## How to Solve Equations with Variables on Both Sides

- Simplify both sides individually
- Add or subtract like terms to cancel out the variable on one of the sides.
- Identify Order of Operations and undo them with inverse operations.

## Examples

Solve for x: \[10x+3=4x-9\]

*Can either side of the equation be simplified?*

\[\green 10x+3=4x-9\]

No, \(\green 10x+3\) is fully simplified and so is \(\green 4x-9\).

*Which term do you need to add or subtract to cancel out the variable on one side of the equation?*

\[\green 10x+3=4x-9\]

I could subtract \(\green 4x\) to cancel the variable on the right side. Or I could subtract \(\green 10x\) to cancel the variable on the left side.

Either way would work, but I prefer to subtract the smaller term because I like to avoid negative numbers. So, I will subtract \(\green 4x\) from both sides of the equation.

\[\begin{array}{ccccc} 10x & +3 & = & 4x & -9 \\ {\green -4x} & & & {\green -4x} & \\ \hline 6x & +3 & = & & -9 \end{array}\]

*If you knew what \(\green x\) was, how would you simplify the equation with order of operations?*

\[\green 6x+3=-9\]

If I knew what \(\green x\) was, I would follow the order of operations like this:

**STEP 1:** Multiply \(\green x\) by \(\green 6\).

**STEP 2:** Add \(\green 3\).

To solve the equation, I will undo each step, working backwards.

I will undo **STEP 2** by subtracting \(\red 3\) from both sides of the equation.

\[\begin{array}{cccc} 6x & +3 & = & -9 \\ & {\red -3} & & {\red -3} \\ \hline 6x & & = & -12 \end{array}\]

Then I will undo **STEP 1** by dividing both sides of the equation by \(\yellow 6\).

\[\begin{array}{ccc} 6x & = & -12 \\ {\yellow \div \, 6} & & {\yellow \div \, 6} \\ \hline x & = & -2 \end{array}\]

\[\green x=-2\]

I can check my answer by substituting \(\green x=-2\) into the original equation and simplifying to see if it makes a true statement.

\[10({\green -2})+3=4({\green -2})-9\]

\[-20+3=-8-9\]

\[-17=-17\]

This is a true statement, so \(\green x=-2\) is a solution to \(\green 10x+3=4x-9\).

Solve for x: \[2(3x-5)+8=11(x-2)\]

*Can either side of the equation be simplified?*

\[\purple 2(3x-5)+8=11(x-2)\]

Yes, the left side \(\purple 2(3x-5)+8\) can be simplified by distributing the \(\purple 2\) and combining like terms.

\[\purple 2(3x-5)+8\]

\[\purple 6x-10+8\]

\[\purple 6x-2\]

The right side \(\purple 11(x-2)\) can also be simplified by distributing the \(\purple 11\).

\[\purple 11(x-2)\]

\[\purple 11x-22\]

So, the simplified equation is…

\[\purple 6x-2=11x-22\]

*Which term do you need to add or subtract to cancel out the variable on one side of the equation?*

\[\purple 6x-2=11x-22\]

I could subtract \(\purple 11x\) to cancel the variable on the right side. Or I could subtract \(\purple 6x\) to cancel the variable on the left side.

Either way would work, but I prefer to subtract the smaller term because I like to avoid negative numbers. So, I will subtract \(\purple 6x\) from both sides of the equation.

\[\begin{array}{ccccc} 6x & -2 & = & 11x & -22 \\ {\purple -6x} & & & {\purple -6x} & \\ \hline & -2 & = & 5x & -22 \end{array}\]

*If you knew what \(\purple x\) was, how would you simplify the equation with order of operations?*

\[\purple -2=5x-22\]

If I knew what \(\purple x\) was, I would follow the order of operations like this:

**STEP 1:** Multiply \(\purple x\) by \(\purple 5\).

**STEP 2:** Subtract \(\purple 22\).

To solve the equation, I will undo each step, working backwards.

I will undo **STEP 2** by adding \(\red 22\) to both sides of the equation.

\[\begin{array}{cccc} -2 = & 5x & -22 \\ {\red +22} & & {\red +22} \\ \hline 20 & = & 5x & \end{array}\]

Then I will undo **STEP 1** by dividing both sides of the equation by \(\yellow 5\).

\[\begin{array}{ccc} 20 & = & 5x \\ {\yellow \div \, 5} & & {\yellow \div \, 5} \\ \hline 4 & = & x \end{array}\]

\[\purple x=4\]

I can check my answer by substituting \(\purple x=4\) into the original equation and simplifying to see if it makes a true statement.

\[2(3({\purple 4})-5)+8=11({\purple 4}-2)\]

\[2(12-5)+8=11({\purple 4}-2)\]

\[2(7)+8=11(2)\]

\[14+8=22\]

\[22=22\]

This is a true statement, so \(\purple x=4\) is a solution to \(\purple 2(3x-5)+8=11(x-2)\).

Solve for x: \[(3x)^2-216=(x-4)(x+4)\]

*Can either side of the equation be simplified?*

\[\blue (3x)^2-216=(x-4)(x+4)\]

Yes, the left side \(\blue (3x)^2-216\) can by applying the distributive power of exponents.

\[\blue (3x)^2-216\]

\[\blue 9x^2-216\]

The right side \(\blue (x-4)(x+4)\) can also be simplified by multiplying the binomials and then combining like terms.

I prefer to use the FOIL method, but you can use the box method or the multiplication algorithm if you like.

\[(x-4)(x+4)\]

First Terms: \((x)(x)=x^2\)

Outer Terms: \((x)(4)=4x\)

Inner Terms: \((-4)(x)=-4x\)

Last Terms: \((-4)(4)=-16\)

\[\blue x^2+4x-4x-16\]

\[\blue x^2-16\]

So, the simplified equation is…

\[\blue 9x^2-216=x^2-16\]

*Which term do you need to add or subtract to cancel out the variable on one side of the equation?*

\[\blue 9x^2-216=x^2-16\]

I could subtract \(\blue x^2\) to cancel the variable on the right side. Or I could subtract \(\blue 9x^2\) to cancel the variable on the left side.

Either way would work, but I prefer to subtract the smaller term because I like to avoid negative numbers. So, I will subtract \(\blue x^2\) from both sides of the equation.

\[\begin{array}{ccccc} 9x^2 & -216 & = & x^2 & -16 \\ {\blue -x^2} & & & {\blue -x^2} & \\ \hline 8x^2 & -216 & = & & -16 \end{array}\]

NOTE: If you have an \(x\) term and an \(x^2\) term in your equation, you cannot subtract one of them from the other because they are not like terms. Instead, you will have to solve the equation by following these steps for solving quadratic equations.

*If you knew what \(\purple x\) was, how would you simplify the equation with order of operations?*

\[\blue 8x^2-216=-16\]

If I knew what \(\blue x\) was, I would follow the order of operations like this:

**Step 1:** Raise \(\blue x\) to the exponent of \(\blue 2\).

**STEP 2:** Multiply by \(\blue 8\).

**STEP 3:** Subtract \(\blue 216\).

To solve the equation, I will undo each step, working backwards.

I will undo **STEP 3** by adding \(\red 216\) to both sides of the equation.

\[\begin{array}{cccc} 8x^2 & -216 & = & -16 \\ & {\red +216} & & {\red +216} \\ \hline 8x^2 & = & 200 \end{array}\]

Then I will undo **STEP 2** by dividing both sides of the equation by \(\yellow 8\).

\[\begin{array}{ccc} 8x^2 & = & 200 \\ {\yellow \div \, 8} & & {\yellow \div \, 8} \\ \hline x^2 & = & 25 \end{array}\]

Lastly, I will undo **Step 1** by square rooting both sides of the equation.

\[\begin{array}{ccc} x^2 & = & 25 \\ {\green \sqrt{\navy x^2}} & = & {\green \sqrt{\navy 25}} \\ \hline x & = & \pm 5 \end{array}\]

\[\blue x=\pm 5\]

I can check my answer by substituting \(\blue x=5\) and \(\blue x=-5\) into the original equation and simplifying to see if it makes a true statement.

\[(3({\blue 5}))^2-216=({\blue 5}-4)({\blue 5}+4)\]

\[(15)^2-216=(1)(9)\]

\[(225-216=(1)(9)\]

\[(225-216=9\]

\[(9=9\]

This is a true statement, so \(\blue x=5\) is a solution to \(\blue (3x)^2-216=(x-4)(x+4)\).

\[(3({\blue -5}))^2-216=({\blue -5}-4)({\blue -5}+4)\]

\[(-15)^2-216=(-9)(-1)\]

\[225-216=(-9)(-1)\]

\[225-216=9\]

\[9=9\]

This is also a true statement, so \(\blue x=-5\) is another solution to \(\blue (3x)^2-216=(x-4)(x+4)\).