Question

How do you solve 3x - 10y = -25 and 4x + 40y = 20?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for a common term, in this problem it can be `40y`:

• Equation 1:

`3x - 10y = -25`

`color(red)(-4)(3x - 10y) = color(red)(-4) xx -25`

`(color(red)(-4) xx 3x) + (color(red)(-4) xx -10y) = 100`

`-12x + 40y = 100`

`-12x + color(red)(12x) + 40y = 100 + color(red)(12x)`

`0 + 40y = 100 + 12x`

`40y = 100 + 12x`

• Equation 2:

`4x + 40y = 20`

`4x - color(red)(4x) + 40y = 20 - color(red)(4x)`

`0 + 40y = 20 - 4x`

`40y = 20 - 4x`

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

`100 + 12x = 20 - 4x`

`100 - color(red)(100) + 12x + color(blue)(4x) = 20 - color(red)(100) - 4x + color(blue)(4x)`

`0 + (12 + color(blue)(4))x = -80 - 0`

`16x = -80`

`(16x)/color(red)(16) = -80/color(red)(16)`

`(color(red)(cancel(color(black)(16)))x)/cancel(color(red)(16)) = -5`

`x = -5`

Step 3) Substitute `-5` for `x` in the solution to either equation in Step 1 and solve for `y`:

`40y = 100 + 12x` becomes:

`40y = 100 + (12 * -5)`

`40y = 100 + (-60)`

`40y = 100 - 60`

`40y = 40`

`(40y)/color(red)(40) = 40/color(red)(40)`

`(color(red)(cancel(color(black)(40)))y)/cancel(color(red)(40)) = 1`

`y = 1`

The Solution Is:

`x = -5` and `y = 1`

Or

`(-5, 1)`