How do you solve the system `3x-y=4` and `2x-3y=-9` using substitution?

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

`3x - y = 4`

`-color(red)(3x) + 3x - y = -color(red)(3x) + 4`

`0 - y = -3x + 4`

`-y = -3x + 4`

`-1 xx -y = -1(-3x + 4)`

`y = 3x - 4`

Step 2) Substitute `3x - 4` for `y` in the second equation and solve for `x`:

`2x - 3y = -9` becomes:

`2x - 3(3x - 4) = -9`

`2x - 9x + 12 = -9`

`-7x + 12 - color(red)(12) = -9 - color(red)(12)`

`-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 first equation at the end of Step 1 and calculate `y`:

`y = 3x - 4` becomes:

`y = (3 xx 3) - 4`

`y = 9 - 4`

`y = 5`

The solution is: `x = 3` and `y = 5` or `(3, 5)`

Given:
`3x-y=4" "..............Equation(1)`
`2x-3y=-9" ".........Equation(2)`
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`color(blue)("Determine the value of x")`

Consider `Equation(1)`

Add `color(red)(y)` to both sides

`" "color(green)(3x-ycolor(red)(+y)" "=" "4 color(red)(+y))`

But `-y+y=0` giving:

`" "3x+0=4+y`

`" "3x=4+y`

Subtract `color(red)(4)` from both sides

`" "color(green)(3xcolor(red)(-4)" "=" "4color(red)(-4)+y)`

`" "3x-4=y`

`" "y=3x-4" "......Equation(1_a)`

As we have just used equation(1) we now need to use equation(2)

Using `Equation(1_a)` substitute for `y` in `Equation(2)`

`color(green)(2x-3color(red)(y)=-9" "->" "2x-3(color(red)(3x-4))=-9)`

`" "2x-9x+12=-9`

`" "-7x+12=-9`

Subtract 12 from both sides

`" "color(green)(-7x+12color(red)(-12)" "=" "-9color(red)(-12))`

`" "-7x+0=-21`

Divide both sides by -7

`" "color(green)((-7)/(color(red)(-7))color(white)(.) x=(-21)/(color(red)(-7)))`

`" "+1xx x=+3`

`" "x=3`
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`color(blue)("Determine the value of y")`

I chose equation 1 as it is the most strait forward one to use.

Substitute for x in equation 1 giving:

`3x-y=4" "->" "3(3)-y=4`

`y=5`