# Rules for Manipulating Equations

Hi, and welcome to this video about **manipulating equations**! In this video, we’ll be discussing these three main concepts, seeing what each is, why we need to know it, and how we use it:

- Carrying out the Same Operation on Both Sides of the Equation
- The Advantage of Combining Like Terms
- Canceling Terms on Opposite Sides of an Equation

Let’s get started!

Before we jump into it, let’s quickly review some background info to ensure we’re on the same page. Remember that an equation is a statement of equality. The equation \(2x\) plus 4 equals \(10 [2x + 4 = 10]\) tells us that the expression \(2x\) plus \(4\text{ }2x + 4\) is equivalent to the expression 10. It also tells us there is some value of \(x\) that will make the expression \(2x\) plus 4 actually equal 10 and, therefore, make the statement true. Which in this case, \(x\) would be 3: 2 times 3, plus 4 equals \(10 [2(3) + 4 = 10]\). All other values of \(x\) make the statement false. For instance, if we tried to say that \(x=5\), the left side of the equation would equal 14 and therefore not be equal to 10.

Okay, now let’s get into the first concept: carrying out the same operation on both sides of the equation.

### CARRYING OUT THE SAME OPERATION ON BOTH SIDES OF THE EQUATION

Since the two expressions on either side of the equals sign are in fact equal, we can use any operation we want on one expression as long as we do the exact same thing to the other expression. This is generally known as keeping equations balanced. It produces equivalent expressions and does not change the value of \(x\) that makes the original statement true.

For example, we can subtract 7 from both sides, which would give us \(2x\) minus 3 equals 3.

\( -7\text{ }-7\)

\(2x – 3 = 3\)

The statement is still true when \(x\) equals 3: 2 times 3, minus 3 does in fact equal 3. \(2(3)–3=3\)

Or we can divide all the terms on both sides by 2. So, we have our original again \(2x\) plus 4 equals 10. And if we divide each part by 2, that gives us \(x\) plus 2 equals 5.

\(x+2=5\)

Again, when \(x=3\), the statement is true, and all other values make it false: 3 plus 2 equals 5. \(3+2=5\)

Typically, when given an equation, we want to isolate \(x\) using addition, subtraction, multiplication, or division so we can see its value. So, even though we can manipulate an equation using any terms we want, there are certain operations in every equation that help us reach a solution.

Now that we’ve got that down, let’s discuss our next concept: combining like terms.

### THE ADVANTAGE OF COMBINING LIKE TERMS

Combining like terms is an important equation-solving step that is usually done before any actual solving begins as a way to simplify the equation. To combine like terms, add or subtract all of the terms with the same variable.

For example, consider the equation \(2(x+3)+3(2+x+3)=-4\). The 2 and the 3 in the second set of parentheses are like terms, so we can combine them. Doing so gives us 2 times (\(x\) plus 3), plus 3 times (\(x\) plus 4) equals negative 4. \(2(x+3)+3(x+5)=-4\). Now, we can carry out the multiplications that are implied by the parentheses, distributing the outer 2 and 3 accordingly: 2 times \(x\) is \(2x\), plus 2 times 3 is 6, plus 3 times \(x\) is \(3x\), plus 3 times 5 is 15, equals negative 4.

The \(2x\) and the \(3x\) terms are like terms, so we can add them together to get \(5x\) plus 6, plus 15, equals negative 4. \(5x+6+15=-4\). Now, the constants 6, 15, and \(–4\) are also like terms, and we can combine them as well. Subtracting 6 and 15 from both sides of the equation, will give us this: \(5x\) equals negative 25, which simplifies further (by dividing by 5 on both sides) to \(x=-5\).

Let’s look at another example. In the equation \(2x+4=16–x\), we need to get the \(x\)-terms on one side and the numbers on the other.

We can add x and subtract 4. This will gives us \(3x\) equals 12.

\(2x+4+x=16–x+x\)

\(3x+4–4=16–4\)

\(3x=12\)

Or we can subtract \(2x\) and subtract 16. So, if we try this one again, we have \(2x plus 4 equals 16 minus [latex]x\) and if we subtract \(2x\) from both sides, and subtract 16, that will leave us with negative 12 equals negative \(3x\).

\(2x+4–2x=16–x–2x\)

\(4–16=16–3x–16\)

\(-12=-3x\)

Regardless of the steps to combine the like terms, the solution is the same: \(x\) equals 4. Divide by 3 on both sides and we get that \(x\) equals 4, and here divide by negative 3 on both sides again gives us that \(x\) equals 4 (since dividing a negative by a negative results in a positive.

Now onto the third and final concept we’re going to cover: canceling terms on opposite sides of an equation.

### CANCELING TERMS ON OPPOSITE SIDES OF AN EQUATION

Sometimes, manipulating terms in an equation is called “canceling” or “canceling out”. Two terms on opposite sides of an equation can be canceled if and only if they exactly match each other. They must have the same variable raised to the same power and the same coefficient. Canceling can be done using any of **the four operations**, but there is a more specific way to think about what we’re doing.

When we add or subtract a term to both sides of an equation, we’re making a zero by adding the opposite of a term.

In the equation \(x+6=13\), we would solve by subtracting 6 from both sides.

\(-6\text{ }-6\)

\(x=7\)

By subtracting 6 from both sides, we made 0 on the left side, since 6 minus 6 equals 0, allowing us to solve the equation.

By the same token, in the equation \(x\) minus 5 equals negative \(4x\) \(x–5=-4x\), we might subtract \(x\) from both sides to make a 0 and find the solution. Then we’ll divide by negative 5 on both sides to get that \(x\) equals 1.

\(-x\text{ }-x\)

\(-5=-5x\)

\(1=x\)

With multiplication and division, the goal is to make a 1. In the equation \(3x\) equals 9 \(3x=9\), we need to make that \(3x\) a single \(x\) by dividing both sides by 3.

\(3x/3=9/3\)

\(x=3\)

Since 3 over 3 equals 1 \(3/3=1\), we were able to solve the equation.

Thinking about manipulation in terms of making zeros and making ones will help you determine the path to finding a solution to any equation. Let’s practice a couple before we go!

Solve: x minus 3 over 2 equals 6. \( \frac{x-3}{2}=6\\\)

Make a one by multiplying both sides by 2: \(x\) minus 3 equals 12.

\(x–3=12\)

Make a zero by adding 3 to both sides: \(x\) equals 15.

\(x = 15\)

Solve: \(8x+7=-11–x\)

Make a zero by subtracting 7 from both sides: \(8x\) equals negative 18 minus \(x\)

[latex]8x=-18–x\)

Make another zero by adding x to both sides: 9x equals negative 18.

\(9x=-18\)

Make a one by dividing both sides by 9: x equals negative 2.

\(x=-2\)

I hope that this video increased your equation manipulation skills! Thanks for watching, and happy studying!

## Practice Questions

**Question #1:**

Solve for *x*:

\(3x+6=33\)

**Answer:**

The correct answer is C. We only need to carry out two steps to solve the equation:

Step One: \(3x+6=33\)

\(3x+6-6=33-6\)

Step Two: \(3x=27\)

\(3x\div3=27\div3\)

**Question #2:**

Solve for *x* by cancelling out terms on both sides of the equation:

\(-x+8=10x-14\)

**Answer:**

The correct answer is A. We will need to cancel out a regular, integer term and we will need to cancel out an x-term. We can do this at least two different ways:

\(-x+8=10x-14\)

\(-x+8+x=10x-14+x\)

\(8=11x-14\)

\(8+14=11x-14+14\)

\(22=11x\)

\(22\div11=11x\div11\)

OR

\(-x+8=10x-14\)

\(-x+8-10x=10x-14-10x\)

\(-11x+8=-14\)

\(-11x+8-8=-14-8\)

\(-11x=-22\)

\(-11x\div(-11)=-22\div(-11)\)

**Question #3:**

Solve for *x* by combining like terms and then cancelling out terms on both sides:

\(2(3+x-2)+5(x+5)=132\)

**Answer:**

The correct answer is D. No matter what, the first step that we want to do is combine like terms. We want to make things look as simple and “neat” as possible! Let’s walk through this solution:

\(2(3+x-2)+5(x+5)=132\)

We’ll first combine the 3 and -2 in the left-most set of parentheses:

\(2(x+1)+5(x+5)=132\)

Then, we’ll distribute the 2 and the 5 through their respective parentheses:

\(2x+2+5x+25=132\)

Now we have more like terms to combine!

\(2x+5x+2+25=132\)

\(7x+27=132\)

Finally, subtract 27 from both sides and divide by 7 on both sides.

\(7x+27=132\)

\(7x+27-27=132-27\)

\(7x=105\)

\(7x\div7=105\div7\)

**Question #4:**

Solve for *x*:

\(3(x+3)-2(x-1)=-10x\)

**Answer:**

The correct answer is A. First, we’ll distribute the 3 and the -2 through their respective parentheses. Then, we’ll combine like terms and cancel out certain terms. Let’s walk through that:

\(3(x+3)-2(x-1)=-10x\)

\(3x+9-2x+2=-10x\)

\(x+11=-10x\)

\(11=-11x\)

\(-1=x\)

**Question #5:**

Solve for *x*:

\(10(x+4)-3(x+1)=-3+8x+34\)

**Answer:**

The correct answer is B. Again, we’re first going to combine like terms; then we’ll distribute the 10 and the -3; then combine like terms again; and finally do some cancelling.

\(10(x+4)-3(x+1)=-3+8x+34\)

\(10(x+4)-3(x+1)=8x+31\)

\(10x+40-3x-3=8x+31\)

\(7x+37=8x+31\)

\(37=x+31\)

\(6=x\)