Question

How do you solve the following system: #x + 8y = 15 , 4x + y = -1 #?

Answer 1

The solution is `x=-0.74, y=1.97`.

See the explanation below.

Explanation:

Call the two equations (1) and (2) as follows to make it easier to explain:

`x+8y=15` (1)
`4x+y=-1` (2)

Multiply (1) by 4 and call it (1'):

`4x + 32y = 60` (1')

Subtract (2) from (1') (we are doing this to eliminate `x` so we have only one variable in play)

`4x + 32y = 60` (1') minus
`4x+y=-1` (2)

Result:

`31y=61`

Divide both sides by 31:

#y=61/31 (or 1.97)

Now plug this value back into one of the original equations to find the value of `x`. It's probably simpler to use (1):

`x+8(61/31)=15`
`x=15-8(61/31) = -0.74`

(too messy for me to do it in fractions, but you're in the UK not the US so you can probably handle decimals. ;-))

(you can test these solutions by plugging both into either of the original equations and ensuring that it's true (i.e. that both sides are equal, within rounding error))