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

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

$\frac{3}{2}(x+y)\cdot\frac{2}{3}=6\cdot\frac{2}{3}\leftarrow$ 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

Example 1

$\dfrac{y}{2}-8x=6$

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 $y=ax+b$

$y-16x+16x=12+16x$

Therefore,

$y=16x+12$

Example 2

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

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

$\dfrac{1}{9}(9y)=(-18x+9)\dfrac{1}{9}$

Therefore,

$y=-2x+1$

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.

$y + 5 - 5 = x - 6 - 5$

$y = x - 11$

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.

$y = -x + 4$