Linear Equations (x and y)

Only $19.95/mo. No commitment. Cancel anytime.

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.

For Example:

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

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

Linear Equations (x and y)

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

$y = -x + 4$

Caddell Prep offers online test prep, web-based math assignments, and in-person tutoring and test prep.