Lesson Objectives
  • Demonstrate an understanding of how to solve a linear equation in one variable
  • Learn how to solve a linear equation in one variable with fractions
  • Learn how to solve a linear equation in one variable with decimals
  • Learn how to label a linear equation as conditional, an identity, or a contradiction

How to Solve Linear Equations with Fractions or Decimals


In our last lesson, we reviewed the four-step method for solving a linear equation in one variable:
  • Simplify each side separately
    • Clear parentheses or grouping symbols and combine any like terms on each side
  • Isolate the variable terms on one side of the equation
    • Use the addition property of equality to move all variable terms to one side and all numbers (constants) to the other
  • Isolate the variable
    • Use the multiplication property of equality to get the variable by itself
  • Check
    • Replace the variable with the solution in the original equation; the left and right side should be the same value

Solving Linear Equations in One Variable with Fractions

When we work with a linear equation in one variable and fractions are involved, we can clear the fractions by multiplying both sides of the equation by the LCD of all denominators. Let's take a look at a few examples.
Example 1: Solve each equation $$\frac{19}{5}x + 1 = \frac{-111}{35} + \frac{12}{7}x$$ Recall the LCD is the LCM of the denominators. For this equation, we have denominators of 5, 7, and 35.
LCM(5, 7, 35) = 35
This means we can multiply both sides of the equation by 35 and clear the equation of all fractions: $$35 \cdot \left(\frac{19}{5}x + 1 \right) = 35 \cdot \left(\frac{-111}{35} + \frac{12}{7}x \right)$$ Left Side: $$35 \cdot \frac{19}{5}x + 35 \cdot 1$$ $$\require{cancel}7\cancel{35} \cdot \frac{19}{\cancel{5}}x + 35$$ $$133x + 35$$ Right Side: $$35 \cdot \frac{-111}{35} + 35 \cdot \frac{12}{7}x$$ $$\cancel{35} \cdot \frac{-111}{\cancel{35}} + 5\cancel{35} \cdot \frac{12}{\cancel{7}}x$$ $$-111 + 60x$$ Our transformed equation: $$133x + 35 = -111 + 60x$$ Now we can solve our equation. Let's begin by moving all variable terms to the left side and all constants to the right side. We will subtract (60x) from each side of the equation: $$133x - 60x + 35 = -111 + 60x - 60x$$ $$73x + 35 = -111 + \cancel{60x - 60x}$$ $$73x + 35 = -111$$ Now we can subtract 35 from each side of the equation: $$73x + 35 - 35 = -111 - 35$$ $$73x + \cancel{35 - 35} = -146$$ $$73x = -146$$ We can isolate our variable by dividing both sides of the equation by 73, the coefficient of our variable x: $$\frac{73}{73}x = \frac{-146}{73}$$ $$\frac{\cancel{73}}{\cancel{73}}x = \frac{-2\cancel{146}}{\cancel{73}}$$ $$x = -2$$ We can check our solution by plugging a -2 in for x in the original equation: $$\frac{19}{5}x + 1 = \frac{-111}{35} + \frac{12}{7}x$$ $$\frac{19 \cdot (-2)}{5} + 1 = \frac{-111}{35} + \frac{12 \cdot (-2)}{7}$$ $$\frac{-38}{5} + \frac{5}{5} = \frac{-111}{35} + \frac{-120}{35}$$ $$\frac{-33}{5} = \frac{-231}{35}$$ $$\require{color}\frac{-33}{5} = \frac{-33}{5} \hspace{.5em} \color{green}{✔}$$ Example 2: Solve each equation $$\frac{25}{6} \left(\frac{1}{2}x + \frac{17}{8} \right) = \frac{781}{144} + x$$ When parentheses are involved, clear the parentheses first and then clear the fractions: $$\frac{25}{6} \left(\frac{1}{2}x + \frac{17}{8} \right) = \frac{781}{144} + x$$ $$\frac{25}{6} \cdot \frac{1}{2}x + \frac{25}{6} \cdot \frac{17}{8} = \frac{781}{144} + x$$ $$\frac{25}{12}x + \frac{425}{48} = \frac{781}{144} + x$$ For this equation, we have denominators of 12, 48, and 144.
LCM(12, 48, 144) = 144
This means we can multiply both sides of the equation by 144 and clear the equation of all fractions: $$144 \cdot \left(\frac{25}{12}x + \frac{425}{48} \right) = 144 \cdot \left(\frac{781}{144} + x \right)$$ Left side: $$144 \cdot \left(\frac{25}{12}x + \frac{425}{48} \right)$$ $$144 \cdot \frac{25}{12}x + 144 \cdot \frac{425}{48}$$ $$12\cancel{144} \cdot \frac{25}{\cancel{12}}x + 3\cancel{144} \cdot \frac{425}{\cancel{48}}$$ $$300x + 1275$$ Right Side: $$144 \cdot \left(\frac{781}{144} + x \right)$$ $$144 \cdot \frac{781}{144} + 144x$$ $$\cancel{144} \cdot \frac{781}{\cancel{144}} + 144x$$ $$144x + 781$$ Our Transformed equation: $$300x + 1275 = 144x + 781$$ Now we can solve our equation. Let's begin by moving all variable terms to the left side and all constants to the right side. We will subtract 144x from each side of the equation: $$300x - 144x + 1275 = 144x - 144x + 781$$ $$156x + 1275 = \cancel{144x - 144x} + 781$$ $$156x + 1275 = 781$$ Now we can subtract 1275 from each side of the equation: $$156x + 1275 - 1275 = 781 - 1275$$ $$156x + \cancel{1275 - 1275} = 781 - 1275$$ $$156x = -494$$ We can isolate our variable by dividing both sides of the equation by 156, the coefficient of our variable x: $$\frac{156}{156}x = \frac{-494}{156}$$ $$\frac{\cancel{156}}{\cancel{156}}x = \frac{-19 \cancel{494}}{6\cancel{156}}$$ $$x = \frac{-19}{6}$$ We can check our solution by plugging a -19/6 in for x in the original equation: $$\frac{25}{6} \left(\frac{1}{2}x + \frac{17}{8} \right) = \frac{781}{144} + x$$ $$\frac{25}{6} \left(\frac{1}{2} \cdot \frac{-19}{6} + \frac{17}{8} \right) = \frac{781}{144} + \frac{-19}{6}$$ $$\frac{25}{6} \left(\frac{-19}{12} + \frac{17}{8} \right) = \frac{781}{144} + \frac{-456}{144}$$ $$\frac{25}{6} \left(\frac{-38}{24} + \frac{51}{24} \right) = \frac{325}{144}$$ $$\frac{25}{6} \left(\frac{13}{24} \right) = \frac{325}{144}$$ $$\frac{325}{144} = \frac{325}{144} \hspace{.5em} \color{green}{✔}$$

Solving Linear Equations in One Variable with Decimals

Recall that multiplying by 10 or a positive integer power of 10 (10, 100, 1000,...) means we move our decimal point one place to the right for each zero in the power of ten. When we work with a linear equation in one variable and decimals are involved, we can clear the decimals by multiplying both sides of the equation by the power of ten needed to clear the decimal number with the largest number of decimal places. Let's take a look at a few examples.
Example 3: Solve each equation
5.1x + 7.26 = 7.51 + 5x
If we look at each decimal number, the highest number of decimal places is two. This means we can clear our decimals by multiplying each side of the equation by one hundred:
100(5.1x + 7.26) = 100(7.51 + 5x)
510x + 726 = 751 + 500x
Now we can solve our equation using the normal four-step procedure. Let's begin by moving all variable terms to the left side and all constants to the right side:
510x - 500x = 751 - 726
10x = 25
We can divide each side of the equation by 10, this will isolate our variable x: $$\frac{10}{10}x = \frac{25}{10}$$ $$\frac{\cancel{10}}{\cancel{10}}x = \frac{5\cancel{25}}{2\cancel{10}}$$ $$x = \frac{5}{2} = 2.5$$ We can report our answer as a fraction (5/2) or as a decimal (2.5).
Check:
5.1x + 7.26 = 7.51 + 5x
5.1(2.5) + 7.26 = 7.51 + 5(2.5)
12.75 + 7.26 = 7.51 + 12.5
20.01 = 20.01
Example 4: Solve each equation
-0.69 - 0.55(0.1 - 0.48x) = 0.464x - 3.416
Before we clear our decimals, let's clear the parentheses:
-0.69 - 0.055 + 0.264x = 0.464x - 3.416
Now we can simplify the left side:
-0.745 + 0.264x = 0.464x - 3.416
If we look at each decimal number, the highest number of decimal places is three. This means we can clear our decimals by multiplying each side of the equation by one thousand:
1000(-0.745 + 0.264x) = 1000(0.464x - 3.416)
-745 + 264x = 464x - 3416
Now we can solve our equation using the normal four-step procedure. Let's begin by moving all variable terms to the left side and all constants to the right side:
264x - 464x = -3416 + 745
-200x = -2671
We can divide each side of the equation by (-200), this will isolate our variable x: $$\frac{-200}{-200}x = \frac{-2671}{-200}$$ $$\frac{\cancel{-200}}{\cancel{-200}}x = \frac{\cancel{-1} \cdot 2671}{\cancel{-1} \cdot 200}$$ $$x = \frac{2671}{200} = 13.355$$ Check:
-0.69 - 0.55(0.1 - 0.48x) = 0.464x - 3.416
-0.69 - 0.55(0.1 - 0.48(13.355)) = 0.464(13.355) - 3.416
-0.69 - 0.55(0.1 - 6.4104) = 6.19672 - 3.416
-0.69 - 0.55(-6.3104) = 2.78072
-0.69 + 3.47072 = 2.78072
2.78072 = 2.78072

How to Identify the Type of Equation as: Conditional, Identity, or Contradiction


As we learned in Algebra 1, we can categorize our equations as: conditional, identity, or contradiction.

Conditional Equations

Most equations we work with are conditional equations. A conditional equation is true only under certain conditions. Let's take a look at an example.
Example 5: Determine if the equation is conditional, an identity, or a contradiction
-8x - 38 = -5(1 - 5x)
Let's begin by removing parentheses on the right side:
-8x - 38 = -5 + 25x
Now we can move all the variable terms to the left side and all the constants to the right side:
-8x - 25x = -5 + 38
-33x = 33
Divide each side of the equation by -33, this will isolate x:
(-33/-33)x = (33/-33)
x = -1
Since we have only one solution, we can say this equation is a conditional equation. It is true when (-1) replaces x, but false for any other number.

Contradiction Equation

From time to time, we will run across an equation with no solution. This equation type is known as a contradiction. Let's take a look at an example.
Example 6: Determine if the equation is conditional, an identity, or a contradiction
-4x - 13 = -(8 + 4x)
Let's begin by removing parentheses on the right side:
-4x - 13 = -8 - 4x
Now we can move all the variable terms to the left side and all the constants to the right side:
-4x + 4x = -8 + 13
0 = 5 (false)
When we end up with a false statement, we can stop and say our equation has "no solution". Since we studied sets earlier in our course, we can also use the empty set symbol to show our solution set is empty, meaning it contains no elements:

Identity Equation

Lastly, we will see equations that have an infinite number of solutions. Let's take a look at an example.
Example 7: Determine if the equation is conditional, an identity, or a contradiction
2x + 5(x - 8) = -40 + 7x
Let's begin by removing parentheses on the left side:
2x + 5x - 40 = -40 + 7x
Now we can simplify the left side:
7x - 40 = -40 + 7x
Before we go any further, we should notice that the two sides have exactly the same terms (7x and -40). This means whatever value is plugged in for x will always yield a true statement.
If we continue with our normal procedure, we would next move all the variable terms to the left and all the constants to the right:
7x - 7x = -40 + 40
0 = 0 (true)
When we end up with a true statement and no variable, we can stop and say our equation has "an infinite number of solutions".