How do you solve $5y – x – 6 = 0$ and $-3x + 6y = -9$ ?

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Answer 1 · 2016-05-19T15:42:29

$x=9, y=3$

First Method, using substitution

By rearranging the first equation we have,

$x = 5y -6$

substitute value of $x$ in second equation

$-3*(5y -6) + 6y = -9$

$-15y +18 + 6y = -9$

$-9y +18 = -9$

$rArry-2 = 1$
$or y = 3$

by substituting value of $y$ in either of the equations
we have

$x = 9$

Second Method,
Multiply first equation with 3,

$3*(5y-x-6)=0$

$15y-3x-18=0$

Now subtract LHS of the second equation from the LHS above equation and do the same with the RHS,

$(15y-3x-18)-(-3x+6y) = 0-(-9)$

$15y -3x -18 +3x-6y=9$

$9y -18 =9$
$rArr y =3$

and by substituting $y$ in either of the equations we have

$x=9$

How do you solve 5y – x – 6 = 05y – x – 6 = 0 and -3x + 6y = -9-3x + 6y = -9?