How do you solve the system #2x+5y=-1, 3x-10y=16#?

1 Answer
Jun 8, 2017

Answer:

#(2, -1)#

Explanation:

You have been given two linear equations and are tasked with finding a value for #x# and a value for #y# which makes both of the equations true. There are at least two common ways of accomplishing this feat. Graphing and substitution.

Graphing
You can graph these two equations by solving them for #y# in terms of #x#, and then plotting them on the same graph.

The first equation #2x+5y=-1# can be manipulated as follows:

#5y=-2x-1#

#y=-2/5x-1/5#

The second equation #3x-10y=16#, can also be changed to:

#-10y=-3x+16#

#y=3/10 x-16/10#

Graphing these two linear equations on the same plane show where they intersect. That is the #x# and #y# solution you are looking for.

graph{(2x+5y+1)(3x-10y-16)=0[-1.1,5,-2,0.1]}

You may have to zoom in a bit to find that they meet at the point #(2,-1)#

Substitution
You can also multiply the two equations and add them in such a way that one of the variables gets eliminated. For example, if you multiply the first equation by #2#, you get:

#color(red)(2xx)(2x+5y)=color(red)(2xx)(-1)#

#4x+10y=-2#

Adding this version of the equation with the second equation gives:

#color(white)(a....)4x+10y=-2#
#+color(white)(aa)ul(3x-10y=16)#
#color(white)(aaa.)color(red)(7x+0y=14)#

Or #7x=14#, which means that #x=2#. Then you can plug #x=2# back into either of the original two equations.

#2color(blue)(x)+5y=-1#

#2(color(blue)(2)) +5y=-1#

#4+5y=-1#

#5y=-5#

#y=-1#

Notice the two methods give the same answer: #(2, -1)#!