problem solving equations with fractions lesson 8 7

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

4.9: Solving Equations with Fractions

• Last updated
• Save as PDF
• Page ID 24084

• David Arnold
• College of the Redwoods

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

Undoing Subtraction

We can still add the same amount to both sides of an equation without changing the solution.

Solve for x : \(x - \frac{5}{6} = \frac{1}{3}\).

To “undo” subtracting 5/6, add 5/6 to both sides of the equation and simplify.

\[ \begin{aligned} x - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ x - \frac{5}{6} + \frac{5}{6} = \frac{1}{3} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Add } \frac{5}{6} \text{ to both sides.}} \\ x = \frac{1 \cdot 2}{3 \cdot 2} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Equivalent fractions, LCD = 6.}} \\ x = \frac{2}{6} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Simplify.}} \\ x = \frac{7}{6} ~ & \textcolor{red}{ \text{ Add.}} \end{aligned}\nonumber \]

It is perfectly acceptable to leave your answer as an improper fraction. If you desire, or if you are instructed to do so, you can change your answer to a mixed fraction (7 divided by 6 is 1 with a remainder of 1). That is \(x = 1 \frac{1}{6}\).

Checking the Solution

Substitute 7/6 for x in the original equation and simplify.

\[ \begin{aligned} x - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{7}{6} - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Substitute 7/6 for } x.} \\ \frac{2}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Subtract.}} \\ \frac{1}{3} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Reduce.}} \end{aligned}\nonumber \]

Because the last statement is true, we conclude that 7/6 is a solution of the equation x − 5/6 = 1/3.

Undoing Addition

You can still subtract the same amount from both sides of an equation without changing the solution.

Solve for x : \(x + \frac{2}{3} = - \frac{3}{5}\).

To “undo” adding 2/3, subtract 2/3 from both sides of the equation and simplify.

\[ \begin{aligned} x + \frac{2}{3} = - \frac{3}{5} ~ & \textcolor{red}{ \text{ Original equation.}} \\ x + \frac{2}{3} - \frac{2}{3} = - \frac{3}{5} - \frac{2}{3} ~ & \textcolor{red}{ \text{ Subtract } \frac{2}{3} \text{ from both sides.}} \\ x = - \frac{3 \cdot 3}{5 \cdot 3} - \frac{2 \cdot 5}{3 \cdot 5} ~ & \textcolor{red}{ \text{ Equivalent fractions, LCD = 15.}} \\ x = - \frac{9}{15} - \frac{10}{15} ~ & \textcolor{red}{ \text{ Simplify.}} \\ x = - \frac{19}{15} ~ & \textcolor{red}{ \text{ Subtract.}} \end{aligned}\nonumber \]

Readers are encouraged to check this solution in the original equation.

Solve for x : \(x + \frac{3}{4} = - \frac{1}{2}\)

Undoing Multiplication

We “undo” multiplication by dividing. For example, to solve the equation 2 x = 6, we would divide both sides of the equation by 2. In similar fashion, we could divide both sides of the equation

\[ \frac{3}{5} x = \frac{4}{10}\nonumber \]

by 3/5. However, it is more efficient to take advantage of reciprocals. For convenience, we remind readers of the Multiplicative Inverse Property .

Multiplicative Inverse Property

Let a / b be any fraction. The number b / a is called the multiplicative inverse or reciprocal of a / b . The product of reciprocals is 1.

\[ \frac{a}{b} \cdot \frac{b}{a} = 1.\nonumber \]

Let’s put our knowledge of reciprocals to work.

Solve for x : \(\frac{3}{5}x = \frac{4}{10}\).

To “undo” multiplying by 3/5, multiply both sides by the reciprocal 5/3 and simplify.

\[ \begin{aligned} \frac{3}{5} x = \frac{4}{10} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{5}{3} \left( \frac{3}{5} x \right) = \frac{5}{3} \left( \frac{4}{10} \right) & ~ \textcolor{red}{ \text{ Multiply both sides by 5/3.}} \\ \left( \frac{5}{3} \cdot \frac{3}{5} \right) x = \frac{20}{30} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, use the associative property to regroup.} \\ \text{ On the right, multiply.} \end{array}} \\ 1x = \frac{2}{3} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, } \frac{5}{3} \cdot \frac{3}{5} = 1. \\ \text{ On the right, reduce: } \frac{20}{30} = \frac{2}{3}. \end{array}} \\ x = \frac{2}{3} ~ & \textcolor{red}{ \text{ On the left, } 1x = x.} \end{aligned}\nonumber \]

Substitute 2/3 for x in the original equation and simplify.

\[ \begin{aligned} \frac{3}{5} x = \frac{4}{10} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{3}{5} \left( \frac{2}{3} \right) = \frac{4}{10} ~ & \textcolor{red}{ \text{ Substitute 2/3 for }x.} \\ \frac{6}{15} = \frac{4}{10} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ \frac{2}{5} = \frac{2}{5} ~ & \textcolor{red}{ \text{ Reduce both sides to lowest terms.}} \end{aligned}\nonumber \]

Because this last statement is true, we conclude that 2/3 is a solution of the equation (3/5) x = 4/10.

Solve for y : \( \frac{2}{3} y = \frac{4}{5}\)

Solve for x : \(- \frac{8}{9} x = \frac{5}{18}\).

To “undo” multiplying by −8/9, multiply both sides by the reciprocal −9/8 and simplify.

\[ \begin{aligned} - \frac{8}{9} x = \frac{5}{18} ~ & \textcolor{red}{ \text{ Original equation.}} \\ - \frac{9}{8} \left( - \frac{8}{9} x \right) = - \frac{9}{8} \left( \frac{5}{18} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by } -9/8.} \\ \left[ - \frac{9}{8} \cdot \left( - \frac{8}{9} \right) \right] x = - \frac{3 \cdot 3}{2 \cdot 2 \cdot 2} \cdot \frac{5}{2 \cdot 3 \cdot 3} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, use the associative property to regroup.} \\ \text{ On the right, prime factor.} \end{array}} 1x = \frac{ \cancel{3} \cdot \cancel{3}}{2 \cdot 2 \cdot 2} \cdot \frac{5}{2 \cdot \cancel{3} \cdot \cancel{3}} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, } - \frac{9}{8} \cdot \left( - \frac{8}{9} \right) = 1. \\ \text{ On the right, cancel common factors.} \end{array}} \\ x = - \frac{5}{16} ~ & \textcolor{red}{ \text{ On the left, } 1x = x. \text{ Multiply on the right.}} \end{aligned}\nonumber \]

Solve for z: \(− \frac{2}{7} z = \frac{4}{21}\)

Clearing Fractions from the Equation

Although the technique demonstrated in the previous examples is a solid mathematical technique, working with fractions in an equation is not always the most efficient use of your time.

To clear all fractions from an equation, multiply both sides of the equation by the least common denominator of the fractions that appear in the equation.

Let’s put this idea to work.

In Example 1, we were asked to solve the following equation for x :

\[x - \frac{5}{6} = \frac{1}{3}.\nonumber \]

Take a moment to review the solution technique in Example 1. We will now solve this equation by first clearing all fractions from the equation.

Multiply both sides of the equation by the least common denominator for the fractions appearing in the equation.

\[ \begin{aligned} x - \frac{5}{6}= \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6 \left( x - \frac{5}{6} \right) = 6 \left( \frac{1}{3} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 6.}} \\ 6x - 6 \left( \frac{5}{6} \right) = 6 \left( \frac{1}{3} \right) ~ & \textcolor{red}{ \text{ Distribute the 6.}} \\ 6x-5 = 2 ~ & \textcolor{red}{ \text{ On each side, multiply first.}} \\ ~ & \textcolor{red}{6 \left( \frac{5}{6} \right) = 5 \text{ and } 6 \left( \frac{1}{3} \right) = 2.} \end{aligned}\nonumber \]

Note that the equation is now entirely clear of fractions, making it a much simpler equation to solve.

\[ \begin{aligned} 6x - 5 + 5 = 2 + 5 ~ & \textcolor{red}{ \text{ Add 5 to both sides.}} \\ 6x = 7 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{6x}{6} = \frac{7}{6} ~ & \textcolor{red}{ \text{ Divide both sides by 6.}} \\ x = \frac{7}{6} ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]

Note that this is the same solution found in Example 1.

Solve for t : \(t - \frac{2}{7} = - \frac{1}{4}\)

In Example 4, we were asked to solve the following equation for x .

\[- \frac{8}{9}x = \frac{5}{18}\nonumber \]

Take a moment to review the solution in Example 4. We will now solve this equation by first clearing all fractions from the equation.

Multiply both sides of the equation by the least common denominator for the fractions that appear in the equation.

\[ \begin{aligned} - \frac{8}{9} x = \frac{5}{18} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 18 \left( - \frac{8}{9} x \right) = 18 \left( \frac{5}{18} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 18.}} \\ -16x=5 ~ & \textcolor{red}{ \text{ On each side, cancel and multiply.}} \\ ~ & \textcolor{red}{ 18 \left( - \frac{8}{9} \right) = -16 \text{ and } 18 \left( \frac{5}{18} \right) = 5.} \end{aligned}\nonumber \]

Note that the equation is now entirely free of fractions. Continuing,

\[ \begin{aligned} \frac{-16x}{-16} = \frac{5}{-16} ~ & \textcolor{red}{ \text{ Divide both sides by } -16.} \\ x = - \frac{5}{16} ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]

Note that this is the same as the solution found in Example 4.

Solve for u :

\[ - \frac{7}{9} u = \frac{14}{27}\nonumber \]

Solve for x : \(\frac{2}{3}x + \frac{3}{4} = \frac{1}{2}\).

\[ \begin{aligned} \frac{2}{3} x + \frac{3}{4} = \frac{1}{2} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 \left( \frac{2}{3} x + \frac{3}{4} = \right) = 12 \left( \frac{1}{2} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 12.}} \\ 12 \left( \frac{2}{3}x \right) + 12 \left( \frac{3}{4} \right) = 12 \left( \frac{1}{2} \right) ~ & \textcolor{red}{ \text{ On the left, distribute 12.}} \\ 8x + 9 = 6 ~ & \textcolor{red}{ \text{ Multiply: } 12 \left( \frac{2}{3} x \right) = 8x, ~ 12 \left( \frac{3}{4} \right) = 9,} \\ ~ & \textcolor{red}{ \text{ and } 12 \left( \frac{1}{2} \right) = 6.} \end{aligned}\nonumber \]

Note that the equation is now entirely free of fractions. We need to isolate the terms containing x on one side of the equation.

\[ \begin{aligned} 8x + 9 - 9 = 6 - 9 ~ & \textcolor{red}{ \text{ Subtract 9 from both sides.}} \\ 8x = - 3 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{8x}{8} = \frac{-3}{8} ~ & \textcolor{red}{ \text{ Divide both sides by 8.}} \\ x = - \frac{3}{8} ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

Solve for r : \(\frac{3}{4} r + \frac{2}{3} = \frac{1}{2}\)

Solve for x : \( \frac{2}{3} - \frac{3x}{4} = \frac{x}{2} - \frac{1}{8}.\)

Multiply both sides of the equation by the least common denominator for the fractions in the equation.

\[ \begin{aligned} \frac{2}{3} - \frac{3x}{4} = \frac{x}{2} - \frac{1}{8} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 24 \left( \frac{2}{3} - \frac{3x}{4} \right) = 24 \left( \frac{x}{2} - \frac{1}{8} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 24.}} \\ 24 \left( \frac{2}{3} \right) - 24 \left( \frac{3x}{4} \right) = 24 \left( \frac{x}{2} \right) - 24 \left( \frac{1}{8} \right) ~ & \textcolor{red}{ \text{ On both sides, distribute 24.}} \\ 16 - 18x = 12x - 3 ~ & \textcolor{red}{ \text{ Left: } 24 \left( \frac{2}{3} \right) = 16, ~ 24 \left( \frac{3x}{4} \right) = 18x.} \\ ~ & \textcolor{red}{ \text{ Right: } 24 \left( \frac{x}{2} \right) = 12x, ~ 24 \left( \frac{1}{8} \right) = 3.} \end{aligned}\nonumber \]

\[ \begin{aligned} 16 - 18x - 12x = 12x - 3 - 12x ~ & \textcolor{red}{ \text{ Subtract } 12x \text{ from both sides.}} \\ 16 - 30x = -3 ~ & \textcolor{red}{ \begin{aligned} \text{ Left: } -18x - 12x = -30x. \\ \text{ Right: } 12x - 12x = 0. \end{aligned}} \\ 16 - 30x - 16 = -3 - 16 ~ & \textcolor{red}{ \text{ Subtract 16 from both sides.}} \\ -30x = -19 ~ & \textcolor{red}{ \begin{aligned} \text{ Left: } 16-16=0. \\ \text{ Right: } -3 - 16 = -19. \end{aligned}} \\ \frac{-30x}{-30} = \frac{-19}{-30} ~ & \textcolor{red}{ \text{ Divide both sides by } -30.} \\ x = \frac{19}{30} ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

Solve for s : \( \frac{3}{2} - \frac{2s}{5} = \frac{s}{3} - \frac{1}{5}\).

Add texts here. Do not delete this text first.

Applications

Let’s look at some applications that involve equations containing fractions. For convenience, we repeat the Requirements for Word Problem Solutions .

Requirements for Word Problem Solutions

• Statements such as “Let P represent the perimeter of the rectangle.”
• Labeling unknown values with variables in a table.
• Labeling unknown quantities in a sketch or diagram.
• Set up an Equation . Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement.
• Solve the Equation . You must always solve the equation set up in the previous step.
• Answer the Question . This step is easily overlooked. For example, the problem might ask for Jane’s age, but your equation’s solution gives the age of Jane’s sister Liz. Make sure you answer the original question asked in the problem. Your solution should be written in a sentence with appropriate units. 5. Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it’s possible that your equation incorrectly models the problem’s situation, so you could have a valid solution to an incorrect equation. The important question is: “Does your answer make sense based on the words in the original problem statement.”

In the third quarter of a basketball game, announcers informed the crowd that attendance for the game was 12,250. If this is two-thirds of the capacity, find the full seating capacity for the basketball arena.

We follow the Requirements for Word Problem Solutions .

1. Set up a Variable Dictionary . Let F represent the full seating capacity. Note: It is much better to use a variable that “sounds like” the quantity that it represents. In this case, letting F represent the full seating capacity is much more descriptive than using x to represent the full seating capacity.

2. Set up an Equation . Two-thirds of the full seating capacity is 12,250.

\[ \begin{aligned} \colorbox{cyan}{Two-thirds} & \text{ of } & \colorbox{cyan}{Full Seating Capacity} & \text{ is } & 12,250 \\ \frac{2}{3} & \cdot & F & = & 12,250 \end{aligned}\nonumber \]

Hence, the equation is

\[ \frac{2}{3} F = 12250.\nonumber \]

3. Solve the Equation . Multiply both sides by 3 to clear fractions, then solve.

\[ \begin{aligned} \frac{2}{3} F = 12250 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 3 \left( \frac{2}{3} F \right) = 3(12250) ~ & \textcolor{red}{ \text{ Multiply both sides by 3.}} \\ 2F = 36750 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{2F}{2} = \frac{36750}{2} ~ & \textcolor{red}{ \text{ Divide both sides by 2.}} \\ F = 18375 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

4. Answer the Question . The full seating capacity is 18,375.

5. Look Back . The words of the problem state that 2/3 of the seating capacity is 12,250. Let’s take two-thirds of our answer and see what we get.

\[ \begin{aligned} \frac{2}{3} \cdot 18375 & = \frac{2}{3} \cdot \frac{18375}{1} \\ & = \frac{2}{3} \cdot \frac{3 \cdot 6125}{1} \\ & = \frac{2}{ \cancel{3}} \cdot \frac{ \cancel{3} \cdot 6125}{1} \\ & = 12250 \end{aligned}\nonumber \]

This is the correct attendance, so our solution is correct.

Attendance for the Celtics game was 9,510. If this is 3/4 of capacity, what is the capacity of the Celtics’ arena?

The area of a triangle is 20 square inches. If the length of the base is \(2 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

1. Set up a Variable Dictionary . Our variable dictionary will take the form of a well labeled diagram.

2. Set up an Equation . The area A of a triangle with base b and height h is

\[A = \frac{1}{2} bh.\nonumber \]

Substitute A = 20 and b = \(2 \frac{1}{2}\).

\[20 = \frac{1}{2} \left( 2 \frac{1}{2} \right) h.\nonumber \]

3. Solve the Equation . Change the mixed fraction to an improper fraction, then simplify.

\[ \begin{aligned} 20 = \frac{1}{2} \left( 2 \frac{1}{2} \right) h ~ & \textcolor{red}{ \text{ Original equation.}} \\ 20 = \frac{1}{2} \left( \frac{5}{2} \right) h ~ & \textcolor{red}{ \text{ Mixed to improper: } 2 \frac{1}{2} = \frac{5}{2}.} \\ 20 = \left( \frac{1}{2} \cdot \frac{5}{2} \right) h ~ & \textcolor{red}{ \text{ Associative property.}} \\ 20 = \frac{5}{4} h ~ & \textcolor{red}{ \text{ Multiply: } \frac{1}{2} \cdot \frac{5}{2} = \frac{5}{4}.} \end{aligned}\nonumber \]

Now, multiply both sides by 4/5 and solve.

\[ \begin{aligned} \frac{4}{5} (20) = \frac{4}{5} \left( \frac{5}{4} h \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 4/5.}} \\ 16 = h ~ & \textcolor{red}{ \text{ Simplify: } \frac{4}{5} (20) = 16} \\ ~ & \textcolor{red}{ \text{ and } \frac{4}{5} \cdot \frac{5}{4} = 1.} \end{aligned}\nonumber \]

4. Answer the Question . The height of the triangle is 16 inches.

5. Look Back . If the height is 16 inches and the base is \(2 \frac{1}{2}\) inches, then the area is

\[ \begin{aligned} A & = \frac{1}{2} \left( 2 \frac{1}{2} \right) (16) \\ & = \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{16}{1} \\ & = \frac{5 \cdot 16}{2 \cdot 2} \\ & = \frac{(5) \cdot (2 \cdot 2 \cdot 2 \cdot 2)}{(2) \cdot (2)} \\ & = \frac{5 \cdot \cancel{2} \cdot \cancel{2} \cdot 2 \cot 2}{ \cancel{2} \cdot \cancel{2}} & = 20 \end{aligned}\nonumber \]

This is the correct area (20 square inches), so our solution is correct.

The area of a triangle is 161 square feet. If the base of the triangle measures \(40 \frac{1}{4}\) feet, find the height of the triangle.

1. Is 1/4 a solution of the equation \(x + \frac{5}{8} = \frac{5}{8}\)?

2. Is 1/4 a solution of the equation \(x + \frac{1}{3} = \frac{5}{12}\)?

3. Is −8/15 a solution of the equation \(\frac{1}{4} x = − \frac{1}{15}\)?

4. Is −18/7 a solution of the equation \(− \frac{3}{8} x = \frac{25}{28}\)?

5. Is 1/2 a solution of the equation \(x + \frac{4}{9} = \frac{17}{18}\)?

6. Is 1/3 a solution of the equation \(x + \frac{3}{4} = \frac{13}{12}\)?

7. Is 3/8 a solution of the equation \(x − \frac{5}{9} = − \frac{13}{72}\)?

8. Is 1/2 a solution of the equation \(x − \frac{3}{5} = − \frac{1}{10}\)?

9. Is 2/7 a solution of the equation \(x − \frac{4}{9} = − \frac{8}{63}\)?

10. Is 1/9 a solution of the equation \(x − \frac{4}{7} = − \frac{31}{63}\)?

11. Is 8/5 a solution of the equation \( \frac{11}{14}x = \frac{44}{35}\)?

12. Is 16/9 a solution of the equation \(\frac{13}{18} x = \frac{104}{81}\)?

In Exercises 13-24, solve the equation and simplify your answer.

13. \(2x − 3=6x + 7\)

14. \(9x − 8 = −9x − 3\)

15. \(−7x +4=3x\)

16. \(6x +9= −6x\)

17. \(−2x = 9x − 4\)

18. \(−6x = −9x + 8\)

19. \(−8x = 7x − 7\)

20. \(−6x = 5x + 4\)

21. \(−7x +8=2x\)

22. \(−x − 7=3x\)

23. \(−9x +4=4x − 6\)

24. \(−2x +4= x − 7\)

In Exercises 25-48, solve the equation and simplify your answer.

25. \(x + \frac{3}{2 = \frac{1}{2}\)

26. \(x − \frac{3}{4} = \frac{1}{4}\)

27. \(− \frac{9}{5} x = \frac{1}{2}\)

28. \(\frac{7}{3} x = − \frac{7}{2}\)

29. \(\frac{3}{8} x = \frac{8}{7}\)

30. \(− \frac{1}{9} x = − \frac{3}{5}\)

31. \(\frac{2}{5} x = − \frac{1}{6}\)

32. \(\frac{1}{6} x = \frac{2}{9}\)

33. \(− \frac{3}{2} x = \frac{8}{7}\)

34. \(− \frac{3}{2} x = − \frac{7}{5}\)

35. \(x + \frac{3}{4} = \frac{5}{9}\)

36. \(x − \frac{1}{9} = − \frac{3}{2}\)

37. \(x − \frac{4}{7} = \frac{7}{8}\)

38. \(x + \frac{4}{9} = − \frac{3}{4}\)

39. \(x + \frac{8}{9} = \frac{2}{3}\)

40. \(x − \frac{5}{6} = \frac{1}{4}\)

41. \(x + \frac{5}{2} = − \frac{9}{8}\)

42. \(x + \frac{1}{2} = \frac{5}{3}\)

43. \(− \frac{8}{5} x = \frac{7}{9}\)

44. \(− \frac{3}{2} x = − \frac{5}{9}\)

45. \(x − \frac{1}{4} = − \frac{1}{8}\)

46. \(x − \frac{9}{2} = − \frac{7}{2}\)

47. \(− \frac{1}{4} x = \frac{1}{2}\)

48. \(− \frac{8}{9} x = − \frac{8}{3}\)

In Exercises 49-72, solve the equation and simplify your answer.

49. \(− \frac{7}{3} x − \frac{2}{3} = \frac{3}{4} x + \frac{2}{3}\)

50. \(\frac{1}{2} x − \frac{1}{2} = \frac{3}{2} x + \frac{3}{4}\)

51. \(− \frac{7}{2} x − \frac{5}{4} = \frac{4}{5}\)

52. \(− \frac{7}{6} x + \frac{5}{6} = − \frac{8}{9}\)

53. \(− \frac{9}{7} x + \frac{9}{2} = − \frac{5}{2}\)

54. \(\frac{5}{9} x − \frac{7}{2} = \frac{1}{4}\)

55. \(\frac{1}{4} x − \frac{4}{3} = − \frac{2}{3}\)

56. \(\frac{8}{7} x + \frac{3}{7} = \frac{5}{3}\)

57. \(\frac{5}{3} x + \frac{3}{2} = − \frac{1}{4}\)

58. \(\frac{1}{2} x − \frac{8}{3} = − \frac{2}{5}\)

59. \(− \frac{1}{3} x + \frac{4}{5} = − \frac{9}{5} x − \frac{5}{6}\)

60. \(− \frac{2}{9} x − \frac{3}{5} = \frac{4}{5} x − \frac{3}{2}\)

61. \(− \frac{4}{9} x − \frac{8}{9} = \frac{1}{2} x − \frac{1}{2}\)

62. \(− \frac{5}{4} x − \frac{5}{3} = \frac{8}{7} x + \frac{7}{3}\)

63. \(\frac{1}{2} x − \frac{1}{8} = − \frac{1}{8} x + \frac{5}{7}\)

64. \(− \frac{3}{2} x + \frac{8}{3} = \frac{7}{9} x − \frac{1}{2}\)

65. \(− \frac{3}{7} x − \frac{1}{3} = − \frac{1}{9}\)

66. \(\frac{2}{3} x + \frac{2}{9} = − \frac{9}{5}\)

67. \(− \frac{3}{4} x + \frac{2}{7} = \frac{8}{7} x − \frac{1}{3}\)

68. \(\frac{1}{2} x + \frac{1}{3} = − \frac{5}{2} x − \frac{1}{4}\)

69. \(− \frac{3}{4} x − \frac{2}{3} = − \frac{2}{3} x − \frac{1}{2}\)

70. \(\frac{1}{3} x − \frac{5}{7} = \frac{3}{2} x + \frac{4}{3}\)

71. \(− \frac{5}{2} x + \frac{9}{5} = \frac{5}{8}\)

72. \(\frac{9}{4} x + \frac{4}{3} = − \frac{1}{6}\)

73. At a local soccer game, announcers informed the crowd that attendance for the game was 4,302. If this is 2/9 of the capacity, find the full seating capacity for the soccer stadium.

74. At a local basketball game, announcers informed the crowd that attendance for the game was 5,394. If this is 2/7 of the capacity, find the full seating capacity for the basketball stadium.

75. The area of a triangle is 51 square inches. If the length of the base is \(8 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

76. The area of a triangle is 20 square inches. If the length of the base is \(2 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

77. The area of a triangle is 18 square inches. If the length of the base is \(4 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

78. The area of a triangle is 44 square inches. If the length of the base is \(5 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

79. At a local hockey game, announcers informed the crowd that attendance for the game was 4,536. If this is 2/11 of the capacity, find the full seating capacity for the hockey stadium.

80. At a local soccer game, announcers informed the crowd that attendance for the game was 6,970. If this is 2/7 of the capacity, find the full seating capacity for the soccer stadium.

81. Pirates . About one-third of the world’s pirate attacks in 2008 occurred off the Somali coast. If there were 111 pirate attacks off the Somali coast, estimate the number of pirate attacks worldwide in 2008.

82. Nuclear arsenal . The U.S. and Russia agreed to cut nuclear arsenals of long-range nuclear weapons by about a third, down to 1, 550. How many long-range nuclear weapons are there now? Associated Press-Times-Standard 04/04/10 Nuclear heartland anxious about missile cuts.

83. Seed vault . The Svalbard Global Seed Vault has amassed half a million seed samples, and now houses at least one-third of the world’s crop seeds. Estimate the total number of world’s crop seeds. Associated Press-Times-Standard 03/15/10 Norway doomsday seed vault hits half-million mark.

84. Freight train . The three and one-half mile long Union Pacific train is about 2 1 2 times the length of a typical freight train. How long is a typical freight train? Associated Press-Times-Standard 01/13/10 Unusally long train raises safety concerns.

13. \(− \frac{5}{2}\)

15. \(\frac{2}{5}\)

17. \(\frac{4}{11}\)

19. \(\frac{7}{15}\)

21. \(\frac{8}{9}\)

23. \(\frac{10}{13}\)

25. \(−1\)

27. \(− \frac{5}{18}\)

29. \(\frac{64}{21}\)

31. \(− \frac{5}{12}\)

33. \(− \frac{16}{21}\)

35. \(− \frac{7}{36}\)

37. \(\frac{81}{56}\)

39. \(− \frac{2}{9}\)

41. \(− \frac{29}{8}\)

43. \(− \frac{35}{72}\)

45. \(\frac{1}{8}\)

47. \(−2\)

49. \(− \frac{16}{37}\)

51. \(− \frac{41}{70}\)

53. \(\frac{49}{9}\)

55. \(\frac{8}{3}\)

57. \(− \frac{21}{20}\)

59. \(− \frac{49}{44}\)

61. \(− \frac{7}{17}\)

63. \(\frac{47}{35}\)

65. \(− \frac{14}{27}\)

67. \(\frac{52}{159}\)

69. \(− 2\)

71. \(\frac{47}{100}\)

81. There were about 333 pirate attacks worldwide.

83. 1,500,000

8.4 Solve Equations with Fraction or Decimal Coefficients

Learning objectives.

Solve equations with fraction coefficients

• Solve equations with decimal coefficients

Be Prepared 8.4

Before you get started, take this readiness quiz.

• Multiply: 8 · 3 8 . 8 · 3 8 . If you missed this problem, review Example 4.28
• Find the LCD of 5 6 and 1 4 . 5 6 and 1 4 . If you missed this problem, review Example 4.63
• Multiply: 4.78 4.78 by 100 . 100 . If you missed this problem, review Example 5.18

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions . Let’s solve the same equation again, but this time use the method that clears the fractions.

Example 8.37

Solve: 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

Try It 8.73

Solve: 1 4 x + 1 2 = 5 8 . 1 4 x + 1 2 = 5 8 .

Try It 8.74

Solve: 1 6 y − 1 3 = 1 6 . 1 6 y − 1 3 = 1 6 .

Notice in Example 8.37 that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

Solve equations with fraction coefficients by clearing the fractions.

• Step 1. Find the least common denominator of all the fractions in the equation.
• Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.
• Step 3. Solve using the General Strategy for Solving Linear Equations.

Example 8.38

Solve: 7 = 1 2 x + 3 4 x − 2 3 x . 7 = 1 2 x + 3 4 x − 2 3 x .

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Try It 8.75

Solve: 6 = 1 2 v + 2 5 v − 3 4 v . 6 = 1 2 v + 2 5 v − 3 4 v .

Try It 8.76

Solve: −1 = 1 2 u + 1 4 u − 2 3 u . −1 = 1 2 u + 1 4 u − 2 3 u .

In the next example, we’ll have variables and fractions on both sides of the equation.

Example 8.39

Solve: x + 1 3 = 1 6 x − 1 2 . x + 1 3 = 1 6 x − 1 2 .

Try It 8.77

Solve: a + 3 4 = 3 8 a − 1 2 . a + 3 4 = 3 8 a − 1 2 .

Try It 8.78

Solve: c + 3 4 = 1 2 c − 1 4 . c + 3 4 = 1 2 c − 1 4 .

In Example 8.40 , we’ll start by using the Distributive Property. This step will clear the fractions right away!

Example 8.40

Solve: 1 = 1 2 ( 4 x + 2 ) . 1 = 1 2 ( 4 x + 2 ) .

Try It 8.79

Solve: −11 = 1 2 ( 6 p + 2 ) . −11 = 1 2 ( 6 p + 2 ) .

Try It 8.80

Solve: 8 = 1 3 ( 9 q + 6 ) . 8 = 1 3 ( 9 q + 6 ) .

Many times, there will still be fractions, even after distributing.

Example 8.41

Solve: 1 2 ( y − 5 ) = 1 4 ( y − 1 ) . 1 2 ( y − 5 ) = 1 4 ( y − 1 ) .

Try It 8.81

Solve: 1 5 ( n + 3 ) = 1 4 ( n + 2 ) . 1 5 ( n + 3 ) = 1 4 ( n + 2 ) .

Try It 8.82

Solve: 1 2 ( m − 3 ) = 1 4 ( m − 7 ) . 1 2 ( m − 3 ) = 1 4 ( m − 7 ) .

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, 0.3 = 3 10 0.3 = 3 10 and 0.17 = 17 100 . 0.17 = 17 100 . So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator .

Example 8.42

Solve: 0.8 x − 5 = 7 . 0.8 x − 5 = 7 .

The only decimal in the equation is 0.8 . 0.8 . Since 0.8 = 8 10 , 0.8 = 8 10 , the LCD is 10 . 10 . We can multiply both sides by 10 10 to clear the decimal.

Try It 8.83

Solve: 0.6 x − 1 = 11 . 0.6 x − 1 = 11 .

Try It 8.84

Solve: 1.2 x − 3 = 9 . 1.2 x − 3 = 9 .

Example 8.43

Solve: 0.06 x + 0.02 = 0.25 x − 1.5 . 0.06 x + 0.02 = 0.25 x − 1.5 .

Look at the decimals and think of the equivalent fractions.

0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10 0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10

Notice, the LCD is 100 . 100 .

By multiplying by the LCD we will clear the decimals.

Try It 8.85

Solve: 0.14 h + 0.12 = 0.35 h − 2.4 . 0.14 h + 0.12 = 0.35 h − 2.4 .

Try It 8.86

Solve: 0.65 k − 0.1 = 0.4 k − 0.35 . 0.65 k − 0.1 = 0.4 k − 0.35 .

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

Example 8.44

Solve: 0.25 x + 0.05 ( x + 3 ) = 2.85 . 0.25 x + 0.05 ( x + 3 ) = 2.85 .

Try It 8.87

Solve: 0.25 n + 0.05 ( n + 5 ) = 2.95 . 0.25 n + 0.05 ( n + 5 ) = 2.95 .

Try It 8.88

Solve: 0.10 d + 0.05 ( d − 5 ) = 2.15 . 0.10 d + 0.05 ( d − 5 ) = 2.15 .

ACCESS ADDITIONAL ONLINE RESOURCES

• Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 2: Solve an Equation with Fractions with Variable Terms on Both Sides
• Solving Multiple Step Equations Involving Decimals
• Ex: Solve a Linear Equation With Decimals and Variables on Both Sides
• Ex: Solve an Equation with Decimals and Parentheses

Section 8.4 Exercises

Practice makes perfect.

In the following exercises, solve the equation by clearing the fractions.

1 4 x − 1 2 = − 3 4 1 4 x − 1 2 = − 3 4

3 4 x − 1 2 = 1 4 3 4 x − 1 2 = 1 4

5 6 y − 2 3 = − 3 2 5 6 y − 2 3 = − 3 2

5 6 y − 1 3 = − 7 6 5 6 y − 1 3 = − 7 6

1 2 a + 3 8 = 3 4 1 2 a + 3 8 = 3 4

5 8 b + 1 2 = − 3 4 5 8 b + 1 2 = − 3 4

2 = 1 3 x − 1 2 x + 2 3 x 2 = 1 3 x − 1 2 x + 2 3 x

2 = 3 5 x − 1 3 x + 2 5 x 2 = 3 5 x − 1 3 x + 2 5 x

1 4 m − 4 5 m + 1 2 m = −1 1 4 m − 4 5 m + 1 2 m = −1

5 6 n − 1 4 n − 1 2 n = −2 5 6 n − 1 4 n − 1 2 n = −2

x + 1 2 = 2 3 x − 1 2 x + 1 2 = 2 3 x − 1 2

x + 3 4 = 1 2 x − 5 4 x + 3 4 = 1 2 x − 5 4

1 3 w + 5 4 = w − 1 4 1 3 w + 5 4 = w − 1 4

3 2 z + 1 3 = z − 2 3 3 2 z + 1 3 = z − 2 3

1 2 x − 1 4 = 1 12 x + 1 6 1 2 x − 1 4 = 1 12 x + 1 6

1 2 a − 1 4 = 1 6 a + 1 12 1 2 a − 1 4 = 1 6 a + 1 12

1 3 b + 1 5 = 2 5 b − 3 5 1 3 b + 1 5 = 2 5 b − 3 5

1 3 x + 2 5 = 1 5 x − 2 5 1 3 x + 2 5 = 1 5 x − 2 5

1 = 1 6 ( 12 x − 6 ) 1 = 1 6 ( 12 x − 6 )

1 = 1 5 ( 15 x − 10 ) 1 = 1 5 ( 15 x − 10 )

1 4 ( p − 7 ) = 1 3 ( p + 5 ) 1 4 ( p − 7 ) = 1 3 ( p + 5 )

1 5 ( q + 3 ) = 1 2 ( q − 3 ) 1 5 ( q + 3 ) = 1 2 ( q − 3 )

1 2 ( x + 4 ) = 3 4 1 2 ( x + 4 ) = 3 4

1 3 ( x + 5 ) = 5 6 1 3 ( x + 5 ) = 5 6

In the following exercises, solve the equation by clearing the decimals.

0.6 y + 3 = 9 0.6 y + 3 = 9

0.4 y − 4 = 2 0.4 y − 4 = 2

3.6 j − 2 = 5.2 3.6 j − 2 = 5.2

2.1 k + 3 = 7.2 2.1 k + 3 = 7.2

0.4 x + 0.6 = 0.5 x − 1.2 0.4 x + 0.6 = 0.5 x − 1.2

0.7 x + 0.4 = 0.6 x + 2.4 0.7 x + 0.4 = 0.6 x + 2.4

0.23 x + 1.47 = 0.37 x − 1.05 0.23 x + 1.47 = 0.37 x − 1.05

0.48 x + 1.56 = 0.58 x − 0.64 0.48 x + 1.56 = 0.58 x − 0.64

0.9 x − 1.25 = 0.75 x + 1.75 0.9 x − 1.25 = 0.75 x + 1.75

1.2 x − 0.91 = 0.8 x + 2.29 1.2 x − 0.91 = 0.8 x + 2.29

0.05 n + 0.10 ( n + 8 ) = 2.15 0.05 n + 0.10 ( n + 8 ) = 2.15

0.05 n + 0.10 ( n + 7 ) = 3.55 0.05 n + 0.10 ( n + 7 ) = 3.55

0.10 d + 0.25 ( d + 5 ) = 4.05 0.10 d + 0.25 ( d + 5 ) = 4.05

0.10 d + 0.25 ( d + 7 ) = 5.25 0.10 d + 0.25 ( d + 7 ) = 5.25

0.05 ( q − 5 ) + 0.25 q = 3.05 0.05 ( q − 5 ) + 0.25 q = 3.05

0.05 ( q − 8 ) + 0.25 q = 4.10 0.05 ( q − 8 ) + 0.25 q = 4.10

Everyday Math

Coins Taylor has $2.00 $2.00 in dimes and pennies. The number of pennies is 2 2 more than the number of dimes. Solve the equation 0.10 d + 0.01 ( d + 2 ) = 2 0.10 d + 0.01 ( d + 2 ) = 2 for d , d , the number of dimes.

Stamps Travis bought $9.45 $9.45 worth of 49-cent 49-cent stamps and 21-cent 21-cent stamps. The number of 21-cent 21-cent stamps was 5 5 less than the number of 49-cent 49-cent stamps. Solve the equation 0.49 s + 0.21 ( s − 5 ) = 9.45 0.49 s + 0.21 ( s − 5 ) = 9.45 for s , s , to find the number of 49-cent 49-cent stamps Travis bought.

Writing Exercises

Explain how to find the least common denominator of 3 8 , 1 6 , and 2 3 . 3 8 , 1 6 , and 2 3 .

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

In the equation 0.35 x + 2.1 = 3.85 , 0.35 x + 2.1 = 3.85 , what is the LCD? How do you know?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/prealgebra/pages/1-introduction

• Authors: Lynn Marecek, MaryAnne Anthony-Smith
• Publisher/website: OpenStax
• Book title: Prealgebra
• Publication date: Sep 25, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/prealgebra/pages/1-introduction
• Section URL: https://openstax.org/books/prealgebra/pages/8-4-solve-equations-with-fraction-or-decimal-coefficients

© Feb 9, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Equations involving Fractions Practice Questions

Click here for questions, click here for answers, gcse revision cards.

5-a-day Workbooks

Number Line

• Expanded Form
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions
• reduce\:fraction\:\frac{4}{8}
• \frac{1}{2}+\frac{1}{4}+\frac{3}{4}
• \frac{1}{2}\cdot\frac{8}{7}
• \frac{-\frac{1}{5}}{\frac{7}{4}}
• descending\:order\:\frac{1}{2},\:\frac{3}{6},\:\frac{7}{2}
• decimal\:to\:fraction\:0.35
• What is a mixed number?
• A mixed number is a combination of a whole number and a fraction.
• How can I compare two fractions?
• To compare two fractions, first find a common denominator, then compare the numerators.Alternatively, compare the fractions by converting them to decimals.
• How do you add or subtract fractions with different denominators?
• To add or subtract fractions with different denominators, convert the fractions to have a common denominator. Then you can add or subtract the numerators of the fractions, leaving the denominator unchanged.

fractions-calculator

• My Notebook, the Symbolab way Math notebooks have been around for hundreds of years. You write down problems, solutions and notes to go back...

Please add a message.

Message received. Thanks for the feedback.

Following toggle tip provides clarification

Grades 7 & 8 Mathematics

• Class Homepage

This course covers the topics typically taught in Canadian Grade 7 and 8 Mathematics curricula and, in some instances, extends ideas beyond grade level. Letters are included beside the unit names to help group the units into similar themes.

For more information about the structure and general use of this courseware, see the Course Information unit.

Representing and Comparing Numbers (N)

Operations (n), ratios, rates, and proportions (n), bisectors and properties of shapes (g), area, volume, and angles (g), transformations of shapes (g), representing patterns (a), equations and the pythagorean theorem (a), data collection and graphs (d), data analysis (d), probability (d), course information.

Part A (Lessons 1–7) Topics include representing and comparing positive rational numbers (integers, fractions, and decimals), finding multiples and factors of positive integers, and determining the least common multiple (LCM) and the greatest common factor (GCF) of a pair of positive integers. Part B (Lessons 8–12) Topics include representing negative fractions and negative decimals, comparing the values of any two rational numbers, exponential notation, and using factor trees and prime factorizations to find the LCM or the GCF of a pair of positive integers.

This lesson examines three different number systems: whole numbers, integers, and rational numbers. Connections between different number systems are highlighted to set the groundwork for comparisons and operations.

Mathematicians often use the number line to solve problems. In this lesson, we review the number line, focusing on plotting fractions.

In math, symbols are important for communication. In this lesson, we review the “greater than” and “less than” symbols. In addition, we present two techniques used to compare fractions.

Rational numbers can be written as fractions or decimals. In this lesson, we discuss the connections between fractional representations and decimal representations, specifically, when it comes to plotting numbers on the number line.

In this lesson, we review how to generate a list of multiples of an integer. Using our lists, we identify common multiples of two integers, paying particular attention to the least common multiple (LCM).

Factors, like multiples, have to do with multiplication. In this lesson, we solve problems by identifying factors of positive integers.

Expanding on the factors lesson, we compare the factors of two positive integers to find common factors; specifically, we are often interested in identifying the greatest common factor (GCF). We conclude by solving word problems that require us to apply factors to different contexts.

Fractional quantities can be positive or negative. Similar to negative integers, negative fractions lie to the left of zero on the number line. In this lesson, we plot negative fractions on the number line to help us understand and compare the values of these numbers.

Rational numbers can be written as fractions or decimals. In this lesson, we compare negative decimal numbers by plotting them on the number line. We then compare negative fractions with negative decimals. The decimal equivalents of common fractions are determined and strategies for converting a fraction into a decimal are shown. Finally, we learn how to compare any two rational numbers.

In this lesson, we learn to represent repeated multiplication using exponential notation. Exponential notation is then used to represent whole numbers in expanded form using powers of ten. Square numbers and cube numbers are investigated.

In this lesson, we review prime and composite numbers. We learn how to represent a composite number as a product of its prime factors using a factor tree.

Prime factorizations can be used to determine the greatest common factor (GCF) and the least common multiple (LCM) of a pair of positive integers. We explore how this can be done, and we use these strategies to solve word problems.

Part A (Lessons 1–11) Topics include adding and subtracting rational numbers, multiplying and dividing a whole number by a positive rational number, and evaluating expressions using the order of operations. Part B (Lessons 12–19) Topics include multiplying, and dividing integers, fractions and decimal numbers, approximating square roots of positive integers, and evaluating expressions that include exponents using the order of operations.

Part A (Lessons 1–6) Topics include writing and interpreting ratios; finding equivalent ratios; converting between fractions, decimals, and percents; increasing and decreasing by a percentage; converting between units of measurement; and solving problems involving unit rates. Part B (Lessons 7–11) Topics include recognizing proportional situations in word problems, tables and graphs; connecting unit relates to proportional relationships and their representations in tables, graphs and equations; and fractional percents and percents greater than 100 percent.

Part A (Lessons 1–6) Topics include constructions of angle bisectors and perpendicular bisectors, and the various properties of triangles, quadrilaterals, and more general polygons. In particular, different polygons are classified based on their side lengths and angle measurements. Part B (Lessons 7–10) Topics include quadrilateral diagonals, circle terminology and construction, and applications of circles in the real-world.

Part A (Lessons 1–5) Topics include calculating the area of parallelograms, triangles, trapezoids, and composite shapes; calculating the surface area, volume, and capacity of prisms; and representing 3D objects in different ways. Part B (Lessons 6–10) Topics include calculating the circumference and area of circles; calculating the volume and surface area of cylinders; and properties of angles formed by intersecting lines including parallel lines and transversals.

Part A (Lessons 1–7) Topics include congruence of polygons, triangle congruence rules, plotting points on the Cartesian plane, the image of a polygon on the Cartesian plane under translations, reflections and/or rotations on the Cartesian plane, and tessellations. Part B (Lessons 8–11) Topics include similarity of polygons, triangle similarity rules, dilatations of polygons, and indirect measurements.

Part A (Lessons 1–6) Topics include representing sequences using tables, general terms and graphs, describing patterns using variables and expressions, extending sequences, and solving problems involving unknown quantities. Part B (Lessons 7–11) Topics include equivalent expressions for the general term of a sequence, describing relationships and patterns using equations, and decreasing and naturally occurring sequences.

Part A (Lessons 1–7) Topics include using variables in expressions, equations, and inequalities, identifying and exploring linear relationships, solving equations and inequalities by inspection, trial and error, and using visual models, and simplifying expressions by collecting like terms. Part B (Lessons 8-15) Topics include solving equations and inequalities using algebraic techniques, comparing the differences between evaluating an expression and solving an equation, exploring equations with multiple variables, and the Pythagorean Theorem.

Part A (Lessons 1–5) Topics include different types of data; population, sample and census; bias in data collection arising from question wording, accepted answers and choice of sample group; frequency and relative frequency tables and graphs; reading and creating circle graphs; choosing an appropriate graph type for a data set; bias in data representation arising from the chosen graph type, graph structure and shape, and axis labels and scales. Part B (Lessons 6–9) Topics include organizing continuous data into stem-and-leaf plots and frequency tables with intervals; as well as creating and reading histograms, and scatter plots.

Part A (Lessons 1–4) Topics include determining the mean, median, and mode of data sets; studying the effects of adding data to a data set or removing data from a data set; exploring the effect of outliers on the mean, median, and mode; and practising drawing conclusions and making predictions from data in graphs. Part B (Lessons 5–8) Topics include interpreting data, histograms, and scatter plots and drawing conclusions from these graphs; describing relationships between the two variables in a scatter plot; estimating rates of change associated with scatter plots; making predictions supported by the data in histograms and scatter plots; and using appropriate measures of central tendency to compare two data sets.

Part A (Lessons 1–4) Topics include random experiments, outcomes, and events; calculating theoretical probabilities of single events; comparing probabilities of different events; independent events; experimental probability; and using probabilities to make predictions. Part B (Lessons 5–8) Topics include comparing theoretical probabilities and experimental probabilities; exploring how the number of trials impacts probability estimates; complementary events; setting up and running simulations using probability models; and revisiting independent events.

© 2018 University of Waterloo. Except where noted, all rights reserved.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Course: Algebra 1   >   Unit 2

• Equations with parentheses

Equations with parentheses: decimals & fractions

• Reasoning with linear equations
• Multi-step equations review

• Your answer should be
• a proper fraction, like 1 / 2 ‍   or 6 / 10 ‍  
• an improper fraction, like 10 / 7 ‍   or 14 / 8 ‍  

Get step-by-step solutions to your math problems

Try Math Solver

Get step-by-step explanations

Graph your math problems

Practice, practice, practice

Get math help in your language

• International
• Education Jobs
• Schools directory
• Resources Education Jobs Schools directory News Search

Solving Equations - Full Lesson Sequence

Subject: Mathematics

Age range: 7-11

Resource type: Lesson (complete)

Last updated

17 March 2022

• Share through email
• Share through twitter
• Share through linkedin
• Share through facebook
• Share through pinterest

Full sequence of lessons back-to-back for ‘Solving Equations’

Including -One step equations (addition, subtraction, multiplication, division and mixed problems) -Two step equations -Unknowns on both sides -Equations with brackets

Creative Commons "Sharealike"

Your rating is required to reflect your happiness.

It's good to leave some feedback.

Something went wrong, please try again later.

kirstyturner123

Great resource, really well broken down showing the 3 methods avaiable to solve equations. Thanks.

Empty reply does not make any sense for the end user

Report this resource to let us know if it violates our terms and conditions. Our customer service team will review your report and will be in touch.

Not quite what you were looking for? Search by keyword to find the right resource: