Question

solve this by substitution method​

Answer 1

Answer

$\sf \dashrightarrow 3x - 4y - 1 = 0 \: \: --- (i)$

$\sf \dashrightarrow 2x - \dfrac{8}{3}y + 5 = 0 \: \: --- (ii)$

By first equation,

$\sf \dashrightarrow 3x - 4y - 1 = 0$

$\sf \dashrightarrow 3x - 4y = 1$

$\sf \dashrightarrow 3x = 1 + 4y$

$\sf \dashrightarrow x = \dfrac{1 + 4y}{3}$

Now, we can find the value of y by second equation.

$\sf \dashrightarrow 2x - \dfrac{8}{3}y + 5 = 0$

$\sf \dashrightarrow 2 \bigg( \dfrac{1 + 4y}{3} \bigg) - \dfrac{8}{3}y + 5 = 0$

$\sf \dashrightarrow \dfrac{2 + 8y}{3} - \dfrac{8y}{3} + 5 = 0$

$\sf \dashrightarrow \dfrac{2 + 16y}{3} + 5 = 0$

$\sf \dashrightarrow \dfrac{2 + 16y + 15}{3} = 0$

$\sf \dashrightarrow \dfrac{17 + 16y}{3} = 0$

$\sf \dashrightarrow 17 + 16y = 3$

$\sf \dashrightarrow 16y = 3 - 17$

$\sf \dashrightarrow 16y = -14$

$\sf \dashrightarrow y = \dfrac{-14}{16}$

$\sf \dashrightarrow y = \dfrac{-7}{8}$

Now, we can find the value of x by first equation.

$\sf \dashrightarrow 3x - 4y - 1 = 0$

$\sf \dashrightarrow 3x - 4 \bigg( \dfrac{-7}{8} \bigg) - 1 = 0$

$\sf \dashrightarrow 3x - \dfrac{-28}{8} - 1 = 0$

$\sf \dashrightarrow \dfrac{24x - (-28) - 8}{8} = 0$

$\sf \dashrightarrow \dfrac{24x + 28 - 8}{8} = 0$

$\sf \dashrightarrow \dfrac{24x + 20}{8} = 0$

$\sf \dashrightarrow 24x + 20 = 0$

$\sf \dashrightarrow 24x = -20$

$\sf \dashrightarrow x = \dfrac{-20}{24}$

$\sf \dashrightarrow x = \dfrac{-5}{6}$

Hence, the values of x and y are $\sf \dfrac{-5}{6}$ and $\sf \dfrac{-7}{8}$ respectively.