Question

How do you solve the system of equations `x+ 4y = - 8` and `x + y = - 5`?

Answer 1

We would use elimination process here as I'd outline below.

`x = -4` and `y = -1`

Explanation:

We have `x + 4y = -8` be equation `(1)`
And `x + y = -5` be `(2)`

Then subtracting `(2)` from `(1)`,

`x + 4y - (x + y) = - 8 + 5`
`implies 3y = -3`

Which gives `y = -1`

Substituting the value of `y` in any one equations, say `(1)`

`x - 4 = -8`
Gives, `x = -4`

Answer 2

`x=28/5,y=-3/5`

Please note that this method is not the only way to solve a system of equations! The other answer is just as effective.

Explanation:

Given information
`x+4y=-8`
`x+y=5`

Being a system of equations problem, you can apply a substitution, where you find the "value" of one variable by calculating a new equation for it.

Let's find `x` from the second equation.

Working it out
`x+y=5`
`x=5-y` or `x=-y+5`

Now that we have the "value" for `x`, plug it into the first equation
`\color(red)(x)+4y=-8`

• `\therefore(\color(red)(-y+5))+4y=8`
• `-y+5-4y=8`
• `5-y-4y=8`
• `5-5y=8`
• `-5y=(8-5)`
• `-5y=3`
• `y=3/(-5)`

Based on the resulting value for `y` plugin to the equation found for "value" of `x`
`x=-\color(green)(y)+5`

• `\thereforex=-(\color(green)(3/(-5)))+5`
• `x=3/5+5`
• `x=3/5+25/5`
• `x=28/5`