Solve the following system of simultaneous linear equations by substitution method:

3x - 5y = -2

7x - 3y = -9

Given,

3x - 5y = -2 ………………….(1)

7x - 3y = -9 ………………….(2)

Solving equation (1), we get :

⇒ 3x - 5y = -2

⇒ 3x = -2 + 5y

⇒ x = $\dfrac{-2 + 5y}{3}$

Substituting above value of x in equation (2), we get :

$\Rightarrow 7\Big(\dfrac{-2 + 5y}{3}\Big) - 3y = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 3y \times 3}{3} = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 9y}{3} = -9\\[1em] \Rightarrow 7({-2 + 5y}) - 9y = -27\\[1em] \Rightarrow -14 + 35y - 9y = -27\\[1em] \Rightarrow -14 + 26y = -27\\[1em] \Rightarrow 26y = -27 + 14\\[1em] \Rightarrow 26y = -13\\[1em] \Rightarrow y = -\dfrac{13}{26}\\[1em] \Rightarrow y = -\dfrac{1}{2}.$

Substituting value of y in x = $\dfrac{-2 + 5y}{3}$, we get :

$\Rightarrow x = \dfrac{-2 + 5 \times \dfrac{-1}{2}}{3}\\[1em] \Rightarrow x = \dfrac{-2 - \dfrac{5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{\dfrac{-4 - 5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{-4 - 5}{6}\\[1em] \Rightarrow x = \dfrac{-9}{6}\\[1em] \Rightarrow x = -\dfrac{3}{2}.$

Hence, x = $-\dfrac{3}{2}$ and y = $-\dfrac{1}{2}$.