problem solving with algebraic fractions

Fractions in Algebra

We can add, subtract, multiply and divide fractions in algebra in the same way we do in simple arithmetic.

Adding Fractions

To add fractions there is a simple rule:

(See why this works on the Common Denominator page).

x 2 + y 5 = (x)(5) + (2)(y) (2)(5)

x + 4 3 + x − 3 4 = (x+4)(4) + (3)(x−3) (3)(4)

= 4x+16 + 3x−9 12

Subtracting Fractions

Subtracting fractions is very similar, except that the + is now −

x + 2 x  −  x x − 2  =  (x+2)(x−2) − (x)(x) x(x−2)  

=  (x 2 − 2 2 ) − x 2 x 2 − 2x

=  −4 x 2 − 2x

Multiplying Fractions

Multiplying fractions is the easiest one of all: multiply the tops together, and the bottoms together:

3x x−2  ×  x 3  =  (3x)(x) 3(x−2)  

=  3x 2 3(x−2)  

=  x 2 x−2

Dividing Fractions

To divide fractions first "flip" the fraction we want to divide by, then use the same method as for multiplying:

3y 2 x+1  ÷  y 2  =  3y 2 x+1  ×  2 y  

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

= 6y 2 (x+1)(y)  

=  6y x+1

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Course: Algebra 1   >   Unit 2

• Why we do the same thing to both sides: Variable on both sides
• Intro to equations with variables on both sides
• Equations with variables on both sides: 20-7x=6x-6
• Equations with variables on both sides

Equation with variables on both sides: fractions

• Equations with variables on both sides: decimals & fractions
• Equation with the variable in the denominator

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Video transcript

Algebra: Fraction Problems

Related Topics: More Algebra Word Problems

In these lessons, we will learn how to solve fraction word problems that deal with fractions and algebra. Remember to read the question carefully to determine the numerator and denominator of the fraction.

We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.

Fraction Word Problems using Algebra

Example: 2/3 of a number is 14. What is the number?

Answer: The number is 21.

Example: The numerator of a fraction is 3 less than the denominator. When both the numerator and denominator are increased by 4, the fraction is increased by fraction.

Solution: Let the numerator be x, then the denominator is x + 3, and the fraction is \(\frac{x}{{x + 3}}\) When the numerator and denominator are increased by 4, the fraction is \(\frac{{x + 4}}{{x + 7}}\) \(\frac{{x + 4}}{{x + 7}} - \frac{x}{{x + 3}} = \frac{{12}}{{77}}\) 77(x + 4)(x + 3) – 77x(x+7) = 12(x + 7)(x + 3) 77x 2 + 539x + 924 – 77x 2 – 539x = 12x 2 + 120x + 252 12x 2 + 120x – 672 = 0 x 2 + 10x – 56 = 0 (x – 4)(x + 14) = 0 x = 4 (negative answer not applicable in this case)

How to solve Fraction Word Problems using Algebra? Examples: (1) The denominator of a fraction is 5 more than the numerator. If 1 is subtracted from the numerator, the resulting fraction is 1/3. Find the original fraction. (2) If 3 is subtracted from the numerator of a fraction, the value of the resulting fraction is 1/2. If 13 is added to the denominator of the original fraction, the value of the new fraction is 1/3. Find the original fraction. (3) A fraction has a value of 3/4. When 14 is added to the numerator, the resulting fraction has a value equal to the reciprocal of the original fraction, Find the original fraction.

Algebra Word Problems with Fractional Equations Solving a fraction equation that appears in a word problem Example: One third of a number is 6 more than one fourth of the number. Find the number.

Fraction and Decimal Word Problems How to solve algebra word problems with fractions and decimals? Examples: (1) If 1/2 of the cards had been sold and there were 172 cards left, how many cards were printed? (2) Only 1/3 of the university students wanted to become teachers. If 3,360 did not wan to become teachers, how many university were there? (3) Rodney guessed the total was 34.71, but this was 8.9 times the total. What was the total?

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

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Inverse operations

Solve equations with fractions

Here you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.

Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.

What are equations with fractions?

Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or denominator of a fraction.

The numerator (top number) in a fraction is divided by the denominator (bottom number).

To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.

The inverse operation of addition is subtraction.

The inverse operation of subtraction is addition.

The inverse operation of multiplication is division.

The inverse operation of division is multiplication.

For example,

\begin{aligned} \cfrac{2x+3}{5} \, &= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}

Common Core State Standards

How does this relate to 7th grade and 8th grade math?

• Grade 7: Expressions and Equations (7.EE.A.1) Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
• Grade 8: Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• Grade 8: Expressions and Equations (8.EE.C.7b) Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

How to solve equations with fractions

In order to solve equations with fractions:

Identify the operations that are being applied to the unknown variable.

Apply the inverse operations, one at a time, to both sides of the equation.

Write the final answer, checking that it is correct.

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Solve equations with fractions examples

Example 1: equations with one operation.

Solve for x \text{: } \cfrac{x}{5}=4 .

The unknown is x.

Looking at the left hand side of the equation, the x is divided by 5.

\cfrac{x}{5}

2 Apply the inverse operations, one at a time, to both sides of the equation.

The inverse of “dividing by 5 ” is “multiplying by 5 ”.

You will multiply both sides of the equation by 5.

3 Write the final answer, checking that it is correct.

The final answer is x=20.

You can check the answer by substituting the answer back into the original equation.

\cfrac{20}{5}=20\div5=4

Example 2: equations with one operation

Solve for x \text{: } \cfrac{x}{3}=8 .

Looking at the left hand side of the equation, the x is divided by 3.

\cfrac{x}{3}

The inverse of “dividing by 3 ” is “multiplying by 3 ”.

You will multiply both sides of the equation by 3.

The final answer is x=24.

\cfrac{24}{3}=24\div3=8

Example 3: equations with two operations

Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .

Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).

\cfrac{x \, + \, 1}{2}

First, clear the fraction by multiplying both sides of the equation by 2.

Then, subtract 1 from both sides.

The final answer is x=13.

\cfrac{13 \, +1 \, }{2}=\cfrac{14}{2}=14\div2=7

Example 4: equations with two operations

Solve for x \text{: } \cfrac{x}{4}-2=3 .

Looking at the left hand side of the equation, x is divided by 4 and then 2 is subtracted.

\cfrac{x}{4}-2

First, add 2 to both sides of the equation.

Then, multiply both sides of the equation by 4.

\cfrac{20}{4}-2=20\div4-2=5-2=3

Example 5: equations with three operations

Solve for x \text{: } \cfrac{3x}{5}+1=7 .

Looking at the left hand side of the equation, x is multiplied by 3, then divided by 5 , and then 1 is added.

\cfrac{3x}{5}+1

First, subtract 1 from both sides of the equation.

Then, multiply both sides of the equation by 5.

Finally, divide both sides by 3.

The final answer is x=10.

\cfrac{3 \, \times \, 10}{5}+1=\cfrac{30}{5}+1=6+1=7

Example 6: equations with three operations

Solve for x \text{: } \cfrac{2x-1}{7}=3 .

Looking at the left hand side of the equation, x is multiplied by 2, then 1 is subtracted, and the last operation is divided by 7 (the denominator).

\cfrac{2x-1}{7}

First, multiply both sides of the equation by 7.

Next, add 1 to both sides.

The final answer is x=11.

\cfrac{2 \, \times \, 11-1}{7}=\cfrac{22-1}{7}=\cfrac{21}{7}=3

Example 7: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{24}{x}=6 .

Looking at the left hand side of the equation, x is the denominator. 24 is divided by x.

\cfrac{24}{x}

You need to multiply both sides of the equation by x.

Then, you can divide both sides by 6.

The final answer is x=4.

\cfrac{24}{4}=24\div4=6

Example 8: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{18}{x}-6=3 .

Looking at the left hand side of the equation, x is the denominator. 18 is divided by x , and then 6 is subtracted.

\cfrac{18}{x}-6

First, add 6 to both sides of the equation.

Then, multiply both sides of the equation by x.

Finally, divide both sides by 9.

The final answer is x=2.

\cfrac{18}{2}-6=9-6=3

Teaching tips for solving equations with fractions

• When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.
• Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.
• Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.
• Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.

Easy mistakes to make

• The solution to an equation can be any type of number The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.
• The unknown of an equation can be on either side of the equation The unknown, represented by a letter, is often on the left hand side of the equations; however, it doesn’t have to be. It could also be on the right hand side of an equation.

• Lowest common denominator (LCD) It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Substitution
• Linear equations
• Writing linear equations
• Solving equations
• Identity math
• One step equations

Practice solve equations with fractions questions

1. Solve: \cfrac{x}{6}=3

You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.

The final answer is x = 18.

\cfrac{18}{6}=18 \div 6=3

2. Solve: \cfrac{x \, + \, 4}{2}=7

Then subtract 4 from both sides.

The final answer is x = 10.

\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7

3. Solve: \cfrac{x}{8}-5=1

First, add 5 to both sides of the equation.

Then multiply both sides of the equation by 8.

The final answer is x = 48.

\cfrac{48}{8}-5=48 \div 8-5=1

4. Solve: \cfrac{3x \, + \, 2}{4}=2

First, multiply both sides of the equation by 4.

Next, subtract 2 from both sides.

The final answer is x = 2.

\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2

5. Solve: \cfrac{4x}{7}-2=6

Then multiply both sides of the equation by 7.

Finally, divide both sides by 4.

The final answer is x = 14.

\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6

6. Solve: \cfrac{42}{x}=7

Then you divide both sides by 7.

The final answer is x = 6.

\cfrac{42}{6}=42 \div 6=7

Solve equations with fractions FAQs

Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.

The next lessons are

• Inequalities
• Types of graphs
• Math formulas
• Coordinate plane
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• Algebraic expressions

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Problems on Algebraic Fractions

Here we will learn how to simplify the problems on algebraic fractions to its lowest term.

1.  Reduce the algebraic fractions to their lowest terms:  \(\frac{x^{2}  -  y^{2}}{x^{3}  -  x^{2}y}\)

\(\frac{x^{2}  -  y^{2}}{x^{3}  -  x^{2}y}\)

Factorizing the numerator and denominator separately and cancelling the common factors we get,

=  \(\frac{(x  +  y) (x  -  y)}{x^{2} (x  -  y)} \)

=  \(\frac{x  +  y}{x^{2}}\)

2. Reduce to lowest terms \(\frac{x^{2}  +  x  -  6}{x^{2}  -  4}\)

\(\frac{x^{2}  +  x  -  6}{x^{2} -  4}\)

Step 1: Factorize the numerator x\(^{2}\) + x – 6

                                         = x\(^{2}\) + 3x – 2x – 6

                                         = x(x + 3) – 2(x + 3)

                                         = (x + 3) (x – 2)

Step 2: Factorize the denominator: x\(^{2}\) – 4

                                             = x\(^{2}\) – 2\(^{2}\)

                                             = (x + 2) (x – 2)

Step 3: From steps 1 and 2: \(\frac{x^{2}  +  x  -  6}{x^{2}  -  4}\)

                                     = \(\frac{x^{2}  +  x  -  6}{x^{2}  -  2^{2}}\)

                                     = \(\frac{(x  +  3) (x  -  2)}{(x  +  2) (x  -  2)}\)

                                     = \(\frac{(x  +  3)}{(x  +  2)}\)

3. Simplify the algebraic fractions \(\frac{36x^{2}  -  4}{9x^{2}  +  6x  +  1}\)

\(\frac{36x^{2}  -  4}{9x^{2}  +  6x  +  1}\)

Step 1: Factorize the numerator: 36x\(^{2}\) – 4

                                           = 4(9x\(^{2}\) – 1)

                                           = 4[(3x)\(^{2}\) – (1)\(^{2}\)]

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

Step 2: Factorize the denominator: 9x\(^{2}\) + 6x + 1

                                             = 9x\(^{2}\) + 3x + 3x + 1

                                             = 3x(3x + 1) + 1(3x + 1)

                                             = (3x + 1) (3x + 1)

Step 3: Simplification of the given expression after factorizing the numerator and the denominator:

= \(\frac{4(3x  +  1)(3x  -  1)}{(3x  +  1)(3x  +  1)}\)

= \(\frac{4(3x  -  1)}{(3x  +  1)}\)

4.  Reduce and simplify:  \(\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )\)

\(\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )\)

=  \(\frac{8x^{3}y^{2}z}{2xy^{3}} of \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \times \frac{35x^{2}yz^{3}}{7xy^{2}}\)

=  \(\frac{4x^{3}y^{2}z}{xy^{3}} \left ( \frac{x^{5}y^{2}z^{2}}{xy^{3}z} \times \frac{x^{2}yz^{3}}{xy^{2}} \right )\)

= 4x\(^{10 - 3}\) ∙ y\(^{-3}\) ∙ z\(^{5}\)

=  \(\frac{4x^{7}\cdot z^{5}}{y^{3}}\)

5.  Simplify:  \(\frac{2x^{2}  -  3x  -  2}{x^{2}  +  x  -  2} \div \frac{2x^{2}  +  3x  +  1}{3x^{2}  +  3x  -  6}\)

\(\frac{2x^{2}  -  3x  -  2}{x^{2}  +  x   -   2} \div \frac{2x^{2}  +  3x  +  1}{3x^{2}  +  3x  -  6}\)

Step 1: First factorize each of the polynomials separately:

2x\(^{2}\) – 3x – 2 = 2x\(^{2}\) – 4x + x – 2

                 = 2x(x – 2) + 1 (x – 2)

                 = (x – 2) (2x + 1)

x\(^{2}\) + x – 2 = x\(^{2}\) + 2x - x – 2

              = x(x + 2) - 1 (x + 2)

              = (x + 2) (x - 1)

2x\(^{2}\) + 3x + 1 = 2x\(^{2}\) + 2x + x + 1

                 = 2x(x + 1) + 1 (x + 1)

                 = (x + 1) (2x + 1)

3x\(^{2}\) + 3x – 6 = 3[x\(^{2}\) + x – 2]

                 = 3[x\(^{2}\) + 2x - x – 2]

                 = 3[x(x + 2) – 1(x + 2)]                   

                 = 3[(x + 2) (x - 1)]

                 = 3(x + 2) (x - 1)

Step 2: Simplify the given expressions by substituting with their factors

\(\frac{2x^{2}  -  3x  -  2}{x^{2}  +  x  -  2} \div \frac{2x^{2}  +  3x  +  1}{3x^{2}  +  3x  -  6}\)

=  \(\frac{2x^{2}  -  3x  -  2}{x^{2}  +  x  -  2} \times \frac{3x^{2}  +  3x  -  6}{2x^{2}  +  3x  +  1}\)

=  \(\frac{(x  -  2) (2x  +  1)}{(x  +  2) (x  -  1)}\times\frac{3(x  +  2) (x  -  1)}{(x  +  1) (2x  +  1)}\)

=  \(\frac{3(x  -  2)}{(x  +  1)}\)

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1.26: Solving Fractional Equations

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A fractional equation is an equation involving fractions which has the unknown in the denominator of one or more of its terms.

Example 24.1

The following are examples of fractional equations:

a) \(\frac{3}{x}=\frac{9}{20}\)

b) \(\frac{x-2}{x+2}=\frac{3}{5}\)

c) \(\frac{3}{x-3}=\frac{4}{x-5}\)

d) \(\frac{3}{4}-\frac{1}{8 x}=0\)

e) \(\frac{x}{6}-\frac{2}{3 x}=\frac{2}{3}\)

The Cross-Product property can be used to solve fractional equations.

Cross-Product Property

If \(\frac{A}{B}=\frac{C}{D}\) then \(A \cdot D=B \cdot C\).

Using this property we can transform fractional equations into non-fractional ones. We must take care when applying this property and use it only when there is a single fraction on each side of the equation. So, fractional equations can be divided into two categories.

I. Single Fractions on Each Side of the Equation

Equations a), b) and c) in Example 24.1 fall into this category. We solve these equations here.

a) Solve \(\frac{3}{x}=\frac{9}{20}\)

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 20=9 \cdot x \\ \text{Linear Equation} & 60=9 x \\ \text{Divide by 9 both sides} & \frac{60}{9}=x \end{array}\nonumber\]

The solution is \(x=\frac{60}{9}=\frac{20}{3}\).

\[\begin{array}{ll} \text{Cross-Product} & 5 \cdot(x-2)=3 \cdot(x+2) \\ \text{Remove parentheses} & 5 x-10=3 x+6 \\ \text{Linear Equation: isolate the variable} & 5 x-3 x=10+6 \\ & 2 x=16 \\ \text{Divide by 2 both sides} & \frac{2 x}{2}=\frac{16}{2}\end{array}\nonumber\]

the solution is \(x=8\).

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot(x-5)=4 \cdot(x-3) \\ \text{Remove parentheses} & 3 x-15=4 x-12 \\ \text{Linear Equation: isolate the variable} & 3 x-4 x=15-12 \\ & -x=3 \\ \text{Divide by 2 both sides} & \frac{-x}{-1}=\frac{3}{-1}\end{array}\nonumber\]

The solution is \(x=-3\)

Note: If you have a fractional equation and one of the terms is not a fraction, you can always account for that by putting 1 in the denominator. For example:

\[\frac{3}{x}=15\nonumber\]

We re-write the equation so that all terms are fractions.

\[\frac{3}{x}=\frac{15}{1}\nonumber\]

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 1=15 \cdot x \\ \text{Linear Equation: isolate the variable} & 3=15 x \\ \text{Divide by 15 both sides} & \frac{3}{15}=\frac{15 x}{15} \end{array}\nonumber\]

The solution is \(x=\frac{3}{15}=\frac{3 \cdot 1}{3 \cdot 5}=\frac{1}{5}\).

II. Multiple Fractions on Either Side of the Equation

Equations d) and e) in Example 24.1 fall into this category. We solve these equations here.

We use the technique for combining rational expressions we learned in Chapter 23 to reduce our problem to a problem with a single fraction on each side of the equation.

d) Solve \(\frac{3}{4}-\frac{1}{8 x}=0\)

First we realize that there are two fractions on the LHS of the equation and thus we cannot use the Cross-Product property immediately. To combine the LHS into a single fraction we do the following:

\[\begin{array}{ll} \text{Find the LCM of the denominators} & 8 x \\ \text{Rewrite each fraction using the LCM} & \frac{3 \cdot 2 x}{8 x}-\frac{1}{8 x}=0 \\ \text{Combine into one fraction} & \frac{6 x-1}{8 x}=0 \\ \text{Re-write the equation so that all terms are fractions} & \frac{6 x-1}{8 x}=\frac{0}{1} \\ \text{Cross-Product} & (6 x-1) \cdot 1=8 x \cdot 0 \\ \text{Remove parentheses} & 6 x-1=0 \\\text{Linear Equation: isolate the variable} & 6 x=1 \\ \text{Divide by 6 both sides} & \frac{6 x}{6}=\frac{1}{6} \end{array}\nonumber\]

The solution is \(x=\frac{1}{6}\).

e) Solve \(\frac{x}{6}+\frac{2}{3 x}=\frac{2}{3}\)

\[\begin{array}{ll} \text{Find the LCM of the denominators of LHS} & 6x \\ \text{Rewrite each fraction on LHS using their LCM} & \frac{x \cdot x}{6 x}+\frac{2 \cdot 2}{6 x}=\frac{2}{3} \\ \frac{x^{2}+4}{6 x}=\frac{2}{3} \text{Combine into one fraction} & \left(x^{2}+4\right) \cdot 3=6 x \cdot 2 \\ \text{Cross-Product} & 3 x^{2}+12=12 x \\ \text{Remove parentheses} & 3 x^{2}-12 x+12=0 \\ \text{Quadratic Equation: Standard form} & 3 x^{2}-12 x+12=0 \\\text{Quadratic Equation: Factor} & 3 \cdot x^{2}-3 \cdot 4 x+3 \cdot 4=0 \\ & 3\left(x^{2}-4 x+4\right)=0 \\ & 3(x-2)(x-2)=0 \\ \text{Divide by 3 both sides} & \frac{3(x-2)(x-2)}{3}=\frac{0}{3} \\ & (x-2)(x-2)=0 \\ \text{Quadratic Equation: Zero-Product Property} & (x-2)=0 \text { or }(x-2)=0 \end{array}\nonumber\]

Since both factors are the same, then \(x-2=0\) gives \(x=2\). The solution is \(x=2\)

Note: There is another method to solve equations that have multiple fractions on either side. It uses the LCM of all denominators in the equation. We demonstrate it here to solve the following equation: \(\frac{3}{2}-\frac{9}{2 x}=\frac{3}{5}\)

\[\begin{array} \text{Find the LCM of all denominators in the equation} & 10x \\ \text{Multiply every fraction (both LHS and RHS) by the LCM} & 10 x \cdot \frac{3}{2}-10 x \cdot \frac{9}{2 x}=10 x \cdot \frac{3}{5} \\ & \frac{10 x \cdot 3}{2}-\frac{10 x \cdot 9}{2 x}=\frac{10 x \cdot 3}{5} \\ \text{Simplify every fraction} & \frac{5 x \cdot 3}{1}-\frac{5 \cdot 9}{1}=\frac{2 x \cdot 3}{1} \\ \text{See how all denominatiors are now 1, thus can be disregarded} & 5 x \cdot 3-5 \cdot 9=2 x \cdot 3 \\ \text{Solve like you would any other equation} & 15 x-45=6 x \\ \text{Linear equation: islolate the variable} & 15 x-6 x=45 \\ & 9 x=45 \\ & x=\frac{45}{9} \\ & x=5 \end{array} \nonumber\]

The solution is \(x=5\)

Exit Problem

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

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SAT Math : Algebraic Fractions

Study concepts, example questions & explanations for sat math, all sat math resources, example questions, example question #1 : algebraic fractions.

Another way to represent this question is:

The one carries to the ten's column.

One way to simplify this complex fraction is to find the least common multiple of all the denominators, i.e. the least common denominator (LCD). If we find this, then we can multiply every fraction by the LCD and thereby be left with only whole numbers. This will make more sense in a little bit.

The denominators we are dealing with are 2, 3, 4, 5, and 6. We want to find the smallest multiple that these numbers have in common. First, it will help us to notice that 6 is a multiple of both 2 and 3. Thus, if we find the least common multiple of 4, 5, and 6, it will automatically be a multiple of both 2 and 3. Let's list out the first several multiples of 4, 5, and 6.

4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

The smallest multiple that 4, 5, and 6 have in common is 60. Thus, the least common multiple of 4, 5, and 6 is 60. This also means that the least common multiple of 2, 3, 4, 5, and 6 is 60. Therefore, the LCD of all the fractions is 60.

Let's think of the expression we want to simplify as one big fraction. The numerator contains the fractions 1/4, 1/3, and –1/5. The denominator of the fraction is 1/2, –1/6 and 1. Remember that if we have a fraction, we can multiply the numerator and denominator by the same number without changing the value of the fraction. In other words, x / y = ( xz )/( yz ). This will help us because we can multiply the numerator (which consists of 1/4, 1/3, and –1/5) by 60, and then mutiply the denominator (which consists of 1/2, –1/6, and 1) by 60, thereby ridding us of fractions in the numerator and denominator. This process is shown below:

This means that a / b = 23/80. We are told that a and b are both positive and that their greatest common factor is 1. In other words, a / b must be the simplified form of 23/80. When a fraction is in simplest form, the greatest common factor of the numerator and denominator equals one. Since 23/80 is simplified, a = 23, and b = 80. The sum of a and b is thus 23 + 80 = 103.

The answer is 103.

Change to a mixed number 

To convert from a fraction to a mixed number we must find out how many times the denominator goes into the numerator using division and the remainder becomes the new fraction. 

To average, we have to add the values and divide by two. To do this we need to find a common denomenator of 6. We then add and divide by 2, yielding 4.5/6. This reduces to 3/4.

Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4

None of the answers are correct

This problem is solved the same way ½ + 1/3 is solved.  For example,  ½ + 1/3 = 3/6 + 2/6 = 5/6.  Find a common denominator then convert each fraction into an equivalent fraction using that common denominator.  The final step is to add the two new fractions and simplify.

Example Question #1 : How To Simplify A Fraction

Set up the conversions as fractions and solve:

Can't be simplified

To simplify exponents which are being divided, subtract the exponents on the bottom from exponents on the top.  Remember that only exponents with the same bases can be simplified

Example Question #841 : Algebra

x 2  –  y 2 can be also expressed as ( x + y )( x –  y ).

Therefore, the fraction now can be re-written as ( x + y )( x  –  y )/( x + y ).

This simplifies to ( x –  y ).

Example Question #6 : Algebraic Fractions

Now multiply both the numerator and denominator by the conjugate of the denominator:

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Chapter 8: Rational Expressions

8.6 Solving Complex Fractions

When solving two or more equated fractions, the easiest solution is to first remove all fractions by multiplying both sides of the equations by the LCD. This strategy is shown in the next examples.

Example 8.6.1

Solve [latex]\dfrac{x+3}{4}=\dfrac{2}{3}[/latex].

For these two fractions, the LCD is 3 × 4 = 12. Therefore, we multiply both sides of the equation by 12:

[latex]12\left(\dfrac{x+3}{4}\right)=\left(\dfrac{2}{3}\right)12[/latex]

This reduces the complex fraction to:

[latex]3(x+3)=2(4)[/latex]

Multiplying this out yields:

[latex]3x + 9 = 8[/latex]

Now just isolate and solve for [latex]x[/latex]:

[latex]\begin{array}{rrrrr} 3x&+&9&=&8 \\ &-&9&&-9 \\ \hline &&3x&=&-1 \\ \\ &&x&=&-\dfrac{1}{3} \end{array}[/latex]

Example 8.6.2

Solve [latex]\dfrac{2x-3}{3x+4} = \dfrac{2}{5}[/latex].

For these two fractions, the LCD is [latex]5(3x + 4)[/latex]. Therefore, both sides of the equation are multiplied by [latex]5(3x + 4)[/latex]:

[latex]5(3x+4)\left(\dfrac{2x-3}{3x+4}\right)=\left(\dfrac{2}{5}\right)5(3x+4)[/latex]

[latex]5(2x-3)=2(3x+4)[/latex]

[latex]10x - 15 = 6x + 8[/latex]

Now isolate and solve for [latex]x[/latex]:

[latex]\begin{array}{rrrrrrr} 10x&-&15&=&6x&+&8 \\ -6x&+&15&&-6x&+&15 \\ \hline &&4x&=&23&& \\ \\ &&x&=&\dfrac{23}{4}&& \end{array}[/latex]

Example 8.6.3

Solve [latex]\dfrac{k+3}{3}= \dfrac{8}{k-2}[/latex].

For these two fractions, the LCD is [latex]3(k-2)[/latex]. Therefore, multiply both sides of the equation by [latex]3(k-2)[/latex]:

[latex]3(k-2)\left(\dfrac{k+3}{3}\right)=\left(\dfrac{8}{k-2}\right)3(k-2)[/latex]

[latex](k - 2) (k + 3) = 8 (3)[/latex]

This multiplies out to:

[latex]k^2 + k - 6 = 24[/latex]

Now subtract 24 from both sides of the equation to turn this into an equation that can be easily factored:

[latex]\begin{array}{rrrrrrr} k^2&+&k&-&6&=&24 \\ &&&-&24&&-24 \\ \hline k^2&+&k&-&30&=&0 \end{array}[/latex]

This equation factors to:

[latex](k + 6)(k - 5) = 0[/latex]

The solutions are:

[latex]k = -6[/latex] and [latex]k=5[/latex]

Solve each of the following complex fractions.

• [latex]\dfrac{m-1}{5}=\dfrac{8}{2}[/latex]
• [latex]\dfrac{8}{2}=\dfrac{8}{x-8}[/latex]
• [latex]\dfrac{2}{9}=\dfrac{10}{p-4}[/latex]
• [latex]\dfrac{9}{n+2}=\dfrac{3}{9}[/latex]
• [latex]\dfrac{3}{10}=\dfrac{a}{a+2}[/latex]
• [latex]\dfrac{x+1}{3}=\dfrac{x+3}{4}[/latex]
• [latex]\dfrac{2}{p+4}=\dfrac{p+5}{3}[/latex]
• [latex]\dfrac{5}{n+1}=\dfrac{n-4}{10}[/latex]
• [latex]\dfrac{x+5}{5}=\dfrac{6}{x-2}[/latex]
• [latex]\dfrac{4}{x-3}=\dfrac{x+5}{5}[/latex]
• [latex]\dfrac{m+3}{4}=\dfrac{11}{m-4}[/latex]
• [latex]\dfrac{x-5}{8}=\dfrac{4}{x-1}[/latex]

Answer Key 8.6

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Mathematics > Algebraic Geometry

Title: positive moments forever: undecidable and decidable cases.

Abstract: Is there an algorithm to determine attributes such as positivity or non-zeroness of linear recurrence sequences? This long-standing question is known as Skolem's problem. In this paper, we study the complexity of an equivalent problem, namely the (generalized) moment membership problem for matrices. We show that this problem is decidable for orthogonal, unitary and real eigenvalue matrices, and undecidable for matrices over certain commutative and non-commutative polynomial rings. Our results imply that the positivity problem for simple unitary linear recurrence sequences is decidable, and is undecidable for linear recurrence sequences over the ring of commutative polynomials. As a byproduct, we prove a free version of Polya's theorem.

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