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(Video) Solving Equations with Variables on Both Sides

by Friendly Math 101

Here's a video by Friendly Math 101 showing you how to solve equations when there's variables on both sides.

$\large{\text{Summary}}$

The steps to solving these problems are to:

1. Distribute.
2. Combine like terms.
3. Get all the variables to one side.
4. Solve for your variable.

The first problem she looks at is: $$4x-7=18+2x$$ To solve this, we want to get all the variable terms over to one side and everything else on the other side. $$\begin{align} 4x-7&=18+2x\\ 4x\,{\color{#8925AE}{-\,2x}}-7&=18+2x\,{\color{#8925AE}{-\,2x}}\\ 2x-7&=18\\ 2x-7\,\color{#8925AE}{+\,7}&=18\,\color{#8925AE}{+\,7}\\ 2x&=25\\ \frac{2x}{\color{#8925AE}{2}}&=\frac{25}{\color{#8925AE}{2}}\\ x&=\frac{25}2 \end{align}$$

Remember the basic rule of creating equivalent equations: whatever you do to one side of the equation, you have to do to the other. Use the Addition and Subtraction Properties of Equality and the Multiplication and Division Properties of Equality.

First, we're going to subtract $7$ from $3$ on the left side and distribute the $4$ across $(2x+2)$ on the right. $$\begin{align} 2x+3 -7 &= 4(2x+2)\\ 2x-4&=8x+8\end{align}$$ Now, we are going to isolate the variables onto one side of the equation. $$\begin{align}2x-4 \, {\color{#8925AE}{- \, 2x}}&= 8x+8 \, {\color{#8925AE}{- \, 2x}}\\ -4 \, {\color{#8925AE}{- \, 8}} &= 6x + 8 \, {\color{#8925AE}{- \, 8}}\\ -12 &= 6x \end{align}$$ Finally, we can divide both sides by $6$ to solve the equation. $$\begin{align} \frac{-12}{\color{#8925AE}{6}} &= \frac{6x}{\color{#8925AE}{6}}\\ -2 &= x \end{align}$$

(Videos) Solving Linear Equations with Variables on Both Sides

by mathman1024

This video by mathman1024 looks at some more complicated problems with variables on both sides.

$\large{\text{Summary}}$

Make sure you write out all your work when solving because it's easy to make mistakes when there's a lot of terms. The goal is to still get variables on one side and constants on the other.

The first problem he looks at is: $$\frac{x-5}{4}=3x+2$$ This is a multi-step equation. First, we want to remove the denominator by multiplying both sides by $4$. $$\begin{align} \frac{x-5}{4}&=3x+2\\ {\color{#8925ae}{4}}\left(\frac{x-5}{4}\right)&={\color{#8925ae}{4}}\left(3x+2\right)\\ x-5&=12x+8 \end{align}$$ Next, we want to get all the variables to one side and all the constants to the other. $$\begin{align} x-5&=12x+8\\ x\,{\color{#8925ae}{-\,x}}-5&=12x\,{\color{#8925ae}{-\,x}}+8\\ -5&=11x+8\\ -5\,{\color{#8925ae}{-\,8}}&=11x+8\,{\color{#8925ae}{-\,8}}\\ -13&=11x \end{align}$$ Finally, divide. It'll look like the equations have a common denominator. $$\begin{align} -13&=11x\\ \frac{-13}{\color{#8925ae}{11}}&=\frac{11x}{\color{#8925ae}{11}}\\ -\frac{13}{11}&=x \end{align}$$

Don't forget to check your answers! To do that, plug in the $x$ value you just solved for back into the original equation. If both sides of the equations match, you know your answer is correct!

(Videos) Solving Linear Equations

by mathman1024

Here's another video with some more problems. Try these on your own first and see if you can get the same answers.

1. $3x+10=5-12x$
2. $7x-11=3x-11$

Make sure you're mindful of the subtraction and addition properties & the multiplication and division properties.

Despite looking a little cluttered, equations with variables on both sides can be solved in much the same way as equations with variables on only one side.

Think of any of these equations like a messy room with piles of stuff everywhere. We've probably all seen a room a little like this.

The key to working with equations that have variables on both sides is to re-organize them so that similar things are together on the same side.

This explanation includes:

• Simple Equations with Variables on Both Sides
• Equations with Fractions and Variables on Both Sides (rational expressions)
• Equations with Variables on Both Sides Requiring Factoring

Simple Equation with Variables on Both Sides

$$-3x + 4 = 7x - 16$$

Clean-up time! Just like we'd do in the messy room, we need to group similar things together. Let's start by putting everything with an $x$ on one side of the equation and everything without an $x$ on the other side. Use addition and subtraction to move the terms around so that everything's on the right side.

$$\begin{align} -3x + 4 &= 7x - 16\\ -3x + 4\,\color{#8925AE}{ -\, 4} &= 7x - 16\,\color{#8925AE}{ -\, 4}\\ -3x &= 7x - 20\\ -3x\,\color{#8925AE}{ - \,7x} &= 7x - 20 \,\color{#8925AE}{-\, 7x}\\ -10x &= -20\\ \end{align}$$

Now that we've reorganized the equation by making our "piles" of like terms, all that's left is to solve for $x$ just like we would in any other problem.

Good! If we divide $-20$ by $-10$, we are left with $x=2$.

$$\begin{align} -10x &= -20\\ \frac{-10x}{\color{#8925AE}{-10}} &= \frac{-20}{\color{#8925AE}{-10}}\\ x &= 10 \end{align}$$

Equations with Fractions and Variables on Both Sides

It's also possible that you'll see an equation with rational expressions (fractions with polynomials) with variables on both sides. Although this might look more complicated, the general idea of reorganizing the problem still works. $$\frac{x}{2} + \frac{5}{6} = \frac{x}{3}$$

Before we can reorganize our equation, we need to make sure that the bottom number of each fraction is the same. The least common denominator is what we want: it's the smallest number that is a multiple of all of the denominators we're working with.

Now we need to multiply each term by a fraction equal to $1$ (so, one that has the same number in the numerator and the denominator) so that every single denominator is $6$. $$\begin{align} \left(\frac{3}{3}\right) \left(\frac{x}{2}\right) + \left(\frac{1}{1}\right)\left(\frac{2}{2}\right) &= \left(\frac{2}{2}\right)\left(\frac{x}{3}\right)\\ \frac{3x}{6} + \frac{5}{6} &= \frac{2x}{6} \end{align}$$

Now that we've got each fraction with the same denominator, we can reorganize our equation. Try this on your own and then check below!

$$\begin{align} \frac{3x}{6} + \frac{5}{6} &= \frac{2x}{6}\\ \frac{3x}{6}+\frac56\,\color{#8925AE}{-\,\frac56} &= \frac{2x}{6}\,\color{#8925AE}{-\,\frac56}\\ \frac{3x}{6}&=\frac{2x}6-\frac56\\ \frac{3x}{6}\,\color{#8925AE}{-\,\frac{2x}{6}}&=\frac{2x}6\,{\color{#8925AE}{-\,\frac{2x}{6}}}-\frac56\\ \frac{x}{6} &= - \frac{5}{6}\\ \end{align}$$

Now, let's finish solving the equation.

$$\begin{align} \frac{x}{6} &= - \frac{5}{6}\\ 6\left(\frac{x}{6}\right) &= 6\left(-\frac{5}{6}\right)\\ x &= -5\\ \end{align}$$

Quadratic Equations with Variables on Both Sides

Note: These require factoring which you might not have seen yet.

Sometimes, you might find equations with quadratic terms (like anything squared or cubed) that also have variables on both sides. You'll still need to reorganize the equation into "piles."

This time, though, we want ALL of the terms on the same side of the equation, leaving $0$ on the other side. $$2x - 5 = x^2 - 8x + 16$$

$$\begin{align} 2x - 5 &= x^2 - 8x + 16\\ 2x &= x^2 - 8x + 16 + 5\\ 0 &= x^2 - 8x + 16 + 5 - 2x\\ 0 &= x^2 - 10x + 21\\ \end{align}$$

Now that we've reorganized the equation, all you need to do is factor and solve for $x$.

$$0 = (x-7)(x-3)\\ x = 7, 3$$

The key to solving equations with variables on both sides is to move all the terms with the variable to one side, and everything else on the other. You can move the terms using addition or subtraction, multiplication, and division (or some combination therein). Then solve for the variable.

Example

Let's solve the following equation: $$4x-3=2x+7$$ The first step is to move all the $x$ terms to the left of the equation and all the constants (the normal numbers) to the other: $$\begin{align} 4x-3&=2x+7\\ 4x-3\,\color{#8925AE}{+\,3}&=2x+7\,\color{#8925AE}{+\,3}\\ 4x&=\ \ \ 2x+10\\ 4x\,\color{#8925AE}{-\,2x}&=2x\,{\color{#8925AE}{-\,2x}}+10\\ 2x&=10 \end{align}$$

Now we've got all the terms on the right sides of the equals sign, so we just have to finish solving for $x$. In this case, we'll use division. $$\begin{align} 2x&=10 \\ \frac{2x}{\color{#8925AE}{2}}&=\frac{10}{\color{#8925AE}{2}}\\ x&=5 \end{align}$$

Done!

We can also look at this equation through modeling.