How do you solve $x + 3 y = - 1$ $x + 3y = -1$ and $6 x + 4 y = - 6$ $6x + 4y = -6$ ?

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Answer 1 · 2018-05-27T11:29:14

Solution : $x = - 1, y = 0$ $x =-1 , y=0$

$x + 3 y = - 1; (1), 6 x + 4 y = - 6; (2)$ $x+3 y = -1 ; (1) , 6 x +4 y = -6 ; (2)$ Multiplying equation (1)

by $6$ $6$ on both sides we get , $6 x + 18 y = - 6; (3)$ $6 x+18 y = -6 ; (3)$ Subtracting

equation (2) from equation (3 ) we get ,

$(6 x + 18 y) - (6 x + 4 y) = - 6 - (- 6)$ $(6 x+18 y) - (6 x +4 y ) = -6 - (-6)$ or

$cancel (6 x)+18 y - cancel (6 x) -4 y = cancel(-6) +cancel 6$ or

$14 y =0 :. y=0$ . Putting $y=0$ in equation (1) we get,

$x+0=-1 :. x =-1$

Solution $x =-1 , y=0$ [Ans]

How do you solve x + 3y = -1x+3y=−1x + 3y = -1 and 6x + 4y = -66x+4y=−66x + 4y = -6?