# Standard Form

A linear equation in **standard form** is an equation that looks like

*ax* + * by* = *c*

where *a*, *b*, and *c* are real numbers and *a* and *b* aren't both zero. But *c* can be zero if it wants. It's the favorite child, so it gets special privileges.

If only *a* = 0, the equation can be rewritten to look like this:

*y* = (some number)

If only *b* = 0, the equation can be rewritten, too:

*x* = (some number)

For example, the equation 8*y* = 3 is equivalent to the equation , which is also in standard form (with *b* = 1).

Meanwhile, the equation 2*x* = 4 is equivalent to the equation *x* = 2, which is also in standard form (with *a* = 1).

If either *a* or *b* is zero, we know how to graph the equation and how to read off an equation from a graph. You probably suspect there will be some cases where it won't be so easy, and neither *a* *nor* *b* will be zero. You suspect right.

Okay, now what if an equation throws us a curveball? Should we sacrifice our bodies and take our base?

If neither *a* nor *b* is zero, we can most easily graph the linear equation by finding its intercepts.

### Sample Problem

Graph the linear equation *x* + 4*y* = 8.

Let's find the intercepts. To find the *x*-intercept, let *y* = 0 and solve for *x*, since the *x*-intercept will be at a point of the form (something, 0).

*x* + 4(0) = 8

So *x* = 8 is the *x*-intercept.

For the *y*-intercept, let *x* = 0 and solve for *y*.

0 + 4*y* = 8

And *y* = 2 is the *y*-intercept. Sweet, we've tracked down both intercepts. Who needs *a* or *b* to be zero? Not us.

Now we can plot the intercepts:

Connect the dots to get the line:

### Sample Problem

Write, in standard form, the linear equation graphed below:

The *x* intercept is at (-1, 0), which means whatever *a, b, * and *c* are, our equation looks like this:

*a*(-1) + *b*(0) = *c*

Let's make life easy on ourselves and let *a* = 1. That's right...we're going to dip this equation in a bucket of A-1 sauce.

1(-1) + *b*(0) = *c*

-1 = *c*

To find *b*, the remaining coefficient, we look at the *y*-intercept: *y* = -2. At that point, *x* will be 0, and we've already decided that *c* = -1, so we find:

0 + *b*(-2) = -1

Therefore, . We now know all the coefficients and can write the equation:

If we want to make things pretty, we can multiply both sides of the equation by 2 and write the resulting equation, which has integer coefficients. If we want to make things *really* pretty, we can dress the equation up in a sequined ball gown and give it a makeover. Let's start small, though:

2*x* + *y* = -2

### Sample Problem

Write, in standard form, the linear equation graphed below:

The *x* intercept is -2, which means whatever *a, b, * and *c* are, our standard-form equation is:

*a*(-2) + *b*(0) = *c*

We can let *a* = 1, so:

-2 = *c*

To find *b* we look at the *y*-intercept, which occurs at (0, 4). And since we've decided *c* = -2, we find:

0 + *b*(4) = -2

This means . We now know all the coefficients. Not on a first-name basis, but well enough to get by. We can now write the equation.

To make things pretty, we can multiply both sides of the equation by 2 to get an equivalent equation with integer coefficients:

2*x* – *y* = -4

Now for that makeover.