Question

Solve the equation by substitution method.Find X and Y.​

Answer 1

$\frac{4}{x } +3y=8\\\\3y=8-\frac{4}{x}\\\\ 3y=\frac{8x-4}{x}\\\\ y=\frac{8x-4}{3x}-----(1)\\\\\frac{6}{x}-4y=-5\\\\\frac{6}{x}-4(\frac{8x-4}{3x})=-5\\\\\frac{6}{x}-\frac{32x-16}{3x}=-5\\\\\frac{18x-32x-16}{3x}=-5\\\\-14x-16=-15x\\ x=16$

Putting value of x in (1) equation.

$y=\frac{(8*16)-4}{3*16} \\\\y=\frac{128-4}{48}\\ \\y=\frac{31}{12}$

To check whether the solution is correct you can put the value of x and y in the equation 4/x + 3y = 8.

If LHS comes to be 8 then the values are correct.