Question

How do you solve the following system?: `8x - 2y = - 4 , 2x - 3y = 5`

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for `8x`:

• Equation 1:

`8x - 2y = -4`

`8x - 2y + color(red)(2y) = -4 + color(red)(2y)`

`8x - 0 = -4 + 2y`

`8x = -4 + 2y`

• Equation 2:

`2x - 3y = 5`

`color(red)(4)(2x - 3y) = color(red)(4) xx 5`

`(color(red)(4) xx 2x) - (color(red)(4) xx 3y) = 20`

`8x - 12y = 20`

`8x - 12y + color(red)(12y) = 20 + color(red)(12y)`

`8x - 0 = 20 + 12y`

`8x = 20 + 12y`

Step 2) Because the left side of both equations are the same we can equate the right side of both equations and solve for `y`:

`-4 + 2y = 20 + 12y`

`-4 - color(blue)(20) + 2y - color(red)(2y) = 20 - color(blue)(20) + 12y - color(red)(2y)`

`-24 + 0 = 0 + (12 - color(red)(2))y`

`-24= 10y`

`-24/color(red)(10)= (10y)/color(red)(10)`

`-12/5 = (color(red)(cancel(color(black)(10)))y)/cancel(color(red)(10))`

`-12/5 = y`

`y= -12/5`

Step 3) Substitute `-12/5` for `y` in either of the equations in Step 1 and calculate `x`:

`8x = -4 + 2y` becomes:

`8x = -4 + (2 xx -12/5)`

`8x = -4 + (-24/5)`

`8x = -4 - 24/5`

`8x = (5/5 xx -4) - 24/5`

`8x = -20/5 - 24/5`

`8x = -44/5`

`8x xx 1/8 = -44/5 xx 1/8`

`8/8x = -44/40`

`x = -11/10`

The Solution Is:

`x = -11/10` and `y= -12/5`

Or

`(-11/10, -12/5)`