Solve the following pair of linear equations by the elimination method and the substitution
(i) $x + y = 5$ and $2x – 3y = 4$
(ii) $3x + 4y = 10$ and $2x – 2y = 2$
(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$.

To do:

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

Solution:

(i) By elimination method:

$x + y = 5$.....(i)

$2x – 3y = 4$.......(ii)

Multiplying (i) by 2 and subtracting (ii) from it, we get,

$2x+2y=10$

$2x-3y=4$

---------------

$      5y=6$

$y=\frac{6}{5}$

Putting the value of $y$ in (i), we get,

$x+\frac{6}{5}=5$

$x=5-\frac{6}{5}$

$x=\frac{5\times5-6}{5}$

$x=\frac{25-6}{5}$

$x=\frac{19}{5}$

By substitution method:

$x+y=5$

This implies,

$x=5-y$.....(i)

$2x-3y=4$

$2(5-y)-3y=4$            [From (i)]

$2(5)-2(y)-3y=4$

$10-2y-3y=4$

$10-4=5y$

$5y=6$

$y=\frac{6}{5}$

Therefore,

$x=5-\frac{6}{5}$

$x=\frac{5\times5-6}{5}$

$x=\frac{25-6}{5}$

$x=\frac{19}{5}$

The values of $x$ and $y$ are $\frac{19}{5}$ and $\frac{6}{5}$ respectively.

(ii) By elimination method:

$3x + 4y = 10$.....(i)

$2x – 2y = 2$.......(ii)

Multiplying (ii) by 2 and adding (i) to it, we get,

$3x+4y=10$

$4x-4y=4$

---------------

$7x=14$

$x=2$

Putting the value of $x$ in (i), we get,

$3(2)+4y=10$

$4y=10-6$

$y=\frac{4}{4}$

$y=1$

By substitution method:

$3x+4y=10$

This implies,

$x=\frac{10-4y}{3}$.....(i)

$2x-2y=2$

$2(\frac{10-4y}{3})-2y=2$            [From (i)]

$\frac{20-8y}{3}-2y=2$

$\frac{20-8y-6y}{3}=2$

$20-14y=2(3)$

$20-14y=6$

$14y=20-6$

$y=\frac{14}{14}$

$y=1$

Therefore,

$x=\frac{10-4(1)}{3}$

$x=\frac{10-4}{3}$

$x=\frac{6}{3}$

$x=2$

The values of $x$ and $y$ are $2$ and $1$ respectively.

(iii) By elimination method:

$3x - 5y = 4$.....(i)

$9x – 2y = 7$.......(ii)

Multiplying (i) by 3 and subtracting (ii) from it, we get,

$9x-15y=12$

$9x-2y=7$

---------------

$-13y=5$

$y=\frac{-5}{13}$

Putting the value of $y$ in (i), we get,

$3x-5(\frac{-5}{13})=4$

$3x+\frac{25}{13}=4$

$3x=4-\frac{25}{13}$

$3x=\frac{13(4)-25}{13}$

$3x=\frac{52-25}{13}$

$3x=\frac{27}{13}$

$x=\frac{9}{13}$

By substitution method:

$3x-5y=4$

This implies,

$x=\frac{4+5y}{3}$.....(i)

$9x-2y=7$

$9(\frac{4+5y}{3})-2y=7$            [From (i)]

$3(4+5y)-2y=7$

$12+15y-2y=7$

$13y=7-12$

$13y=-5$

$y=\frac{-5}{13}$

Therefore,

$x=\frac{4+5(\frac{-5}{13})}{3}$

$x=\frac{4-\frac{25}{13}}{3}$

$x=\frac{52-25}{13(3)}$

$x=\frac{27}{13(3)}$

$x=\frac{9}{13}$

The values of $x$ and $y$ are $\frac{9}{13}$ and $\frac{-5}{13}$ respectively.

(iv) By elimination method:

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

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

$3x+4y=-6$.....(i)

$x – \frac{y}{3} = 3$

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

$3x-y=9$.......(ii)

Subtracting (ii) from (i), we get,

$3x+4y=-6$

$3x-y=9$

---------------

$5y=-15$

$y=\frac{-15}{5}$

$y=-3$

Putting the value of $y$ in (i), we get,

$3x+4(-3)=-6$

$3x-12=-6$

$3x=-6+12$

$3x=6$

$x=\frac{6}{3}$

$x=2$

By substitution method:

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

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

$3x+4y=-6$.....(i)

$x – \frac{y}{3} = 3$

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

$3x-y=9$.......(ii)

This implies,

From (i),

$x=\frac{-6-4y}{3}$.....(iii)

$3x-y=9$

$3(\frac{-6-4y}{3})-y=9$            [From (iii)]

$-6-4y-y=9$

$5y=-6-9$

$5y=-15$

$y=-3$

Therefore,

$x=\frac{-6-4(-3)}{3}$

$x=\frac{-6+12}{3}$

$x=\frac{6}{3}$

$x=2$

The values of $x$ and $y$ are $2$ and $-3$ respectively.

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Updated on: 10-Oct-2022

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