How do you solve the system #x+2y=13# and #-2x-3y=-18# using substitution?

2 Answers
May 29, 2017

Answer:

#x=-3# and #y=8#

Explanation:

Starting with #x+2y=13#, get #x# by itself on the left-hand side by subtracting #2y# from both sides.

#x=-2y+13#

This gives us a new name for #x#, which we can plug back in to

#-2color(red)(x)-3y=-18#

#-2(color(red)(-2y+13))-3y=-18#

#4y-26-3y=-18#

Combine the terms with #y#

#y-26=-18#

Add #26# to both sides.

#y=8#

Now plug #y=8# back into one of the two equations with both #x# and #y# still in it from above.

#x=-2color(red)(y)+13#

#x=-2(color(red)(8))+13#

#x=-16+13#

#x=-3#

ANSWER: #x=-3# and #y=8#

May 29, 2017

Answer:

#x = -3#

#y = 8#

Explanation:

Since,

#x + 2y = 13#

Then,

#x = 13 - 2y#

Substuting #[x = 13 - 2y]# in the other equation, #(-2x - 3y = -1)#, It becomes

#-2(13 - 2y) - 3y = -18#

Simplify,

#-26 + 4y - 3y = -18#

Which is the same as:

#-26 + y = -18#

Put #y# on the left side of the equal sign and actual numbers on the right

#y = -18 + 26#

#y = 8#

Go back and solve #[x = 13 - 2y]# for #x#, since you now know what #y# is:

#x = 13 - 2(8)#

#x = 13 - 16#

#x = -3#

Ta da! Please do correct if I'm wrong...
Cheers and all the best!