# Tips for Solving Equations With Variables on Both Sides ••• SARINYAPINNGAM/iStock/GettyImages

When you first start solving algebraic equations, you're given relatively easy examples like ​x​ = 5 + 4 or ​y​ = 5(2 + 1). But as time creeps on you'll be faced with harder problems that have variables on both sides of the equation; for example, 3​x​ = ​x​ + 4 or even the scary-looking ​y2 = 9 – 3​y2.​ When this happens, don't panic: You're going to use a series of simple tricks to help make sense of those variables.

Your first step is to group the variables on one side of the equal sign – usually on the left. Consider the example of 3​x​ = ​x​ + 4. If you add the same thing to both sides of the equation you won't change its value, so you're going to add the additive inverse of ​x​, which is ​−x​, to both sides (this is the same as subtracting ​x​ from both sides). This gives you:

3x - x = x + 4 - x

Which in turn simplifies to:

2x = 4

#### Tips

• When you add a number to its additive inverse, the result is zero – so you are effectively zeroing out the variable on the right.

Now that your variable expressions are all on one side of the expression, it's time to solve for the variable by stripping away any non-variable expressions on that side of the equation. In this case, you need to remove the coefficient 2 by performing the inverse operation (dividing by 2). As before, you must perform the same operation on both sides. This leaves you with:

\frac{2x}{2} = \frac{4}{2}

Which in turn simplifies to:

x = 2

### Another Example

Here's another example, with the added wrinkle of an exponent; consider the equation

y^2 = 9 - 3y^2

You'll apply the same process you used without the exponents:

Don't let the exponent intimidate you. Just as with a "normal" variable of the first order (without an exponent), you'll use the additive inverse to "zero out" −3​y2 from the right side of the equation. Add 3​y2 to both sides of the equation. This gives you:

y^2 + 3y^2 = 9 - 3y^2 + 3y^2

Once simplified, this results in:

4y^2 = 9

Now it's time to solve for ​y​. First, to strip away any non-variables from that side of the equation, divide both sides by 4. This gives you:

\frac{4y^2}{4} = \frac{9}{4}

Which in turn simplifies to:

y^2 = \frac{9}{4} \text{ or } y^2 = \frac{9}{4}

Now you have only variable expressions on the left side of the equation, but you're solving for the variable ​y​, not ​y2. So you have one more step remaining.

Cancel out the exponent on the left side by applying a radical of the same index. In this case, that means taking the square root of both sides:

\sqrt{y^2} = \sqrt{\frac{9}{4}}

Which then simplifies to:

y = \frac{3}{2}

### A Special Case: Factoring

What if your equation has a mix of variables of different degrees (e.g., some with exponents and some without, or with different degrees of exponents)? Then it's time to factor, but first, you'll start the same way you did with the other examples. Consider the example of

x^2 = -2 - 3x

As before, group all the variable terms on one side of the equation. Using the additive inverse property, you can see that adding 3​x​ to both sides of the equation will "zero out" the ​x​ term on the right side.

x^2 + 3x = -2 - 3x + 3x

This simplifies to:

x^2 + 3x = -2

As you can see, you have, in effect, moved the ​x​ over to the left side of the equation.

Here's where the factoring comes in. It's time to solve for ​x​, but you can't combine ​x2 and 3​x​. So instead, some examination and a little logic might help you recognize that adding 2 to both sides zeroes out the right side of the equation and sets up an easy-to-factor form on the left. This gives you:

x^2 + 3x + 2 = -2 + 2

Simplifying the expression on the right results in:

x^2 + 3x + 2 = 0

Now that you've set yourself up to make it easy, you can factor the polynomial on the left into its component parts:

(x + 1)(x + 2) = 0

Because you have two variable expressions as factors, you have two possible answers for the equation. Set each factor, (​x​ + 1) and (​x​ + 2), equal to zero and solve for the variable.

Setting (​x​ + 1) = 0 and solving for ​x​ gets you ​x​ = −1.

Setting (​x​ + 2) = 0 and solving for ​x​ gets you ​x​ = −2.

You can test both solutions by substituting them into the original equation:

(-1)^2 + 3 × (-1) = -2

simplifies to

1 - 3 = -2 \text{ or } -2 = -2

which is true, so this ​x​ = −1 is a valid solution.

(-2)^2 + 3 × (-2) = -2

simplifies to

4 - 6 = -2 \text{ or, again } -2 = -2

Again you have a true statement, so ​x​ = −2 is a valid solution as well.

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