Question

How do you solve `6x+y=13` and `y-x= -8` using substitution?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `y`:

`y - x = -8`

`y - x + color(red)(x) = -8 + color(red)(x)`

`y - 0 = -8 + x`

`y = -8 + x`

Step 2) Substitute `(-8 + x)` for `y` in the first equation and solve for `x`:

`6x + y = 13` becomes:

`6x + (-8 + x) = 13`

`6x - 8 + x = 13`

`6x - 8 + 1x = 13`

`6x + 1x - 8 = 13`

`(6 + 1)x - 8 = 13`

`7x - 8 = 13`

`7x - 8 + color(red)(8) = 13 + color(red)(8)`

`7x - 0 = 21`

`7x = 21`

`(7x)/color(red)(7) = 21/color(red)(7)`

`(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 3`

`x = 3`

Step 3) Substitute `3` for `x` in the solution to the second equation at the end of Step 1 and calculate `y`:

`y = -8 + x` becomes:

`y = -8 + 3`

`y = -5`

The Solution Is: `x = 3` and `y = -5` or `(3, -5)`

Answer 2

`(x,y)to(3,-5)`

Explanation:

`6x+color(red)(y)=13to(1)`

`color(red)(y)-x=-8to(2)`

`"from " (2)color(white)(x)color(red)(y)=x-8to(3)`

`"substitute "y=x-8" in "(1)`

`rArr6x+x-8=13`

`rArr7x-8=13`

`"add 8 to both sides"`

`7xcancel(-8)cancel(+8)=13+8`

`rArr7x=21`

`"divide both sides by 7"`

`(cancel(7) x)/cancel(7)=21/7`

`rArrx=3`

`"substitute this value in "(3)" and evaluate y"`

`rArry=3-8=-5`

`color(blue)"As a check"`

`"substitute these values into "(1)`

`(6xx3)+(-5)=13larr" True"`

`rArr"point of intersection "=(3,-5)`
graph{(y+6x-13)(y-x+8)((x-3)^2+(y+5)^2-0.04)=0 [-16.02, 16.02, -8.01, 8.01]}