How do you solve 3x=13+5y and 4x-7y-17=0?

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Answer 1 · 2018-05-04T11:45:09

Solution: $x = 6, y = 1$ $x=6 , y=1$

$3 x = 13 + 5 y; 4 x - 7 y - 17 = 0$ $3 x=13+5 y ; 4 x -7 y -17 =0$ or

$3 x - 5 y = 13 (1); 4 x - 7 y = 17 (2)$ $3 x- 5 y = 13 (1) ; 4 x -7 y =17 (2)$ Multiplying equation (1)

by $4$ $4$ and equation (2) by $3$ $3$ we get ,

$12 x - 20 y = 52 (3); 12 x - 21 y = 51 (4)$ $12 x- 20 y = 52 (3) ; 12 x -21 y =51 (4)$ Subtracting equation

(4) from equation (3) we get, $y = 1$ $y = 1$ Putting $y = 1$ $y=1$ in equation

(1) we get, $3 x - 5 * 1= 13 or 3 x = 18 :. x =6$

Solution: $x=6 , y=1$ [Ans]