Question

How do you solve `-x-3y= 15` and `2x+7y= -36` using substitution?

Answer 1

`x = 3`

Explanation:

`-x - 3y = 15" "("Solve for "x)`

`3y = 15 + -x`

`-x = 15+3y`

`(-x)/-1 = (15+3y)/-1`

`x = (15+3y)/-1`

`(15+3y)/-1xx(-1 )/-1 =( -15 + -3y)/1`

`2x + 7y = -36`

Substitute `-15 + -3y` for x

`2*(-15+ -3y)+7y =-36"(Distribute)"`

`(2*- 15) + (2 *3y) +7y = -36`

`-30+ -6y +7y = -36`

`-30 + y = -36`

`y = -6`

Plug in the y's value

`-x - 3 * -6 = 15`

`-x - (-18) = 15`

`-x + 18 = 15`

`-x = 15-18`

`-x = -3`

`-(-x) = -(-3)`

`x = 3`

But an easier approach would be using elimination method.

Answer 2

`3,-6)`

Explanation:

`-color(red)(x)-3y=15to(1)`

`2color(red)(x)+7y=-36to(2)`

`"from " (1)" we can make x the subject"`

`rArrcolor(red)(x)=-3y-15to(3)`

`"substitute this value into " (2)`

`rArr2(-3y-15)+7y=-36`

`rArr-6y-30+7y=-36larr" distributing"`

`rArry-30=-36larr" simplifying left side"`

`"add 30 to both sides"`

`ycancel(-30)cancel(+30)=-36+30`

`rArry=-6`

`"substitute into " (3)" and evaluate for x"`

`x=(-3xx-6)-15=18-15=3`

`rArr"point of intersection "=(3,-6)`