Solving Equations with Variables on Both Sides

Up until now, the equations you've solved have had the variable on just one side. But what happens when you see something like \(5x + 3 = 2x + 12\)? Now the variable appears on both sides of the equation. Don't worry—this isn't as complicated as it looks. You just need one additional step: getting all the variable terms on one side and all the constant terms on the other.

The Basic Idea

Think of the equation as a game where you're trying to collect all the \(x\) terms on one side and all the numbers on the other side. You can move terms to either side—whichever makes the math easier—as long as you remember the golden rule: whatever you do to one side, you must do to the other.

Let's jump right into an example.

Example: Solve \(5x + 3 = 2x + 12\).

We have \(5x\) on the left and \(2x\) on the right. Let's move the \(2x\) to the left side by subtracting \(2x\) from both sides.

$$5x + 3 - 2x = 2x + 12 - 2x$$ $$3x + 3 = 12$$

Now the variable is only on the left side. We can solve this like a regular multi-step equation.

Subtract 3 from both sides: $$3x = 9$$

Divide by 3: $$x = 3$$

Let's verify this is correct by substituting \(x = 3\) into the original equation:

Left side: \(5(3) + 3 = 15 + 3 = 18\)

Right side: \(2(3) + 12 = 6 + 12 = 18\)

Both sides equal 18, so \(x = 3\) is correct.

Choosing Which Side for the Variable

You can move the variable terms to either side. Sometimes one choice leads to easier numbers than the other.

Example: Solve \(7x - 4 = 3x + 8\).

Option 1: Move \(3x\) to the left (subtract \(3x\) from both sides). $$7x - 4 - 3x = 3x + 8 - 3x$$ $$4x - 4 = 8$$ $$4x = 12$$ $$x = 3$$

Option 2: Move \(7x\) to the right (subtract \(7x\) from both sides). $$7x - 4 - 7x = 3x + 8 - 7x$$ $$-4 = -4x + 8$$ $$-12 = -4x$$ $$x = 3$$

Both approaches work! The first one is a bit cleaner because we end up with positive coefficients, but you get the same answer either way. In general, it's often easier to move the smaller variable term and keep the larger one where it is, so you avoid negative coefficients.

Equations That Need Simplification First

Sometimes you need to simplify one or both sides before you can start moving terms around.

Example: Solve \(4x + 2x - 5 = 3x + 7\).

First, combine the like terms on the left side: $$6x - 5 = 3x + 7$$

Now subtract \(3x\) from both sides: $$3x - 5 = 7$$

Add 5 to both sides: $$3x = 12$$

Divide by 3: $$x = 4$$

Check: Left side: \(4(4) + 2(4) - 5 = 16 + 8 - 5 = 19\). Right side: \(3(4) + 7 = 12 + 7 = 19\). ✓

Using the Distributive Property

If there are parentheses on one or both sides, distribute first.

Example: Solve \(3(x + 4) = 2x + 18\).

Distribute the 3 on the left: $$3x + 12 = 2x + 18$$

Subtract \(2x\) from both sides: $$x + 12 = 18$$

Subtract 12: $$x = 6$$

Check: \(3(6 + 4) = 3(10) = 30\) and \(2(6) + 18 = 12 + 18 = 30\). ✓

Example: Solve \(5(y - 2) = 3(y + 4)\).

Here we have parentheses on both sides. Distribute on both sides first.

Left side: \(5(y - 2) = 5y - 10\)

Right side: \(3(y + 4) = 3y + 12\)

Now we have: $$5y - 10 = 3y + 12$$

Subtract \(3y\) from both sides: $$2y - 10 = 12$$

Add 10: $$2y = 22$$

Divide by 2: $$y = 11$$

Check: \(5(11 - 2) = 5(9) = 45\) and \(3(11 + 4) = 3(15) = 45\). ✓

When the Answer is a Fraction or Decimal

Not every answer will be a whole number.

Example: Solve \(3x + 5 = x + 9\).

Subtract \(x\) from both sides: $$2x + 5 = 9$$

Subtract 5: $$2x = 4$$

Divide by 2: $$x = 2$$

That one worked out nicely. But what about this one?

Example: Solve \(4x + 7 = x + 13\).

Subtract \(x\): $$3x + 7 = 13$$

Subtract 7: $$3x = 6$$

Divide by 3: $$x = 2$$

Still a whole number. Let's try one more.

Example: Solve \(5x + 2 = 2x + 11\).

Subtract \(2x\): $$3x + 2 = 11$$

Subtract 2: $$3x = 9$$

Divide by 3: $$x = 3$$

Okay, here's one that actually gives us a fraction:

Example: Solve \(3x + 4 = x + 7\).

Subtract \(x\): $$2x + 4 = 7$$

Subtract 4: $$2x = 3$$

Divide by 2: $$x = \frac{3}{2}$$

That's the exact answer. You could also write it as 1.5 if you prefer decimals, or as the mixed number \(1\frac{1}{2}\).

Negative Coefficients

When you end up with a negative coefficient in front of your variable, don't panic.

Example: Solve \(2x - 8 = 5x + 4\).

Let's subtract \(5x\) from both sides: $$2x - 8 - 5x = 5x + 4 - 5x$$ $$-3x - 8 = 4$$

Add 8: $$-3x = 12$$

Divide by -3: $$x = -4$$

When you divide a positive by a negative, you get a negative.

Check: \(2(-4) - 8 = -8 - 8 = -16\) and \(5(-4) + 4 = -20 + 4 = -16\). ✓

Try These Yourself

1. \(6x + 5 = 2x + 17\)
2. \(9y - 3 = 5y + 13\)
3. \(3(m + 2) = m + 14\)
4. \(4a - 7 = a + 8\)
5. \(2(p - 3) = 4(p + 1)\)
6. \(7x + 4 = 3x + 20\)

Answers:

1. Subtract \(2x\): \(4x + 5 = 17\), subtract 5: \(4x = 12\), divide by 4: \(x = 3\)
2. Subtract \(5y\): \(4y - 3 = 13\), add 3: \(4y = 16\), divide by 4: \(y = 4\)
3. Distribute: \(3m + 6 = m + 14\), subtract \(m\): \(2m + 6 = 14\), subtract 6: \(2m = 8\), divide by 2: \(m = 4\)
4. Subtract \(a\): \(3a - 7 = 8\), add 7: \(3a = 15\), divide by 3: \(a = 5\)
5. Distribute: \(2p - 6 = 4p + 4\), subtract \(2p\): \(-6 = 2p + 4\), subtract 4: \(-10 = 2p\), divide by 2: \(p = -5\)
6. Subtract \(3x\): \(4x + 4 = 20\), subtract 4: \(4x = 16\), divide by 4: \(x = 4\)

Where Students Struggle

When you subtract a variable term from both sides, make sure you're actually subtracting it—not adding. If you have \(5x = 2x + 9\) and you subtract \(2x\) from the left, you get \(3x\), not \(7x\). Subtraction means taking away, not combining.

Sign errors become even more common when variables are on both sides. Keep careful track of whether you're adding or subtracting, and whether the coefficients are positive or negative. One flipped sign can throw off the entire answer.

Not simplifying first can lead to unnecessary complications. If you see \(3x + 2x + 1 = 4x + 6\), combine the \(3x\) and \(2x\) into \(5x\) before you start moving terms around. It'll save you headaches.

Some students forget the final step of actually solving for the variable. Getting to \(3x = 12\) is great progress, but you're not done until you divide and get \(x = 4\). Don't stop one step early.

Always, always check your answer. Substitute the coordinates back into both original equations. It only takes a few seconds and catches most errors before you turn in your work.