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

1 Answer
Jan 30, 2016

Answer:

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))