Question

How do you solve the following system: `4x+3y= -7, 2x-5y=-19`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for `4x`:

Equation 1:

`4x + 3y - color(red)(3y) = -7 - color(red)(3y)`

`4x + 0 = -7 - 3y`

`4x = -7 - 3y`

Equation 2:

`2x - 5y + color(red)(5y) = -19 + color(red)(5y)`

`2x - 0 = -19 + 5y`

`2x = -19 + 5y`

`color(red)(2) xx 2x = color(red)(2)(-19 + 5y)`

`4x = (color(red)(2) xx -19) + (color(red)(2) xx 5y)`

`4x = -38 + 10y`

Step 2) Because the left sides of each equation is equal, we can equate the right sides of each equation and solve for `y`:

`-7 - 3y = -38 + 10y`

`color(blue)(38) - 7 - 3y + color(red)(3y) = color(blue)(38) - 38 + 10y + color(red)(3y)`

`31 - 0 = 0 + (10 + color(red)3)y`

`31 = 13y`

`31/color(red)(13) = (13y)/color(red)(13)`

`31/13 = (color(red)(cancel(color(black)(13)))y)/cancel(color(red)(13))`

`31/13 = y`

`y = 31/13`

Step 3) Substitute `31/13` for `y` in the solution to either equation in Step 1 and calculate `x`:

`4x = -7 - 3y` becomes:

`4x = -7 - (3 xx 31/13)`

`4x = -7 - 93/13`

`4x = -(13/13 xx 7) - 93/13`

`4x = -91/13 - 93/13`

`4x = -184/13`

`color(red)(1/4) xx 4x = color(red)(1/4) xx -184/13`

`4/color(red)(4)x = color(red)(1/color(black)(cancel(color(red)(4)))) xx -(color(red)(cancel(color(black)(184)))46)/13`

`x = -46/13`

The Solution Is: `x = -46/13` and `y = 31/13` or `(-46/13, 31/13)`