In this video, we will be learning how to solve for one variable in terms of another in linear equations, such as solving for y in terms of x. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

Example of Solving a Linear Equation


\frac{3}{2}(x+y)\cdot\frac{2}{3}=6\cdot\frac{2}{3}\leftarrow Multiply both sides by \frac{2}{3} to isolate x and y

 (x+y)-x=4-x\leftarrow Subtract x from both sides to isolate y

y=-x+4\leftarrow Write in y=mx+b form

Linear Equations (x and y)

Example 1


First, we must get rid of \dfrac{1}{2}. So, we multiply it by its reciprocal.

2(\dfrac{y}{2}-8x)=(6)(2) y-16x=12

Let’s add 16x to both sides to keep the format y=ax+b




Example 2


First, distribute 3 to each of the terms inside the parenthesis

 3(6x)+3(3y)=9 18x+9y=9

Then, subtract 18x to both sides

18x+9y-18x=9-16x 9y=-18x+9

We have to get rid of 9 in 9y to keep the format y=ax+b

So, we multiply it by its reciprocal




Video-Lesson Transcript

In this lesson, we’ll go over linear equations also known as first degree equations.

The basic format of a linear equation is

y = ax + b, where a and b are numbers.

The first thing is that we have two variables y and x.

A very important reminder about linear equation is why it’s called first-degree equation.

It’s because the exponent for the variables – y and x – is 1.

So if we have this equation:

y = ax^2+b

The end of the equation doesn’t matter at this point.

As soon as you see an exponent that isn’t 1 it’s not a linear equation.

The exponent has to be one.

So, what we really want to do in a linear equation is to solve for one of the variables.

So what we want is some variable such as y equals everything else. Whatever that everything else is.

For example:

y + 5 = x - 6

What you need to do is to isolate y by itself.

So, we have to do inverse operations or reverse PEMDAS to get y by itself.

Don’t worry about the fact that there’s x here. It’s just a number minus another number. This doesn’t matter.

The only thing that we want to focus on is y.

So let’s isolate y by subtracting 5 from both sides of the equation.

y + 5 - 5 = x - 6 - 5

Our final answer is

y = x - 11

Let’s try another one.

\dfrac{3}{2} (x + y) = 6

What we’re going to do here is to get rid of \dfrac{3}{2}.

To do this, we just have to multiply it by it’s reciprocal. It cancels out perfectly.

\dfrac{2}{3} \dfrac{3}{2} (x + y) = 6 \times \dfrac{2}{3}

A side note:

The reason why we multiplied it by its reciprocal is because we want to divide the fraction to cancel it out.

What is \dfrac{3}{2} \div \dfrac{3}{2}?

We have to do keep-change-flip. So we have

\dfrac{3}{2} \times \dfrac{2}{3}

So by multiplying by reciprocal, I actually am dividing.

Let’s move on solving

x + y = \dfrac{6}{1} \times \dfrac{2}{3}

Maybe you want to put a denominator to 6, so we can cross cancel.

So now we have

x + y = 4 y = 4 - x

This is okay but we want to write x-term first.

So our final answer is

y = -x + 4