How do solve the following linear system?: # -7x+y=-19 , -2x+3y=-1 #?

1 Answer
Feb 27, 2017

#(x,y)=(56/19,31/19)#

Explanation:

#{(-7x+y=-19),(-2x+3y=-1):}#

Notice that the single variable #y# is easily isolated in the first equation.

Taking the first equation #-7x+y=-19#, we can rewrite this by adding #7x# to both sides to see that

#color(blue)(y=7x-19)#

We now have an expression equal to #y# completely in terms of #x#. We can then take the equation we haven't used yet and replace #y# with #7x-19#, since for this system we know these are equivalent expressions.

#-2x+3color(blue)y=-1" "=>" "-2x+3color(blue)((7x-19))=-1#

We can now solve this equation, since it's entirely in terms of #x#. Distributing the #3# into the parentheses gives

#-2x+(21x-57)=-1#

Combining the #x# terms then adding #57# to both sides gives

#19x-57=-1#

#19x=56#

Then

#color(red)(x=56/19#

Now we can plug this into either equation to find the value of #y#:

#-2color(red)x+3y=-1#

#-2color(red)((56/19))+3y=-1#

#-112/19+3y=-1#

Multiplying everything by #19# yields

#-112+57y=-19#

So

#57y=93#

#y=93/57#

Both of these are divisible by #3#:

#y=(31xx3)/(19xx3)#

#color(green)(y=31/19)#

So the solution is the point #(color(red)x,color(green)y)=(color(red)(56/19),color(green)(31/19))#.

#" "#

The graphs of the the lines #-7x+y=-19# and #-2x+3y=-1# should intersect at the point #(56/19,31/19)approx(2.95,1.63)#:

graph{(-7x+y+19)(-2x+3y+1)=0 [-10.61, 17.87, -4.8, 9.44]}

They do!