are $9$ and $6$ respectively.

(iii) $3x – y = 3$........(i)

This implies,

$y = 3x-3$

Putting the value of $y$ in equation $9x-3y=9$, we get,

$9x -3(3x-3) = 9$

$9x-9x+9 = 6$

$9 = 9$

This implies,

$x$ and $y$ can have infinite real values.

(iv) The given system of equations can be written as,

$0.2x+0.3y=1.3$

Multiplying by $10$ on both sides, we get,

$2x+3y=13$---(i)

$0.4x+0.5y=2.3$

$\Rightarrow 0.4x=2.3-0.5y$

Multiplying by $10$ on both sides, we get,

$4x=23-5y$

$\Rightarrow x=\frac{23-5y}{4}$----(ii)

Substitute $x=\frac{23-5y}{4}$ in equation (i), we get,

$2(\frac{23-5y}{4})+3y=13$

$\frac{2(23-5y)}{4}+3y=13$

Multiplying by $2$ on both sides, we get,

$2(\frac{23-5y}{2})+2(3y)=2(13)$

$23-5y+6y=26$

$y=26-23$

$y=3$

Substituting the value of $y=3$ in equation (ii), we get,

$x=\frac{23-5(3)}{4}$

$x=\frac{8}{4}$

$x=2$

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

(v) The given pair of equations are:

$\sqrt{2}x\ +\ \sqrt{3}y\ =\ 0$................(i)

$\sqrt{3}x\ −\ \sqrt{8}y\ =\ 0$…………(ii)

From equation (i) $x = -\sqrt{\frac{3}{2}}y$ …………….(iii)

Substituting this value in equation (ii) we obtain

$\sqrt{3}x\ −\ \sqrt{8}y\ =\ 0$

$\Rightarrow \sqrt{3}\left(-\sqrt{\frac{3}{2}}\right)y -\sqrt{8} y=0$

$\Rightarrow \frac{3}{-\sqrt{2}}y-\sqrt{8}y=0$

$\Rightarrow (\sqrt2) \frac{3}{-\sqrt{2}}y-\sqrt{8}y=(\sqrt2)0$

$\Rightarrow -3y-4y = 0$

$\Rightarrow y=0$

Now, substituting y in equation (iii) we obtain,

$x = -\sqrt{\frac{3}{2}}(0)$

$\Rightarrow x = 0$

This implies,

The value of $x$ and $y$ obtained are $0$ and $0$respectively.

(vi) Given pair of equations are:

$\frac{3 x}{2}-\frac{5 y}{3}=-2$  .......... $( i)$

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$  .......$(ii)$

On multiplying equation $(i)$ by $\frac{1}{3}$

$\frac{3 x}{2}\times\frac{1}{3}-\frac{5 y}{3}\times\frac{1}{3}=-2\times\frac{1}{3}$

$\frac{3}{6}x-\frac{5}{9}y=-\frac{2}{3}$  .......... $(iii)$

And multiply $(ii)$ by $\frac{3}{2}$

$\frac{x}{3}\times\frac{3}{2}+\frac{y}{2}\times\frac{3}{2}=\frac{13}{6}\times\frac{3}{2}$

$\frac{3}{6}x+\frac{3}{4}y=\frac{39}{12}$  ......... $(iv)$

Subtract $(iii)$ from $(iv)$

$\frac{3}{6}x+\frac{3}{4}y-\frac{3}{6}x+\frac{5}{9}y=\frac{39}{12}-( -\frac{2}{3})$

$\Rightarrow \frac{3}{4}y+\frac{5}{9}y=\frac{39}{12}+\frac{2}{3}$

$\Rightarrow \frac{27+20}{36}y=\frac{117+24}{36}$

$\Rightarrow \frac{47}{36}y=\frac{141}{36}$

$\Rightarrow 47y=141$

$\Rightarrow y=\frac{141}{47}$

$\Rightarrow y=3$

On putting $y=3$, in $( i)$

$\frac{3 x}{2}-\frac{5}{3}\times3=-2$

$\Rightarrow \frac{3 x}{2}-\frac{15}{3}=-2$

$\Rightarrow \frac{3x}{2}=-2+\frac{15}{3}$

$\Rightarrow \frac{3x}{2}=\frac{-6+15}{3}$

$\Rightarrow \frac{3x}{2}=\frac{9}{3}$

$\Rightarrow 9x=18$

$\Rightarrow x=\frac{18}{9}$

$\Rightarrow x=2$

This implies,

The value of $x$ and $y$ obtained are $2$ and $3$respectively.

Updated on 10-Oct-2022 13:19:45