How do you solve the following system of equations #x - 2y = 5# and #2x - 4y = 10#?

2 Answers
Mar 29, 2015

Since #2x - 4y = 10# is #(x-2y=5)# multiplied by #2#
They are really the same equation with an infinite number of solutions
.
Any pair #(x,(x-5)/2)# is a solution.

Mar 29, 2015

We could use substitution or the addition/subtraction method.

Let's use substitution:
#x-2y=5#
#2x-4y=10#

We'll solve the first equation for #x# (because we get to choose, and that looks easiest)

#x=5+2y#
Substitute in the other equation, to get:

#2(5+2y) - 4y = 10#, now we'll solve for #y#.

#10+4y-4y=10#
so we want #10=10#.

What happened?

The way substitution works is to suppose we have an #x# and a #y# that make the first equation true. Then we can solve for one variable in terms of the other variable. And, still assuming that we have a solution, try to also make the second equation true.

When we get #10=10# (or #5=5#, or #0=0# or anything like that), then what has happened is:

Every solution to the first equation already is a solution to the second. There is no additional requirement.

If you think about the lines we'd get if we graphed these 2 equations, you'll see that they are the same line. Every point on one line is a point of the other line.
OK there's only one line so it might be better to say: every solution to one equation is a solution to the other.

There are a couple of ways to write the solution:
we can say "the system is dependent"
of, course that doesn't really say what the solutions are.

Every solution to the first equation, #x-2y=5#, has #x=5+2y#, so we could write: the solutions are all pairs of the form: #(5+2y, y)#

Often, we like to do things by first choosing #x# rather than #y# so we'll solve #x-2y=5# for #y# in terms of #x#.

#x-2y-x=5-x#

#-2y=5-x#

#(-1/2)(-2y) = (-1/2)(5-x)#
so
#y=-5/2+x/2# (Or, better yet: #y=-1/2x+5/2#)

We can now write the solutions: #(x, -1/2x+5/2)#

Final note
If you write both equations in slope-intercept form: #y=mx+b#,
you'll get #y=-1/2x+5/2#. And you can see that they are both equations for the same line.
(That's one reason we like slope-intercept form.)