Question

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

Answer 1

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.

Answer 2

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.)