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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 ## 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. $y = -x + 4$