Question

How do you solve the following system: # 1/2x - 2y = 4 , 4x + y = 2 #?

Answer 1

See a solution process below:

Explanation:

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

`1/2x - 2y = 4`

`color(red)(2)(1/2x - 2y) = color(red)(2) * 4`

`(color(red)(2) * 1/2x) - (color(red)(2) * 2y) = 8`

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

`1x - 4y = 8`

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

`x - 0 = 8 + 4y`

`x = 8 + 4y`

Step 2) Substitute `(8 + 4y)` for `x` in the second equation and solve for `y`:

`4x + y = 2` becomes:

`4(8 + 4y) + y = 2`

`(4 * 8) + (4 * 4y) + y = 2`

`32 + 16y + y = 2`

`32 + 16y + 1y = 2`

`32 + (16 + 1)y = 2`

`32 + 17y = 2`

`32 - color(red)(32) + 17y = 2 - color(red)(32)`

`0 + 17y = -30`

`17y = -30`

`(17y)/color(red)(17) = -30/color(red)(17)`

`(color(red)(cancel(color(black)(17)))y)/cancel(color(red)(17)) = -30/17`

`y = -30/17`

Step 3) Substitute `-30/17` for `y` in the solution to the first equation at the end of Step 1 and calculate `x`:

`x = 8 + 4y` becomes:

`x = 8 + (4 * -30/17)`

`x = 8 + (-120/17)`

`x = 8 - 120/17`

`x = (17/17 * 8) - 120/17`

`x = 136/17 - 120/17`

`x = 16/17`

The Solution Is:

`x = 16/17` and `y = -30/17`

Or

`(16/17, -30/17)`