Question

3x-5y+8=0,7x+6y-9=0 linear pair of equation

Answer 1

Answer

$\sf \dashrightarrow 3x - 5y + 8 = 0 \: \: --- (i)$

$\sf \dashrightarrow 7x + 6y - 9 = 0 \: \: --- (ii)$

By first equation,

$\sf \dashrightarrow 3x - 5y + 8 = 0$

$\sf \dashrightarrow 3x - 5y = -8$

$\sf \dashrightarrow 3x = -8 + 5y$

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

Now, let's find the value of y by second equation.

$\sf \dashrightarrow 7x + 6y - 9 = 0$

$\sf \dashrightarrow 7 \bigg( \dfrac{-8 + 5y}{3} \bigg) + 6y - 9 = 0$

$\sf \dashrightarrow \dfrac{-56 + 35y}{3} + 6y - 9 = 0$

$\sf \dashrightarrow \dfrac{-56 + 35y + 18y}{3} - 9 = 0$

$\sf \dashrightarrow \dfrac{-56 + 53y}{3} - 9 = 0$

$\sf \dashrightarrow \dfrac{-56 + 53y - 27}{3} = 0$

$\sf \dashrightarrow \dfrac{-83 + 53y}{3} = 0$

$\sf \dashrightarrow -83 + 53y = 0$

$\sf \dashrightarrow 53y = 83$

$\sf \dashrightarrow y = \dfrac{83}{53}$

Now, let's find the value of x by first equation.

$\sf \dashrightarrow 3x - 5y + 8 = 0$

$\sf \dashrightarrow 3x - 5 \bigg( \dfrac{83}{53} \bigg) + 8 = 0$

$\sf \dashrightarrow 3x - \dfrac{415}{53} + 8 = 0$

$\sf \dashrightarrow \dfrac{159x - 415 + 424}{53} = 0$

$\sf \dashrightarrow \dfrac{159x + 9}{53} = 0$

$\sf \dashrightarrow 159x + 9 = 0$

$\sf \dashrightarrow 159x = -9$

$\sf \dashrightarrow x = \dfrac{-9}{159}$

$\sf \dashrightarrow x = \dfrac{-3}{53}$

Hence, the values of x and y are -3/53 and 83/53 respectively.