If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.

Given,

Equations : 10y = 7x - 4 and 12x + 18y = 1

⇒ 10y = 7x - 4

⇒ y = $\dfrac{7x - 4}{10}$ ………(1)

Substituting value of y from equation (1) in 12x + 18y = 1, we get :

$\Rightarrow 12x + 18 \times \Big(\dfrac{7x - 4}{10}\Big) = 1 \\[1em] \Rightarrow 12x + \dfrac{126x - 72}{10} = 1 \\[1em] \Rightarrow \dfrac{120x + 126x - 72}{10} = 1\\[1em] \Rightarrow 246x - 72 = 10 \\[1em] \Rightarrow 246x = 10 + 72 \\[1em] \Rightarrow 246x = 82 \\[1em] \Rightarrow x = \dfrac{82}{246} = \dfrac{1}{3}.$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{7 \times \dfrac{1}{3} - 4}{10}\\[1em] = \dfrac{\dfrac{7}{3} - 4}{10} \\[1em] = \dfrac{\dfrac{7 - 12}{3}}{10} \\[1em] = \dfrac{-5}{3 \times 10} \\[1em] = \dfrac{-5}{30} \\[1em] = -\dfrac{1}{6}.$

Substituting value of x and y in 4x + 6y and 8y - x, we get :

$\Rightarrow 4x + 6y = 4 \times \dfrac{1}{3} + 6 \times -\dfrac{1}{6} \\[1em] = \dfrac{4}{3} + (-1) \\[1em] = \dfrac{4 - 3}{3} \\[1em] = \dfrac{1}{3}. \\[1em] \Rightarrow 8y - x = 8 \times -\dfrac{1}{6} - \dfrac{1}{3} \\[1em] = -\dfrac{4}{3} - \dfrac{1}{3} \\[1em] = \dfrac{-4 - 1}{3} \\[1em] = -\dfrac{5}{3}.$

Hence, $4x + 6y = \dfrac{1}{3} \text{ and } 8y - x = -\dfrac{5}{3}$.