How do you solve by substitution #3x + 2y = 6# and #x + 2y = -6#?

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Answer 1 · 2015-06-13T22:03:48

#x=6#
#y=-6#

Try this:
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Answer 2 · 2015-06-13T22:05:53

#x=6#
#y=-6#

To solve a system by substitution, you need to replace one unknown by a function of the other, using one of the equations.

Example:
#x+2y=-6#
becomes
#x=-2y-6#

Then, in the other equation, you replace #x# with its #y#-depending expression:

#3x+2y=6#
becomes
#2y-6y-18=6#
#rarr -4y=24#
#rarr y=-6#

Now that you have the value of #y#, you can find the value of #x#:

#x=-2y-6#
becomes
#x=-2*(-6)-6#
#rarr x=12-6=6#