How do you solve the system of equations $y= 2x + 8$ and $3x + 5y = 1$ ?

Question

2356 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-26T10:51:42

By arranging the equations

Arrange the equations (multiply the 1st by 3 and the 2nd by 2):

$3y - 6x = 24$
$6x + 10y = 2$

Combine these:
$13 y = 26$

$y= 26/13 = 2$

Put this in the first original equation:
$2x = y-8$

$2x = 2-8$

$2x = -6$

$x=-3$

Answer 2 · 2018-06-26T10:54:24

$x=-3$ and $y=2$

After putting first equation into second,

$3x+5*(2x+8)=1$

$13x+40=1$

$13x=-39$ , so $x=-3$

Thus, $y=2*(-3)+8=2$

How do you solve the system of equations y= 2x + 8y= 2x + 8 and 3x + 5y = 13x + 5y = 1?