So far, you’ve been dealing only with expressions. Now you’re going to be dealing with equations. An important structural difference between equations and expressions is that an equation consists of two expressions separated by an equals sign, while an expression lacks an equals sign altogether. An equation is a sentence: “Something equals something else.” The somethings are each expressions.
Pretty much everything you will be doing with equations is related to one basic principle: You can do anything you want to one side of the equation, as long as you also do the same thing to the other side of the equation. Take the equation 3 + 5 = 8. You want to subtract 5 from the left side of the equation, but you still want the equation to be true. All you have to do is subtract 5 from the right side as well, and you can be confident that your new equation will still be valid:
Note that this would also work if you had variables in your equation:
Next, you’re going to see some of the many ways you can apply this principle to solving algebra problems.
What does it mean to solve an equation? What are you really doing when you manipulate algebraic equations?
A solution to an equation is a number that, when substituted in for the variable, makes the equation true. Remember, an equation is a sentence: “Something equals something else.” In general, a sentence like this can be true or false. You want a way to make it true.
Take the equation 2x + 7 = 15. You are looking for the value of x that will make this equation true. What if you plugged in 3 for x? If you replaced x with the number 3, you would get 2(3) + 7 = 15. This equation can be simplified to 6 + 7 = 15, which further simplifies to 13 = 15. However, 13 does not equal 15, so when x = 3, the equation is not true. Therefore, x = 3 is not a solution to the equation.
Now, if you replaced x with the number 4, you would get 2(4) + 7 = 15. This equation can be simplified to 8 + 7 = 15. Simplify it further, and you get 15 = 15, which is a true statement.
That means that when x = 4, the equation is true. So x = 4 is a solution to the equation.
Now the question becomes, what is the best way to find these solutions? If you had to use trial and error, or guessing, the process could take a very long time. The following sections will talk about how you can efficiently and accurately manipulate equations so that solutions become easier to find.
You know that you can make a change to an equation as long as you make the same change to both sides. This is called the Golden Rule. Now, look at the various changes you can make. Try to solve the following problem:
If 5(x − 1)3 − 30 = 10, then x = ?
To solve for a variable, you need to get it by itself on one side of the equals sign. To do that, you need to change the appearance of the equation, without changing the fact that it’s true. The good news is that all of the changes you need to make to this equation to solve for x will actually be very familiar to you—PEMDAS operations!
To get x by itself, you want to move every term that doesn’t include the variable to the other side of the equation. The easiest thing to move at this stage is the 30, so start there. If 30 is being subtracted on the left side of the equation, and you want to move it to the other side, then you need to do the opposite operation to cancel it out. So you’re going to add 30 to both sides, like this:
Now you’ve only got one term on the left side of the equation. The x is still inside the parentheses, and the expression in the parentheses is being multiplied by 5, so the next step will be to move that 5 over to the other side of the equation. Once again, you want to perform the opposite operation, so divide both sides of the equation by 5:
At this point, you could cube (x − 1), but that is going to involve a whole lot of multiplication. Instead, you can get rid of the exponent by performing the opposite operation. Roots are the opposite of exponents. So if the left side of the equation is raised to the third power, you can undo that by taking the third root of both sides, also known as the cube root, as shown here:
Now that nothing else is being done to the parentheses, you can just get rid of them. The equation is:
x − 1 = 2
Then, add 1 to both sides, and you get x = 3. This would have been hard to guess! If you plug 3 back in for x in the original equation (a step often worth doing, especially as you’re learning), you’ll find that this value makes the original equation true.
Take a look at the steps you took to isolate x. Do you notice anything? You added 30, then you divided by 5, then you got rid of the exponent, and then you simplified the parentheses. You did PEMDAS backward! And, in fact, when you’re isolating a variable, it turns out that the simplest way to do so is to reverse the order of PEMDAS when deciding what order you will perform your operations. Start with addition/subtraction, then multiplication/division, then exponents, and finish with terms in parentheses.
Now that you know the best way to isolate a variable, go through one more example. Try it on your own first, then an explanation will follow.
If
then x = ?
A/S
M/D
E
P
The equation you’re simplifying is
. If there’s anything to add or subtract, that will be the easiest first step. There is, so first get rid of the 7 by subtracting 7 from both sides:
Now you want to see if there’s anything being multiplied or divided by the term containing an x. The square root that contains the x is being multiplied by 4, so, your next step will be to get rid of the 4. You can do that by dividing both sides of the equation by 4:
Now that you’ve taken care of multiplication and division, it’s time to check for exponents. That means you need to check for exponents and roots, because they’re so intimately related. There are no exponents in the equation, but the x is inside a square root. To cancel out a root, you can use an exponent. Squaring a square root will cancel it out, so your next step is to square both sides:
The final step is to add 6 to both sides, and you end up with x = 15.
Solve for x in the following equations.
3(x + 4)3 − 5 = 19
You’ve covered the basic operations that you’ll be dealing with when solving equations. But what would you do if you were asked to solve for x in the following equation?
Now x appears in multiple parts of the equation, and your job has become more complicated. In addition to your PEMDAS operations, you also need to be able to simplify, or clean up, your equation. There are different ways you can clean up this equation. First, notice how you have an x in the denominator (the bottom of the fraction) on the left side of the equation. You’re trying to find the value of x, not of some number divided by x. So your first clean-up move is to always get variables out of denominators. The way to do that is to multiply both sides of the equation by the entire denominator. Watch what happens:
If you multiply a fraction by its denominator, you can cancel out the entire denominator. Now you’re left with:
5x − 3(4 − x) = 20x
No more fractions! What should you do next? At some point, if you want the value of x, you’re going to have to get all the terms that contain an x together. Right now, however, that x sitting inside the parentheses seems pretty tough to get to. To make that x more accessible, you should simplify grouped terms within the equation. That 3 on the outside of the parentheses wants to multiply the terms inside, so you need to distribute it. What that means is you’re going to multiply the 3 by each term inside, one at a time: 3 times 4 is 12, and 3 times −x is −3x. The equation becomes:
5x − (12 − 3x) = 20x
Now, if you subtract what’s in the parentheses from 5x, you can get rid of the parentheses altogether. Just as you multiplied the 3 by both terms inside the parentheses, you also have to subtract both terms:
Remember, subtracting a negative number is the same as adding a positive number; the negative signs cancel out.
You’re very close. You’re ready to make use of your final clean-up move—combine like terms. “Like terms” are terms that can be combined into one term. For example, “3x” and “5x” are like terms because they can be combined into “8x.” Ultimately, all the PEMDAS operations and clean-up moves have one goal: to get a variable by itself so you can determine its value. At this point, you have four terms in the equation: 5x, −12, 3x, and 20x. You want to get all the terms with an x on one side of the equation, and all the terms that only contain numbers on the other side.
First, combine 5x and 3x, because they’re on the same side of the equation. That gives you:
8x − 12 = 20x
Now you want to get the 8x together with the 20x, but which one should you move? The best move to make here is to move the 8x to the right side of the equation, because that way one side of the equation will have terms that contain only numbers (−12) and the right side will have terms that contain variables (8x and 20x). So now it’s time for the PEMDAS operations again. To find x:
Before moving on to the next topic, review what you’ve learned:
Solve for x in the following equations.
