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

Mar 29, 2015

Since $2 x - 4 y = 10$ is $\left(x - 2 y = 5\right)$ multiplied by $2$
They are really the same equation with an infinite number of solutions
.
Any pair $\left(x , \frac{x - 5}{2}\right)$ is a solution.

Mar 29, 2015

We could use substitution or the addition/subtraction method.

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

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

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

$2 \left(5 + 2 y\right) - 4 y = 10$, now we'll solve for $y$.

$10 + 4 y - 4 y = 10$
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, $x - 2 y = 5$, has $x = 5 + 2 y$, so we could write: the solutions are all pairs of the form: $\left(5 + 2 y , y\right)$

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

$x - 2 y - x = 5 - x$

$- 2 y = 5 - x$

$\left(- \frac{1}{2}\right) \left(- 2 y\right) = \left(- \frac{1}{2}\right) \left(5 - x\right)$
so
$y = - \frac{5}{2} + \frac{x}{2}$ (Or, better yet: $y = - \frac{1}{2} x + \frac{5}{2}$)

We can now write the solutions: $\left(x , - \frac{1}{2} x + \frac{5}{2}\right)$

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