How do you solve #x=y+4# and #5x+3y=-20# using substitution?

1 Answer
Mar 20, 2016

Answer:

#y = -5#
#x = -1#

Explanation:

First we substitute in our value for #x# (which is #y+4#) into the other equation so that we only have one unknown, #y#.
Then you expand the bracket and simplify it to find #y#.

#5x+3y = -20#
#5(y+4)+3y=-20#
#5y + 20 +3y= -20#
#8y+20=-20)#
#8y = -40#
#y=-40/8#
#y=-5#

Now we use our value for #y# in the original equation to find #x#.

#x = y+4#
#x = -5+4#
#x=-1#

We can then check our answer by substituting both values into the other equation:

#(5xx-1)+(3xx-5) = -20#
#-5+ (-15) = -20)#
#-5 - 15 = -20#
#-20 = -20#
So we know it's right!
It's really important to check your answers, as in a test, it could save you a whole load of marks despite only taking a few seconds:)

Hope this helps; let me know if I can do anything else:)