How do you solve #15x-10y=-5# and #15y=5+10x# using substitution?

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Answer 1 · 2018-06-27T01:15:40

#x=-1/5, y=1/5#

Since every term in both equations is divisible by #5#, let's divide every term by this. We now have the equations

#3x-2y=-1#

#3y=1+2x#

In the second equation, let's divide all terms by #3# to solve for #y#. We get

#color(blue)(y=2/3x+1/3)#

This is where the substitution comes in...let's plug this value for #y# into the first equation.

#3x-2color(blue)((2/3x+1/3))=-1#

Distributing the #-2#, we get

#3x-4/3x-2/3=-1#

Let's make all terms have a common denominator:

#9/3x-4/3x-2/3=-3/3#

Now, we can simplify this further:

#5/3x-2/3=-3/3#

#=>5/3x=-1/3#

#=>5x=-1#

#=>color(darkviolet)(x=-1/5)#

Let's solve for #y# now. We can plug this #x# value into our blue equation:

#y=2/3color(darkviolet)((-1/5))+1/3#

#=>y=-2/15+1/3#

#=>y=-2/15+5/15=3/15=1/5#

#x=-1/5, y=1/5#

Hope this helps!