Question

Question 1 (iii)
Solve the following pair of linear equations by the elimination method and the substitution method:
3x - 5y - 4 = 0 and 9x = 2y + 7

Answer 1

3x - 5y - 4 = 0 and 9x = 2y + 7

By elimination method:

$3x-5y-4=0$
$\Rightarrow 3x-5y=4...\left(i\right)$
$9x=2y+7$
$\Rightarrow 9x-2y=7...\left(ii\right)$
Multiplying equation (i) by 3, we get
$9x-15y=12...\left(iii\right)$

And we have
$9x-2y=7...\left(ii\right)$

Subtracting equation (ii) from equation (iii), we get
-13y = 5
$y=\frac{-5}{13}$

Putting value in equation (i), we get
3x - 5y = 4 ... (i)
$3x-5\left(\frac{-5}{13}\right)=4$

Multiplying by 13, we get
39x + 25 = 52
39x = 27

$x=\frac{27}{39}=\frac{9}{13}$

Therefore, $x=\frac{9}{13}$ and $y=\frac{-5}{13}$

By substitution method:

3x – 5y = 4 ... (i)
Adding 5y on both sides, we get
3x = 4 + 5y
Dividing by 3, we get,
$x=\frac{4+5y}{3}$ ... (iv)
Putting this value in equation (ii) we get,
9x - 2y = 7 ... (ii)
$\Rightarrow 9\frac{(4+5y)}{3}-2y=7$

$\Rightarrow 3(4+5y)-2y=7$
$\Rightarrow 12+15y-2y=7$
$\Rightarrow 13y=-5$

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

Substituting the value in equation (iv), we obtain:

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

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

$\therefore x=\frac{9}{13},y=\frac{-5}{13}$

Hence, we get $x=\frac{9}{13}$ and $y=\frac{-5}{13}$