Question

the pair of equations 2x-5y+4=0 and 2x+y-8=0​

Answer 1

Answer

$\sf \dashrightarrow 2x - 5y + 4 = 0 \: \: --- (i)$

$\sf \dashrightarrow 2x + y - 8 = 0 \: \: --- (ii)$

By first equation,

$\sf \dashrightarrow 2x - 5y + 4 = 0$

$\sf \dashrightarrow 2x - 5y = -4$

$\sf \dashrightarrow 2x = -4 + 5y$

$\sf \dashrightarrow x = \dfrac{-4 + 5y}{2}$

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

$\sf \dashrightarrow 2x + y - 8 = 0$

$\sf \dashrightarrow 2x + y = -8$

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

$\sf \dashrightarrow \dfrac{-8 + 10y}{2} + y = -8$

$\sf \dashrightarrow \dfrac{-8 + 10y + 2y}{2} = -8$

$\sf \dashrightarrow \dfrac{-8 + 12y}{2} = -8$

$\sf \dashrightarrow -8 + 12y = -8 \times 2$

$\sf \dashrightarrow -8 + 12y = -16$

$\sf \dashrightarrow 12y = -16 + 8$

$\sf \dashrightarrow 12y = -8$

$\sf \dashrightarrow y = \dfrac{-8}{12}$

$\sf \dashrightarrow y = \dfrac{-2}{3}$

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

$\sf \dashrightarrow 2x - 5y + 4 = 0$

$\sf \dashrightarrow 2x - 5y = -4$

$\sf \dashrightarrow 2x - 5 \bigg( \dfrac{-2}{3} \bigg) = -4$

$\sf \dashrightarrow 2x - \bigg(\dfrac{-10}{3} \bigg) = -4$

$\sf \dashrightarrow \dfrac{6x - (-10)}{3} = -4$

$\sf \dashrightarrow \dfrac{6x + 10}{3} = -4$

$\sf \dashrightarrow 6x + 10 = -4 \times 3$

$\sf \dashrightarrow 6x + 10 = -12$

$\sf \dashrightarrow 6x = -12 - 10$

$\sf \dashrightarrow 6x = -22$

$\sf \dashrightarrow x = \dfrac{-22}{6}$

$\sf \dashrightarrow x = \dfrac{-11}{3}$

Hence, the values of x and y are $\sf \dfrac{-11}{3}$ and $\sf \dfrac{-2}{3}$ respectively.