How do you solve the system #5x-10y=15# and #3x-2y=3# by multiplication?

1 Answer
Jun 4, 2018

Answer:

See explanation.

Explanation:

The system is:

#{(5x-10y=15),(3x-2y=3):}#

First we can see that all terms in the first equation are divisible by #5#. So we can divide both sides bu #5# to get the lower numbers:

#{(x-2y=3),(3x-2y=3):}#

Now the coefficients of #y# are the same in both equations #(-2)#, so if we multiply any of the equations by #-1# we get the opposite coefficients:

#{(-x+2y=-3),(3x-2y=3):}#

Now if we add both sides of both equations we get an equation with one unknown only:

#2x=0#

#x=0#

Now we can substitute the calculated alue of #x# into any of the previous equations to calculate #y#:

#3*0-2y=3#

#-2y=3#

#y=-3/2#

Now we can write the answer:

The system has one solution:

#{(x=0),(y=-3/2):}#