How do you solve x=y+4x=y+4 and 5x+3y=-205x+3y=−20 using substitution?

1 Answer
Sarah C
Mar 20, 2016

y = -5y=−5
x = -1x=−1

Explanation:

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

5x+3y = -205x+3y=−20
5(y+4)+3y=-205(y+4)+3y=−20
5y + 20 +3y= -205y+20+3y=−20
8y+20=-20)8y+20=−20)
8y = -408y=−40
y=-40/8y=−408
y=-5y=−5

Now we use our value for yy in the original equation to find xx.

x = y+4x=y+4
x = -5+4x=−5+4
x=-1x=−1

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

(5xx-1)+(3xx-5) = -20(5×−1)+(3×−5)=−20
-5+ (-15) = -20)−5+(−15)=−20)
-5 - 15 = -20−5−15=−20
-20 = -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:)

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