Download Section 5E – Steps to Solving General Linear Equations

Section 5E – Steps to Solving General Linear Equations
A linear equation is an equation where the variable is to the first power. If a variable has an
exponent of 2 (square) or 3 (cube) or higher, then it will require more advanced methods to
solve the problem. In this chapter we are focusing on solving linear equations, but first we
need to talk about the different types of equations.
3 Types of Equations
There are 3 types of equations: conditional, contradiction and identity.
Conditional equations are only true sometimes. Look at the equation x  4  11 . This type of
equation is a conditional equation because it is only true if x = 7 and is not true if x is any other
number. When an equation has a finite number of solutions, it is conditional. Most equations
are conditional. Our job is to find the number we can plug in to make the equation true.
Contradiction equations are never true. Look at the equation w+3 = w+5 . This equation is
never true. No matter what number we replace w with, we always get a false statement.
When an equation has no solution it is called a contradiction equation. When asked to solve a
contradiction, something weird will happen. When we subtract w from both sides we are left
with 3 = 5 . This is not true. So you know the equation has “no solution”. Look at another
example 3b  7  3b  2 . A technique in solving equations is to bring the variable terms to one
side. But if we subtract 3b from both sides we get +7 = -2. That is never true!! This tells us that
the equation is never true no matter what. Hence this is a contradiction equation and the
answer is “No Solution”.
The third type of equation is an identity equation. This is one that is always true. Look at y+4 =
y+4. We can plug in any number we want for y and it will be true. For example we could plug
in 70 and see that 70+4=70+4 (true). We could plug in -549 and see that -549+4 = -549+4.
Every time we plug in any number we get a true statement. The solution to an identity
equation is “all real numbers”, since it has infinitely many solutions. When solving an identity
equation like y+4 = y+4 , we will subtract y from both sides, but then all the variables are gone
and we are left with the true statement 4=4. When that happens you know the answer is “all
real numbers”. Look at another example 5a  1  5a  1 . Did you notice the two sides are
exactly the same? If not, we can try to bring the variable terms to one side. But if we subtract
5a from both sides we get +1 = +1. That is always true!! This tells us that the equation is true
no matter what we plug in for a. Hence this is an identity equation and the answer is “All Real
Numbers”.
General Equation Solving
We have seen that we can solve equations by guessing the answer. If we cannot guess we can
use the multiplication and addition properties to help us figure out the answer. Let’s now look
at some more complicated equations and the steps to solving them.
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Steps to Solving a Linear Equation (It is Vital to Memorize These!!)
1. Eliminate parenthesis by using the distributive property.
2. Eliminate fractions by multiplying both sides of the equation by the LCD.
3. Eliminate decimals by multiplying both sides of the equation by a power
of 10 (10, 100, 1000…)
4. Use the addition property to eliminate variable terms so that there are only
variables on one side of the equation.
5. Use the addition property to eliminate constants so that there are only constants
on one side of the equation. The constants should be on the opposite side of
the variables.
6. Use the multiplication property to multiply or divide both sides of the equation in
order to isolate the variable by creating a coefficient of 1 for the variable.
7. Check your answer by plugging it into the original equation and see if the two
sides are equal.
Note: After each step, always add or subtract like terms that lie on the same side of the
equation.
Note: Remember that an equation can have “no solution” or “all real numbers” as a solution in
the cases of contradiction and identity equations.
Let’s look at another example equation 5 x  7  4 x  3 . When dealing with an equation like
this, our goal is to bring letters to one side and the constant numbers to the other side. If we
want to eliminate the 4x on the right side, we can subtract 4x from both sides.
5x  7  4 x  3
4 x
 4x
x 7  03
x  7  3
Notice that there are only x variables on the right side. Can you guess the answer now? If not
we can get rid of the 7 by subtracting 7 (adding -7) to both sides.
x  7  3

7

7
x  0  10
x  10
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Notice the number we can plug in for x that makes the equation true is -10. Check if that is the
correct answer. Plugging in -10 into the original equation we get the following:
5  10   7  4  10   3
50  7  40  3
43  43
So when we plug in -10, we do get a true statement. Hence -10 is the solution. Note that the
two sides were equal and both equal to -43, but -43 is not the solution. The solution is the
number we replaced the letter with that made the two sides equal. Also notice this was a
conditional equation. It was only true when x = -10 and false otherwise.
Let’s look at an example 6  2 w  8   4 w  3w  (7 w  9)
Step 1: Our first step is to distribute and eliminate parenthesis, so we will multiply the -6 times
both the 2w and the -8. We will also distribute the negative to the 7w and the 9 and eliminate
that parenthesis as well. We should only distribute to the terms in the parenthesis. For
example we do not distribute the -6 to the 4w since the 4w is not in the parenthesis.
6  2 w  8   4 w  3w  (7 w  9)
12w  48  4 w  3w  7 w  9
Always look to simplify after each step. For example in this problem the -12w and 4w are like
terms on the same side. Also the 3w and -7w are also like terms on the same side. If you
struggle with negatives, you can convert the -7w to adding the opposite. Be careful. Do not
add or subtract terms on opposite sides of the equation.
6  2w  8   4w  3w  (7 w  9)
12w  48  4w  3w  7 w  9
8w  48  4 w  9
Step 2 and 3: There are no fractions or decimals so we proceed directly to step 4.
Step 4: We need to bring the variables to one side. You can bring variables to either side, but
many students like to bring the variables to the left side only. So we will need to eliminate the
-4w on the right side. Hence we will add the opposite +4w to both sides.
8w  48  4 w  9
4w
 4w
4w  48  0  9
4w  48  9
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Step 5: We need to bring the constants to the opposite side as the variables. So we need to get
rid of the +48. Hence we will subtract 48 (add -48) to both sides. We are then left with
-4w= -57
4 w  48  9
 48  48
4 w  0  57
 4 w  57
Step 6: We now need to get the w by itself. Since the w is being multiplied by -4, we will divide
both sides by -4 to get our answer of 57/4. Notice the answer can be written three ways and all
1
 57

are equally correct  or14 or14.25 
4
 4

4w  57
1
4
57
w
4
4
1w  14.25
w  14.25
Step 7: Let’s check our answer by plugging into the original equation. Notice that all of the w’s
have to be replaced with 14.25 and don’t forget to use the order of operations when simplifying
each side. As you can see, checking your answer can be just as much work as solving the
equation.
 6  2 w  8   4 w  3w  (7 w  9)
6  2 14.25  8   4 14.25  3 14.25  (7 14.25  9)
 6  28.5  8   4 14.25  3 14.25  (99.75  9)
 6  20.5   4 14.25  3 14.25  (108.75)
 123  57  42.75  (108.75)
 66  66
Let’s try another example. Look at
1
3 1
c  c4
3
5 2
Step 1: There are no parenthesis so we proceed to step 2.
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Step 2: To eliminate fractions we find the LCD. Since the denominators are 3,5 and 2 the LCD is
30. Hence we will multiply everything on both sides by 30. This will eliminate the fractions.
Remember all the terms must be multiplied by 30. When multiplying a whole number (30) by
fractions it is good to write the whole number as a fraction (30/1).
30  1
3  30  1

 c     c  4
1 3
5 1  2

30 1
30 3 30 1
30
 c    c 4
1 3
1 5 1 2
1
30
90 30
120
c
 c
3
5
2
1
10c  18  15c  120
Notice we are now left with an equation without fractions.
Step 3: There are no decimals, so we proceed to step 4.
Step 4: We bring all the variables to the left side by subtracting 15c (adding -15c) to both sides.
10c  18  15c  120
15c
 15c
 5c  18  0  120
 5c  18  120
Step 5: We bring all the constants to the opposite side. We can eliminate the -18 by adding
+18 to both sides.
5c  18  120
 18  18
 5c  0  138
 5c  138
Step 6: Isolate the variable. Since we are multiplying the w by -5, we divide both sides by -5.
5c  138
1
5
138
c
5
5
138
1c  
5
3
c  27 or  27.6
5
Step 7: Check your answer. By plugging 27.6 into the original equation, the two sides are
equal.
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Let’s try a third example: 0.24  3m  1  0.82m  0.24  0.1m
Step 1: We will distribute the 0.24 to the 3m and -1 to eliminate the parenthesis. We will make
sure to combine any like terms that are on the same side. Notice 0.82m and -0.1m are like
terms on the same side, so we can add them.
0.24  3m  1  0.82 x  0.24  0.1m
0.24  3m  0.24  1  0.82m  0.24  0.1m
0.72m  0.24  0.72m  0.24
Step 2: There are no fractions, so we proceed to step 3.
Step 3: Since the most decimal places to the right is two (hundredths place), we will multiply
everything on both sides by 100. If one of the decimals had ended in the thousandths place, we
would multiply by 1000 and so on. Notice all the decimals are gone.
0.72m  0.24  0.72m  0.24
100  0.72m  0.24   100  0.72m  0.24 
100  0.72m  100   0.24  100  0.72m  100   0.24
72m  24  72m  24
Step 4: Bring the variables to one side by subtracting 72m from both sides. Notice all variables
cancel and we are left with -24 = -24 which is a true statement.
72m  24  72m  24
72m
 72m
0  24  0  24
 24  24
Since we are left with a true statement and all the variables are gone, we need go no further.
This is an always true equation (identity). So the answer is “All Real Numbers”.
Try to solve the following equations with your instructor.
Example 1: 6d  8  5d  3
Example 2: 3 y  4  3 y
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Example 4:
Example 6:
2d  7  5d  3d  7
1
1 1
5
L  L
4
2 3
6
Example 5: 4(3d  7)  19(d  2)  5d  2
Example 7: 0.25( y  0.4)  0.2  0.15 y  0.5
(Remember to eliminate parenthesis before
eliminating decimals)
Practice Problems Section 5E
Solve the following equations. If your answer is a fraction, be sure to simplify it completely.
For #39, #40 and #47, remember to eliminate parenthesis before eliminating fractions or
decimals.

18  4 p  2
1. 7 y  15  20
2.
3. 12  1a  17
4. 9   7n  8
5. 12 x  7  9 x  11
6. 5 y  14  3 y  18
7. 1a  27  6a  13
8. 2b  13  9b  7
(This section is from Preparing for Algebra and Statistics , Third Edition
by M. Teachout, College of the Canyons, Santa Clarita, CA, USA)
This content is licensed under a Creative Commons
Attribution 4.0 International license
129
9. 6  13v   5  11v
10.

23  5m  19  18m
11. 13  y  1y  17  4
12. 14 g  3   3g  8
13. 6a  5a  7
14. 8b  13  8b  2
15. 17v  16v  9
16.

8b   9b  4

5w   6w  11
17.

8 x   8 x  13  13
18.
19.

9a  13  10a  4
20. 1.3x  2.7  0.3x  3.4
21. 16w  9  15w  w  9
23. 6a 
1
3
 5a 
4
4
25. a  0.004   0.5a  0.053  0.5a 26.
22.

5.2 x  7.3   6.2 x  3.6
24. 0.8b  2.57   0.2b  2.9
8
1 3
1
x  x
5
3 5
3
27. 26 x  19  21x  5 x  17
28.
29. 31m  17  28m  3m  10  27
30. 25  5n  24  3n  18  8n
31. 2  4 x  3  6 x
32. 7  y  3  2 y  6
33. 4(5a  1)  18a  2(3a  2)
34. 11b  (7b  9)  4(b  3)

5.2d  7.3   6.2d  3.6
35.
1
1
m 1   m  2
2
3
36.
2
1 2
n  n
3
5 5
37.
3
1
c  2  1c 
4
2
38.
1
3 1
5
d  d
6
2 2
6
39.
1
1 11

 2 p     p  1
5
3 23

40.
2
1
 w  6    w  8
3
4
41. 0.5 x  0.3  0.8 x  0.9
43. 0.23x  0.4  0.38 x  0.2
45. 0.2m  0.41  0.18m  0.67
42. 0.09 p  0.04  0.09 p  0.06
44. 0.007 y  0.03  0.009 y  0.06
46. 0.6𝑛 − 2 = 𝑛 + 0.3
47. 0.4(2a  1)  0.9a  0.5  0.1a  0.9
130