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})

[egin{array}{ll} ext{Cross-Product} & 3 cdot 20=9 cdot x ext{Linear Equation} & 60=9 x ext{Divide by 9 both sides} & frac{60}{9}=x end{array} onumber]

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

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

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

the solution is (x=8).

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

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

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:

Solve

[frac{3}{x}=15 onumber]

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

[frac{3}{x}=frac{15}{1} onumber]

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

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:

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

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

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

[egin{array}{ll} ext{Find the LCM of the denominators of LHS} & 6x ext{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} ext{Combine into one fraction} & left(x^{2}+4 ight) cdot 3=6 x cdot 2 ext{Cross-Product} & 3 x^{2}+12=12 x ext{Remove parentheses} & 3 x^{2}-12 x+12=0 ext{Quadratic Equation: Standard form} & 3 x^{2}-12 x+12=0 ext{Quadratic Equation: Factor} & 3 cdot x^{2}-3 cdot 4 x+3 cdot 4=0 & 3left(x^{2}-4 x+4 ight)=0 & 3(x-2)(x-2)=0 ext{Divide by 3 both sides} & frac{3(x-2)(x-2)}{3}=frac{0}{3} & (x-2)(x-2)=0 ext{Quadratic Equation: Zero-Product Property} & (x-2)=0 ext { or }(x-2)=0 end{array} onumber]

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})

[egin{array} ext{Find the LCM of all denominators in the equation} & 10x ext{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} ext{Simplify every fraction} & frac{5 x cdot 3}{1}-frac{5 cdot 9}{1}=frac{2 x cdot 3}{1} ext{See how all denominatiors are now 1, thus can be disregarded} & 5 x cdot 3-5 cdot 9=2 x cdot 3 ext{Solve like you would any other equation} & 15 x-45=6 x ext{Linear equation: islolate the variable} & 15 x-6 x=45 & 9 x=45 & x=frac{45}{9} & x=5 end{array} onumber]

The solution is (x=5)

Exit Problem

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

## Fractional calculus

and developing a calculus for such operators generalizing the classical one.

In this context, the term *powers* refers to iterative application of a linear operator *D* to a function *f*, that is, repeatedly composing *D* with itself, as in D n ( f ) = ( D ∘ D ∘ D ∘ ⋯ ∘ D ⏟ n ) ( f ) = D ( D ( D ( ⋯ D ⏟ n ( f ) ⋯ ) ) )

For example, one may ask for a meaningful interpretation of

as an analogue of the functional square root for the differentiation operator, that is, an expression for some linear operator that when applied *twice* to any function will have the same effect as differentiation. More generally, one can look at the question of defining a linear operator

for every real number a in such a way that, when a takes an integer value *n* ∈ ℤ , it coincides with the usual n -fold differentiation D if *n* > 0 , and with the (−*n*) -th power of J when *n* < 0 .

One of the motivations behind the introduction and study of these sorts of extensions of the differentiation operator D is that the sets of operator powers < *D* *a* | *a* ∈ ℝ > defined in this way are *continuous* semigroups with parameter a , of which the original *discrete* semigroup of < *D* *n* | *n* ∈ ℤ > for integer n is a denumerable subgroup: since continuous semigroups have a well developed mathematical theory, they can be applied to other branches of mathematics.

Fractional differential equations, also known as extraordinary differential equations, [1] are a generalization of differential equations through the application of fractional calculus.

## Keywords

**K. Krishnaveni** is Research Scholar in the Department of Mathematics, SASTRA University, Thanjavur, India. Her Research area includes Fractional Differential Equations, Numerical Analysis and Combinatorial Optimization.

**Dr. K. Kannan** is currently working as Professor, SASTRA University, Thanjavur, India. He has been in Academia for the past 25 years. His Research interests are Combinatorial Optimization, Artificial Neural Networks and Hypergraph-based Image Processing, Differential Equations.

**Dr. S. Raja Balachandar** is currently working as an Assistant Professor in the Department of Mathematics, SASTRA University, Thanjavur, India. His research interests are Mathematical modeling, Combinatorial Optimization, Numerical Analysis, Fractional Differential Equations and Wavelet Transforms.

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## Math Unit 3 Test

Which of the equations below could be used to solve the problem?

What was the original number?

Which equation could not be used to find the cost of the drawing tablet?

What is the cost of the drawing tablet?

How many hours did Mark work?

Which of the following equations could you use to solve the problem?

Which equation could you use to find the cost, c, of the shirt before sales tax?

Which equation could you use to find how long it will take to save for the guitar?

How many more weeks will it take her to save enough money?

To the nearest inch, how many inches long is the femur of a male who is 71 inches tall?

Which of the following equations could you use to find how many rides Dylan can go on?

How many rides can he go on?

Let S represent Sara's age.

Which inequality describes Sara's age?

Let f represent the number of Blake's friends.

Which inequality describes the number of Blake's friends?

Let b represent the number of books in the library.

Which inequality describes the number of books?

Which inequality could be used find the number of hours Sophia needs to work to make enough money to buy a new bicycle?

## Fractional Equations

High School Math based on the topics required for the Regents Exam conducted by NYSED.

**How to solve Fractional Equations?**

1. Find the Lowest Common Denominator (LCD).

2. Multiply both sides of the equation by the LCD (to remove the fractions).

3. Solve the equation.

4. Check the solution.

The following diagram gives an example of solving fractional equation. Scroll down the page for more examples and solutions of solving fractional equations.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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

## 1.3 Fractions

A more thorough introduction to the topics covered in this section can be found in the *Elementary Algebra 2e* chapter, Foundations.

### Simplify Fractions

### Fraction

A **fraction** is written a b , a b , where b ≠ 0 b ≠ 0 and

*a* is the **numerator** and *b* is the **denominator**.

Fractions that have the same value are equivalent fractions . The Equivalent Fractions

Property allows us to find equivalent fractions and also simplify fractions.

### Equivalent Fractions Property

If *a*, *b*, and *c* are numbers where b ≠ 0 , c ≠ 0 , b ≠ 0 , c ≠ 0 ,

A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator.

We simplify, or reduce, a fraction by removing the common factors of the numerator and denominator. A fraction is not simplified until all common factors have been removed. If an expression has fractions, it is not completely simplified until the fractions are simplified.

Sometimes it may not be easy to find common factors of the numerator and denominator. When this happens, a good idea is to factor the numerator and the denominator into prime numbers. Then divide out the common factors using the Equivalent Fractions Property.

### Example 1.24

#### How To Simplify a Fraction

#### Solution

We now summarize the steps you should follow to simplify fractions.

### How To

#### Simplify a fraction.

- Step 1. Rewrite the numerator and denominator to show the common factors.

If needed, factor the numerator and denominator into prime numbers first. - Step 2. Simplify using the Equivalent Fractions Property by dividing out common factors.
- Step 3. Multiply any remaining factors.

### Multiply and Divide Fractions

Many people find multiplying and dividing fractions easier than adding and subtracting fractions.

To multiply fractions, we multiply the numerators and multiply the denominators.

### Fraction Multiplication

To multiply fractions, multiply the numerators and multiply the denominators.

When multiplying fractions, the properties of positive and negative numbers still apply, of course. It is a good idea to determine the sign of the product as the first step. In Example 1.25, we will multiply a negative by a negative, so the product will be positive.

When multiplying a fraction by an integer, it may be helpful to write the integer as a fraction. Any integer, *a*, can be written as a 1 . a 1 . So, for example, 3 = 3 1 . 3 = 3 1 .

### Example 1.25

#### Solution

The first step is to find the sign of the product. Since the signs are the same, the product is positive.

Determine the sign of the product. The signs are the same, so the product is positive. |

Write 20x as a fraction. |

Multiply. |

Rewrite 20 to show the common factor 5 and divide it out. |

Simplify. |

Now that we know how to multiply fractions, we are almost ready to divide. Before we can do that, we need some vocabulary. The reciprocal of a fraction is found by inverting the fraction, placing the numerator in the denominator and the denominator in the numerator. The reciprocal of 2 3 2 3 is 3 2 . 3 2 . Since 4 is written in fraction form as 4 1 , 4 1 , the reciprocal of 4 is 1 4 . 1 4 .

To divide fractions, we multiply the first fraction by the reciprocal of the second.

### Fraction Division

To divide fractions, we multiply the first fraction by the **reciprocal** of the second.

### Example 1.26

Find the quotient: − 7 18 ÷ ( − 14 27 ) . − 7 18 ÷ ( − 14 27 ) .

#### Solution

To divide, multiply the first fraction by the reciprocal of the second. |

Determine the sign of the product, and then multiply. |

Rewrite showing common factors. |

Remove common factors. |

Simplify. |

The numerators or denominators of some fractions contain fractions themselves. A fraction in which the numerator or the denominator is a fraction is called a complex fraction .

### Complex Fraction

A **complex fraction** is a fraction in which the numerator or the denominator contains a fraction.

Some examples of complex fractions are:

To simplify a complex fraction, remember that the fraction bar means division. For example, the complex fraction 3 4 5 8 3 4 5 8 means 3 4 ÷ 5 8 . 3 4 ÷ 5 8 .

### Example 1.27

#### Solution

### Add and Subtract Fractions

When we multiplied fractions, we just multiplied the numerators and multiplied the denominators right straight across. To add or subtract fractions, they must have a common denominator.

### Fraction Addition and Subtraction

To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.

The least common denominator (LCD) of two fractions is the smallest number that can be used as a common denominator of the fractions. The LCD of the two fractions is the least common multiple (LCM) of their denominators.

### Least Common Denominator

The **least common denominator** (LCD) of two fractions is the least common multiple (LCM) of their denominators.

After we find the least common denominator of two fractions, we convert the fractions to equivalent fractions with the LCD. Putting these steps together allows us to add and subtract fractions because their denominators will be the same!

### Example 1.28

#### How to Add or Subtract Fractions

#### Solution

### How To

#### Add or subtract fractions.

- Step 1. Do they have a common denominator?
- Yes—go to step 2.
- No—rewrite each fraction with the LCD (least common denominator).
- Find the LCD.
- Change each fraction into an equivalent fraction with the LCD as its denominator.

- Step 2. Add or subtract the fractions.
- Step 3. Simplify, if possible.

We now have all four operations for fractions. Table 1.3 summarizes fraction operations.

When starting an exercise, always identify the operation and then recall the methods needed for that operation.

### Example 1.29

#### Solution

First ask, “What is the operation?” Identifying the operation will determine whether or not we need a common denominator. Remember, we need a common denominator to add or subtract, but not to multiply or divide.

What is the operation? The operation is subtraction. | |

Do the fractions have a common denominator? No. | 5 x 6 − 3 10 5 x 6 − 3 10 |

Find the LCD of 6 and 10 | The LCD is 30. |

6 = 2 · 3 10 = 2 · 5 ___________ LCD = 2 · 3 · 5 LCD = 30 6 = 2 · 3 10 = 2 · 5 ___________ LCD = 2 · 3 · 5 LCD = 30 | |

Rewrite each fraction as an equivalent fraction with the LCD. | 5 x · 5 6 · 5 − 3 · 3 10 · 3 5 x · 5 6 · 5 − 3 · 3 10 · 3 |

25 x 30 − 9 30 25 x 30 − 9 30 | |

Subtract the numerators and place the difference over the common denominators. | 25 x − 9 30 25 x − 9 30 |

Simplify, if possible. There are no common factors. The fraction is simplified. |

### Use the Order of Operations to Simplify Fractions

The fraction bar in a fraction acts as grouping symbol. The order of operations then tells us to simplify the numerator and then the denominator. Then we divide.

### How To

#### Simplify an expression with a fraction bar.

- Step 1. Simplify the expression in the numerator. Simplify the expression in the denominator.
- Step 2. Simplify the fraction.

Where does the negative sign go in a fraction? Usually the negative sign is in front of the fraction, but you will sometimes see a fraction with a negative numerator, or sometimes with a negative denominator. Remember that fractions represent division. When the numerator and denominator have different signs, the quotient is negative.

### Placement of Negative Sign in a Fraction

For any positive numbers *a* and *b*,

### Example 1.30

Simplify: 4 ( − 3 ) + 6 ( − 2 ) − 3 ( 2 ) − 2 . 4 ( − 3 ) + 6 ( − 2 ) − 3 ( 2 ) − 2 .

#### Solution

The fraction bar acts like a grouping symbol. So completely simplify the numerator and the denominator separately.

Simplify: 8 ( − 2 ) + 4 ( − 3 ) − 5 ( 2 ) + 3 . 8 ( − 2 ) + 4 ( − 3 ) − 5 ( 2 ) + 3 .

Simplify: 7 ( − 1 ) + 9 ( − 3 ) − 5 ( 3 ) − 2 . 7 ( − 1 ) + 9 ( − 3 ) − 5 ( 3 ) − 2 .

Now we’ll look at complex fractions where the numerator or denominator contains an expression that can be simplified. So we first must completely simplify the numerator and denominator separately using the order of operations. Then we divide the numerator by the denominator as the fraction bar means division.

### Example 1.31

#### How to Simplify Complex Fractions

#### Solution

### How To

#### Simplify complex fractions.

- Step 1. Simplify the numerator.
- Step 2. Simplify the denominator.
- Step 3. Divide the numerator by the denominator. Simplify if possible.

### Example 1.32

Simplify: 1 2 + 2 3 3 4 − 1 6 . 1 2 + 2 3 3 4 − 1 6 .

#### Solution

It may help to put parentheses around the numerator and the denominator.

Simplify: 1 3 + 1 2 3 4 − 1 3 . 1 3 + 1 2 3 4 − 1 3 .

Simplify: 2 3 − 1 2 1 4 + 1 3 . 2 3 − 1 2 1 4 + 1 3 .

### Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

### Example 1.33

#### Solution

Substitute the values into the expression.

### Media

Access this online resource for additional instruction and practice with fractions.

### Section 1.3 Exercises

#### Practice Makes Perfect

**Simplify Fractions**

In the following exercises, simplify.

**Multiply and Divide Fractions**

In the following exercises, perform the indicated operation.

In the following exercises, simplify.

**Add and Subtract Fractions**

In the following exercises, add or subtract.

**Use the Order of Operations to Simplify Fractions**

In the following exercises, simplify.

7 · 4 − 2 ( 8 − 5 ) 9 · 3 − 3 · 5 7 · 4 − 2 ( 8 − 5 ) 9 · 3 − 3 · 5

9 · 7 − 3 ( 12 − 8 ) 8 · 7 − 6 · 6 9 · 7 − 3 ( 12 − 8 ) 8 · 7 − 6 · 6

9 ( 8 − 2 ) − 3 ( 15 − 7 ) 6 ( 7 − 1 ) − 3 ( 17 − 9 ) 9 ( 8 − 2 ) − 3 ( 15 − 7 ) 6 ( 7 − 1 ) − 3 ( 17 − 9 )

8 ( 9 − 2 ) − 4 ( 14 − 9 ) 7 ( 8 − 3 ) − 3 ( 16 − 9 ) 8 ( 9 − 2 ) − 4 ( 14 − 9 ) 7 ( 8 − 3 ) − 3 ( 16 − 9 )

**Mixed Practice**

In the following exercises, simplify.

**Evaluate Variable Expressions with Fractions**

In the following exercises, evaluate.

#### Writing Exercises

Why do you need a common denominator to add or subtract fractions? Explain.

How do you find the LCD of 2 fractions?

Explain how you find the reciprocal of a fraction.

Explain how you find the reciprocal of a negative number.

#### Self Check

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

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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## Abstract

We propose a high-order numerical scheme for nonlinear time fractional reaction-diffusion equations with initial singularity, where

*L*2- 1 σ scheme on graded mesh is used to approximate Caputo fractional derivative and Legendre spectral method is applied to discrete spatial variable. We give the priori estimate, existence and uniqueness of numerical solution. Then the unconditional stability and convergence are proved. The rate of convergence is O ( M − min < r α , 2 >+ N − m ) , which is obtained without extra regularity assumption on the exact solution. Numerical results are given to confirm the sharpness of error analysis.

## Clearing Equations of Decimals

**To clear an equation of decimals, multiply each term on both sides by the power of ten that will make all the decimals whole numbers.**In our example above, if we multiply .25 by 100, we will get 25, a whole number. Since each decimal only goes to the hundredths place, 100 will work for all three terms.So let's multiply each term by 100 to clear the decimals:

(100) 0.25x + (100) 0.35 = (100) (-0.29)

Now we can solve the equation as normal:

x = -2.56 Since the original was in decimal form, the answer should most likely also be in decimal form.

We have to think a little more carefully about what multiple of ten to use here. 6.2 only needs to be multiplied by 10, but 1.25 needs 100, so we will multiply every term by 100. Don't forget to multiply the 4 by 100 as well.

We had to be extra careful as we multiplied by 100. Now we can solve the equation as normal:

**Practice:**Clear each equation of decimals, then solve. Round each answer to the hundredths place.

## Publications of Jie Shen

(with Duo Cao and Jie Xu) Computing interface with quasiperiodicity. J. Comput. Phys. 424:109863, 2021.

(with Fukeng Huang and Zhiguo Yang) A highly efficient and accurate new SAV approach for gradient flows. SIAM J. Sci. Comput. 42:A2514-A2536, 2020.

(with Xiaofeng Yang) The IEQ and SAV approaches and their extensions for a class of highly nonlinear gradient flow systems. Proceedings celebrating 75 years of Mathematics of Computation, Contemp. Math. 754:217-245, 2020.

(with Qing Cheng and Chun Liu) A new Lagrange Multiplier approach for gradient flows. CMAME 367:113070, 2020.

Efficient and accurate structure preserving schemes for complex nonlinear systems. Handbook of Numerical Analysis, V. 20: Processing, Analyzing and Learning of Images, Shapes, and Forms, Part 2, edited by R. Kimmel and X.C. Tai, 647-669, Elsevier, 2019.

(with Yingwei Wang and Jianlin Xia) Fast structured Jacobi-Jacobi transforms. Math. Comp. 88:1743-1772, 2019.

(with Changtao Sheng) Spectral methods for fractional differential equations using generalized Jacobi functions. P. 127-156 in Handbook of Fractional Calculus with Applications, V3: Numerical Methods, edited by G. Karniadakis, De Gruyter, 2019.

(With Weizhu Bao, Xinran Ruan and Changtao Sheng) Fundamental Gaps of the Fractional Schrodinger Operator. Comm. Math. Sci. 17:447-471, 2019.

(with Qingcheng Yang, Arkadz Kirshtein, Yanzhou Ji, Chun Liu, Long-Qing Chen) A Thermodynamically Consistent Phase-Field Model For Viscous Sintering. J. American Ceramic Society, 102:674-685, 2019.