What is the solution to #{:(–x+2y=4),(5x-3y=1):}#?

1 Answer
Jan 13, 2017

Answer:

The solution is #(x,y)=(2,3)#.

Explanation:

Each of these equations represents a line in 2D space. As with any pair of lines, they may cross, they may be parallel, or they may be the same line. Solving a pair of equations simultaneously means finding the #(x, y)# point where the lines cross (if it exists).

We start by assuming there is a point #(x, y)# that works for both equations

#"-"x+2y=4#
#5x-3y=1#

If this is true, then we can rearrange each equation and combine the two equations together to help us narrow in on the coordinates of the #(x,y)# point.

For example, if #"-"x+2y=4#, then we have

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

by solving for #x#. But, if this is the same #x# that works for the other equation, we can substitute this expression for #x# into the other equation like this:

#"      "5color(blue)x"      "-3y=1#
#5(color(blue)(2y-4))-3y=1#

and we end up with an equation with just #y#. Thus, we can solve for #y#:

#10y-20-3y=1#
#color(white)(10y-20-)7y=21#
#color(white)(10y-20-7)color(red)(y=3)#

So, this is the #y#-coordinate of the point that works for both lines. With this, we can now find the matching #x#-coordinate, by plugging in this value for #y# into either of our starting equations:

#"-"x+2color(red)y"   "=4#
#"-"x+2color(red)((3))=4#
#"-"x+6"     "=4#
#"-"x"            "="-"2#

#=>x=2#

That's it—we have found that the lines do cross, and the coordinates of the crossing point are #(x,y)=(2,3)#:

graph{(-x+2y-4)(5x-3y-1)=0 [-10, 10, -2, 8]}