How do you solve #x= -2y+10# and #3x-y=9# using substitution?

1 Answer
Feb 6, 2017

Answer:

#x = -4, y = 3#

Explanation:

When solving systems of equations using the substitution method, the whole goal is to plug one equation into the other to find solutions for variables.

To do this, you want at least one of your equations in the system to be isolated for a variable. For example, compare your first and your second equations. Your first equation, #x=-2y+10# has #x# isolated by itself on one side of the equals sign. Meanwhile, your second equation #3x-y=9# has both variables on one side. This is a usable setup and we can solve for both #x# and #y# with both equations in this format.

To begin, take your equation that has a variable isolated by itself in it (in your case, #x=-2y+10#). Think about what this expression means; anywhere you see an #x#, you can substitute in its place the expression #-2y+10# because they are equal to each other.

This means that for any equations in your system, you can substitute (#-2y+10#) anywhere you see an #x#.

So, let's substitute (#-2y+10#) for #x# in the expression #3x-y=9#.

#3(-2y+10)-y = 9#

Distribute the 3, combine like terms, and solve for the variable.
#-6y+30-y=9#
#-7y+30=9#
#-7y=-21#
#color(red)(y=3)#

This gives you the answer to one of your variables. Now that you have a new expression with an isolated variable, you can substitute it back into either of your original equations to solve for the other unsolved variable.

Plug the answer you got (#y=3#) into your original equation #x=-2y+10#. Everywhere you see a #y#, substitute a 3 in its place.

#x=-2(3)+10#
#x=-6+10#
#color(red)(x=4)#

You can check that the answers you got (#x=4,y=3#) are correct by substituting them back into one of your original expressions and seeing if you get a true statement in return.

Plug 4 in for #x# and 3 in for #y# in the equation #3x-y=9#
#3(4)-3 = 9#
#12-3 = 9#
#color(red)(9=9)#

Thus, we know that our answers are correct because we got a true statement in return. Hope this helps!