How do you solve the following system: #x-4y=2, 5x+2y=20 #?

1 Answer
Jan 18, 2016

Answer:

#x=42/11,y=5/11#

Explanation:

Take the first equation.

#x-4y=2#

Add #4y# to both sides.

#x=4y+2#

Since we know that #x=4y+2#, we can place #x# with #4y+2# in the other equation.

#5color(red)x+2y=20#

#5color(red)((4y+2))+2y=20#

Distribute and solve for #y#.

#20y+10+2y=20#

#22y=10#

#y=10/22=5/11#

Now that we know the value for #y#, plug it into either equation to solve for #x#. I'm choosing the first equation since it's simpler, but the same answer would come from plugging the value into the second equation.

#x-4color(blue)y=2#

#x-4color(blue)((5/11))=2#

#x-20/11=2#

#x=22/11+20/11=42/11#

Since #x=42/11# and #y=5/11#, the two lines should intersect at the point #(42/11,5/11)# or #(3.818,0.455)#.

We can check a graph:

graph{(x-4y-2)(5x+2y-20)=0 [-5.26, 10.54, -3.38, 4.516]}