How do you solve the system $x + y = 4$ $x+y=4$ , $2 x + 3 y = 0$ $2x+3y=0$ ?

Question

3168 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-24T06:38:08

Rearrange the equations and solve by substitution.

$x = - y + 4$ $x=-y+4$ Let this be equation 1.

$2 x + 3 y = 0$ $2x+3y=0$ Let this be equation 2.

Substitute equation 1 into equation 2:

$2 (- y + 4) + 3 y = 0$ $2(color(red)(-y+4))+3y=0$

$- 2 y + 8 + 3 y = 0$ $-2y+8+3y=0$

$y = - 8$ $y=-8$

Substituting $y = - 8$ $y=-8$ into equation 1,

$x = - (- 8) + 4$ $x=-(color(red)(-8))+4$

$x = 12$ $x=12$

How do you solve the system x+y=4x+y=4, 2x+3y=02x+3y=0?