How do you solve the following system?: #5x +5y =27, 9x -5y = 0#

1 Answer
Nov 12, 2015

Answer:

By substitution.

Explanation:

We have two equations, #5x+5y=27# and #9x-5y=0#. In order to use substitution, we need to solve for one variable. The easiest one to solve for would be #9x-5y=0#

#9x-5y=0#
#9x=5y#

Now looking at the equation we need to substitute in for, it would be easier for us to plug something in with a denominator of 5 so they would cancel out and we can do a little less math. So we shall solve for y in this equation.

#y = 9/5x#

Now we plug this y into the other equation.

#5x+5(9/5x)=27# and simplifying this would give us
#5x+9x=27# now we need to combine like terms
#14x=27# and finally
#x=27/14#

Now we plug this into the other equation.

#9(27/14) - 5y = 0#

#243/14 - 5y = 0#

#243/14 = 5y#

#243/70 = y#

My calculator isn't working right now so I can't quite give you decimal values but those fractions should be correct.