Solve Equations with Variables and Constants on Both Sides

Solve an equation with constants on both sides

You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time — so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that started with variables and constants on both sides into the form $ax = b$. We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Example. Solve: $4x + 6 = -14$.

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side.

Since the left side is the variable side, the $6$ is out of place. We must “undo” adding $6$ by subtracting $6$ from both sides:

Simplify:

Now all the $x\text{s}$ are on the left and the constant is on the right. Use the Division Property of Equality to divide both sides by $4$:

Simplify:

Check: substitute $x = -5$ into $4x + 6 = -14$:

Solve an equation with variables on both sides

What if there are variables on both sides of the equation? We will start like we did above — choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.

Example. Solve: $5x = 4x + 7$.

Here the variable, $x$, is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.

We don’t want any variables on the right, so subtract $4x$ from both sides:

Simplify:

We have all the variables on one side and the constants on the other. We have solved the equation.

Check: substitute $7$ for $x$ into $5x = 4x + 7$:

Sometimes the constant is on the left side and the variable on the right, or the coefficients pull the variable toward one side more naturally.

Example. Solve: $5y - 8 = 7y$.

The only constant, $-8$, is on the left side of the equation and the variable, $y$, is on both sides. Let’s leave the constant on the left and collect the variables to the right.

Subtract $5y$ from both sides:

Simplify:

We have the variables on the right and the constants on the left. Divide both sides by $2$:

Simplify:

Rewrite with the variable on the left:

Check: substitute $y = -4$ into $5y - 8 = 7y$:

Example. Solve: $7x = -x + 24$.

The only constant, $24$, is on the right, so let the left side be the variable side.

Remove the $-x$ from the right side by adding $x$ to both sides:

Simplify:

All the variables are on the left and the constants are on the right. Divide both sides by $8$:

Simplify:

Check: substitute $x = 3$ into $7x = -x + 24$:

Solve equations with variables and constants on both sides

The next examples are the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.

Example. Solve: $7x + 5 = 6x + 2$.

Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are $7x$ and $6x$. Since $7$ is greater than $6$, make the left side the variable side and so the right side will be the constant side.

Collect the variable terms to the left side by subtracting $6x$ from both sides:

Simplify:

Now, collect the constants to the right side by subtracting $5$ from both sides:

Simplify:

Check: substitute $x = -3$ into $7x + 5 = 6x + 2$:

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Example. Solve: $6n - 2 = -3n + 7$.

We have $6n$ on the left and $-3n$ on the right. Since $6 > -3$, make the left side the “variable” side.

We don’t want variables on the right side — add $3n$ to both sides to leave only constants on the right:

Combine like terms:

We don’t want any constants on the left side, so add $2$ to both sides:

Simplify:

The variable term is on the left and the constant term is on the right. To get the coefficient of $n$ to be one, divide both sides by $9$:

Simplify:

Check: substitute $1$ for $n$ into $6n - 2 = -3n + 7$:

Example. Solve: $2a - 7 = 5a + 8$.

This equation has $2a$ on the left and $5a$ on the right. Since $5 > 2$, make the right side the variable side and the left side the constant side.

Subtract $2a$ from both sides to remove the variable term from the left:

Combine like terms:

Subtract $8$ from both sides to remove the constant from the right:

Simplify:

Divide both sides by $3$ to make $1$ the coefficient of $a$:

Simplify:

Check: substitute $a = -5$ into $2a - 7 = 5a + 8$:

Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. Choosing the side with the larger coefficient as the variable side helps avoid the negatives.

To solve an equation with fractions, we still follow the same steps to get the solution.

Example. Solve: $\tfrac{3}{2}x + 5 = \tfrac{1}{2}x - 3$.

Since $\tfrac{3}{2} > \tfrac{1}{2}$, make the left side the variable side and the right side the constant side.

Subtract $\tfrac{1}{2}x$ from both sides:

Combine like terms:

Subtract $5$ from both sides:

Simplify:

Check: substitute $x = -8$ into $\tfrac{3}{2}x + 5 = \tfrac{1}{2}x - 3$:

We follow the same steps when the equation has decimals, too.

Example. Solve: $3.4x + 4 = 1.6x - 5$.

Since $3.4 > 1.6$, make the left side the variable side and the right side the constant side.

Subtract $1.6x$ from both sides:

Combine like terms:

Subtract $4$ from both sides:

Simplify:

Use the Division Property of Equality:

Simplify:

Check: substitute $x = -5$ into $3.4x + 4 = 1.6x - 5$:

Solve equations using a general strategy

Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example. Solve: $3(x + 2) = 18$.

Simplify each side of the equation as much as possible using the Distributive Property:

Collect all variable terms on one side of the equation — all $x\text{s}$ are already on the left side.

Collect constant terms on the other side by subtracting $6$ from each side:

Simplify:

Make the coefficient of the variable term equal to $1$. Divide each side by $3$:

Simplify:

Check: substitute $x = 4$ into $3(x + 2) = 18$:

Example. Solve: $-(x + 5) = 7$.

Simplify each side of the equation as much as possible by distributing. The only $x$ term is on the left side, so all variable terms are on the left side of the equation:

Add $5$ to both sides to get all constant terms on the right side of the equation:

Simplify:

Make the coefficient of the variable term equal to $1$ by multiplying both sides by $-1$:

Simplify:

Check: substitute $x = -12$ into $-(x + 5) = 7$:

Example. Solve: $4(x - 2) + 5 = -3$.

Simplify each side of the equation as much as possible. Distribute, then combine like terms:

The only $x$ is on the left side, so all variable terms are on one side of the equation.

Add $3$ to both sides to get all constant terms on the other side of the equation:

Simplify:

Make the coefficient of the variable term equal to $1$ by dividing both sides by $4$:

Simplify:

Check: substitute $x = 0$ into $4(x - 2) + 5 = -3$:

Be careful when distributing a negative sign into parentheses.

Example. Solve: $8 - 2(3y + 5) = 0$.

Simplify — use the Distributive Property:

Combine like terms:

Add $2$ to both sides to collect constants on the right:

Simplify:

Divide both sides by $-6$:

Simplify:

Check: substitute $y = -\tfrac{1}{3}$ into $8 - 2(3y + 5) = 0$:

Example. Solve: $3(x - 2) - 5 = 4(2x + 1) + 5$.

Distribute:

Combine like terms:

Subtract $3x$ to get all the variables on the right since $8 > 3$:

Simplify:

Subtract $9$ to get the constants on the left:

Simplify:

Divide by $5$:

Simplify:

Check: substitute $x = -4$ into $3(x - 2) - 5 = 4(2x + 1) + 5$:

In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.

Example. Solve: $0.24(100x + 5) = 0.4(30x + 15)$.

Distribute:

Subtract $12x$ to get all the $x\text{s}$ to the left:

Simplify:

Subtract $1.2$ to get the constants to the right:

Simplify:

Divide:

Simplify:

Check: substitute $x = 0.4$ into $0.24(100x + 5) = 0.4(30x + 15)$:

Key terms

variable side — the side of an equation chosen to collect all the variable terms. constant side — the side of an equation chosen to collect all the constant terms. general strategy — a five-step method (simplify each side, collect variable terms, collect constant terms, make the coefficient $1$, check) that solves any linear equation, regardless of its starting form.

This section is adapted from Prealgebra 2e, Section 8.3: Solve Equations with Variables and Constants on Both Sides by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the Be Prepared quiz, Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.