Solving Linear Equations
If one process sets algebra apart from basic arithmetic, it is solving linear equations. Figuring out what x means can be frustrating if you don’t have the right tools to figure it out. In this guide, you will learn the processes that you need to solve every equation, every time.
Solving Equations with Addition and Subtraction
When you look at a given equation, you can see that it tells you a lot of information, but not exactly what you want to know. You want to know the solution, the value for the variable that makes the equation true. For example, the equation x + 5 = 12 is saying that x added to 5 is the same as 12. But you want to know what x equals, not what x + 5 equals.
Your goal in solving the equation is to get the x alone on the left side of the equals sign and a number (the solution) on the right side. The 5 that is added to the x is actually in your way and is keeping you from knowing what x equals. It would be good if you could make the left side of the equation change from x + 5 to x, and then x would be alone on the left side.
What would it take to change the 5 into a 0? Subtracting 5 from 5 gives 0. The subtraction property allows you to subtract 5 from the left side as long as you also subtract it from the right side of the equation. Changing the 5 into a 0 will be very helpful because instead of having x + 5 on the left side of the equation, you will have x + 0, which is the same as x.
Let’s go back to the beginning of the sample equation and solve it, showing the steps that lead to the solution:
Solve: x + 5 = 12
x + 5 − 5 = 12 − 5
x + 0 = 7
x = 7
You should always check your solution by substituting the solution into the original equation and verifying that it works. Yes, 7 + 5 = 12, and your solution x = 7 is correct. You probably knew that x = 7 is the solution because the original equation was an easy one to solve mentally. But the steps that include the subtraction property of equality and the identity property of addition show the algebraic process at work.
This process is the same for simple and more complex equations that are not so easily done with mental arithmetic. Learn perform the steps by working on easier problems first, and then the tougher equations will be easier to solve.
See how the subtraction property of equality is applied in the next example:
Solve: a + 3 = −7
a + 3 − 3 = −7 − 3
a = −10
To check the solution, put −10 in for a in the original equation: −10 + 3 = −7; therefore, a = −10 is correct.
Notice that in the solution steps of the previous equation, the step a + 0 = −10 isn’t shown, as x + 0 = 7 was in the previous example. You are free to omit this step in your work as an acceptable shortcut. If you are more comfortable showing the step, that’s fine, too.
The last thing you need to know about solving these types of equations is what to do when the variable is on the right side. Your goal remains to isolate the variable, no matter where it is in the equation. The next example has the variable on the right side:
Solve: 3 = x + 5
3 − 5 = x + 5 − 5
−2 = x or x = −2
To check, put −2 in the original equation for x: 3 = −2 + 5 is true. Therefore, x = −2 is the solution.
Solving Equations with Multiplication or Division
Begin by looking at a true equation from basic arithmetic:
3 + 2 = 1 + 4
It’s true because 5 = 5.
What will happen if you multiply both the left and right sides by 6? Is it true that 6(3 + 2) = 6(1 + 4)? Yes, 6 · 5 = 6 · 5. This is an example of the multiplication property of equality, and more examples from basic arithmetic and algebra follow:
In basic arithmetic, multiplying equal numbers by the same number gives equal products.
Examples: 1 + 5 = 6 is true, so (1 + 5) · 4 = 6 · 4; −3 + 5 = 2 is true, so 3 · (−3 + 5) = 3 · 2
In algebra, multiplying equal expressions by the same amount gives equal products.
Examples: If x = y, then x · z = y · z; and if b = h, then b · d = h · d.
As you might expect, the division property of equality is similar.
Begin again with a true equation from basic arithmetic:
2 + 4 = 6
Would the equation still be true if you divide both sides by 3? Let’s see:
(2 + 4) ÷ 3 = 6 ÷ 3
Yes, both sides of this equation equal 2, and the new equation is true. This is an example of the division property of equality, and more examples follow.
In basic arithmetic, dividing equal numbers by the same number gives equal quotients.
Examples: 1 + 5 = 6 is true, so (1 + 5) ÷ 2 = 6 ÷ 2; −3 + 13 = 10 is true, so (−3 + 13) ÷ 2 = 10 ÷ 2.
In algebra, dividing equal expressions by the same number gives equal quotients.
Examples: If x = y, then x ÷ z = y ÷ z; if b = h, then b ÷ d = h ÷ d.
Now that you know about the multiplication and division properties of equality, you are ready to use them to solve equations. The process is just like solving equations using addition and subtraction.
Consider the following equation. What will it take to find its solution?
Solve: 3x = 15
The equation is telling you that 3 times x is equal to 15. To solve it, you want to know what value of x makes the equation true. You want the x to be isolated on the left side of the equation. The 3 that is multiplied times the x is in the way and is keeping you from knowing what x equals. It would be good if you could make the left side of the equation change from 3 · x to 1 · x because 1 · x = x (identity property of multiplication), and then x would be alone on the left side.
What would it take to change the 3 into a 1? Dividing 3 by 3 gives 1. The division property allows you to divide by 3 on the left side as long as you also divide it on the right side of the equation.
Let’s go back to the beginning of the sample equation and solve it, showing the steps that lead to the solution:
Solve: 3x = 15
3x ÷ 3 = 15 ÷ 3
1x = 5
x = 5
To check, put 5 in for x in the original equation: 3 · 5 = 15 is true. Therefore, x = 5 is the solution. This equation can easily be solved mentally, but just as with the addition and subtraction equations, learning how to use the properties correctly on the easier equations will help you be able to solve the more difficult ones, where you will need to show some or all of the steps.
Now use the multiplication property to solve this sample equation:
Solve: x ÷ 8 = 2
x ÷ 8 · 8 = 2 · 8
x = 16
To check, put 16 in for x in the original equation: 16 ÷ 8 = 2 is true. Therefore, x = 16 is the solution.
The next example has the variable on the right side, but the process is the same:
Solve: 24 = −6x
24 ÷ −6 = −6x ÷ −6
−4 = x or x = −4
To check, put −4 in for x in the original equation: 24 = −6 · −4 is true. Therefore, x = −4 is the solution.
Once you get these processes under your belt you’ll be solving linear equations like a pro. Good luck, and happy calculating!
From Alpha Teach Yourself Algebra 1 in 24 Hours by Jane Warner Cook
