How do you solve the system of equations #-3x-4y=20# and #x-10y=16#?

2 Answers

Answer:

#x=-4 # and #y=-2#

Explanation:

Given system of equations

#-3x-4y=20\ ........(1)#

#x-10y=16\ ........(2)#

Multiplying (2) by #3# & adding to (1) as follows

#-3x-4y+3(x-10y)=20+3\cdot 16#

#-34y=68#

#y=-2#

setting #y=-2# in (1) we get

#x=\frac{-4y-20}{3}#

#=\frac{-4(-2)-20}{3}#

#=-4#

Jul 9, 2018

Answer:

Express #x# as a function of #y# and replace in the second equation.

Explanation:

To get a very fast answer to your problem, just use one of the two equations to express #x# or #y#. In this case, let's do it with #x#.

So, we have the following system:

#1) \ -3x-4y=20#
#2) \ x-10y=16#

If we express #x# in #2)#, then we have:

#x=16+10y#

Then we replace #x# in #1)# with what we obtained with #2)# and we get:

#-3*(16+10y)-4y=20#

Then develop the brackets:

#-48-30y-4y=20#

Solve for #y#:

#-34y=68#

#y=-2#

Then replace #y# in #2)# by what you found:

#x-10*(-2)=16#

Solve for #x#:

#x=-4#

Finished!