how to do algebraic fractions equations

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Algebraic Fractions

Here we will learn about algebraic fractions , including operations with fractions, and solving linear and quadratic equations written in the form of algebraic fractions.

There are also algebraic fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are algebraic fractions?

Algebraic fractions are fractions that contain at least one variable.

The following algebraic expressions are examples of algebraic fractions:

x is the numerator: \quad \quad \quad \frac{x}{12} Both the numerator and the denominator contain an x term: \quad \quad \quad \frac{x+1}{2x}

An expression in terms of x is the denominator: \quad \quad \frac{3}{x+1}\ Both the numerator and the denominator contain an expression with x : \quad \quad \frac{3x+4}{2x-5}

The numerator is a multiple of x : \quad\quad \quad \quad \frac{2x}{15} The numerator and the denominator are quadratic expressions: \quad\quad \quad \quad \frac{(x+3)^{2}}{x^{2}-9}

The main aim of this lesson is to understand how to solve equations that include algebraic fractions.

All the examples above are expressions whereas the examples below are equations as we can find specific values for x for each example to solve the equation.

One step equation: \quad \quad \frac{x}{12}=4 A separate constant term: \quad \frac{x+1}{2x}+4=x

A quadratic equation: \quad\frac{3}{x+1}=x+5 A linear equation: \quad \frac{3x+4}{2x-5}=6

A second fraction: \quad \frac{2x}{15}=\frac{5x}{2} Double brackets, difference of two squares and simultaneous equations: \quad \frac{(x+3)^{2}}{x^{2}-9}=2x-1

It is important to be able to simplify algebraic fractions into their simplest form. If you need to practice this or need a quick refresher, see the lesson on simplifying algebraic fractions for further information.

Step-by-step guide: Simplifying algebraic fractions

$What are algebraic fractions?$

How to solve equations including algebraic fractions

We need to be able to solve equations including algebraic fractions.

Let’s look at a simple example when \frac{8}{x}=2 .

Here, the denominator of the fraction contains the variable, so we first need to get the variable out of the denominator.

Rearranging the equation by multiplying both sides by x and then dividing by 2 , we get the value of x=4 .

We can substitute this into the original equation to prove that the answer is correct.

Here, \frac{8}{4}=2 so we have the correct answer.

We shall now consider more complicated cases when equations involve algebraic fractions.

In order to solve equations including algebraic fractions.

Convert each fraction so they all have a common denominator.

Explain how to solve equations including algebraic fractions

$Explain how to solve equations including algebraic fractions$

Algebraic fractions worksheet

Get your free algebraic fractions worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Algebraic fractions examples

Example 1: equation with one fraction.

Solve the equation

Here, we only have one fraction and so we do not need to convert any other term into a fraction.

2 Multiply the equation throughout by the common denominator .

Multiplying the equation throughout by 3 (the denominator of the fractional term), we get

Make sure that you multiply every term in the equation by 3 .

3 Solve the equation (linear or quadratic) .

You can check your solution by substituting the value for x into the original equation and evaluating it.

Example 2: Equation with two fractions

Here, we have the two fractions with the denominators of 2 and 5 . The lowest common multiple of 2 and 5 is 10 and so we can convert the two fractions so that they have the same denominator.

Remember to use brackets to ensure that you multiply the entire numerator by 5 .

Remember to use brackets to ensure that you multiply the entire numerator by 2 .

We now have the equation

Multiply the equation throughout by the common denominator .

Multiplying the equation throughout by 10 (the denominator of the fractional terms), we get

Solve the equation (linear or quadratic) .

Example 3: Equation with x in the denominator

Here, we have one fraction so we do not need to find a common denominator.

Multiplying the equation throughout by x + 1 (the denominator of the fractional terms), we get

Example 4: Equation with three fractions

Here, we need to find the lowest common multiple of x, 2x, and 3x . As x is the highest common factor, x \times 1 \times 2 \times 3 is the lowest common multiple, which is equal to 6x. .

So by multiplying the numerator and denominator of each fraction by a constant, we can convert each fraction to have the common denominator of 6x: .

We now have an equation which we can immediately simplify.

Multiplying the equation throughout by 6x (the denominator of the fractional terms), we get

Example 5: Denominators are expressions in terms of x

Here, we need to find a common denominator for (x + 2) and (x − 4) .

The easiest way to do this is to multiply the two expressions together.

Top tip: do not expand the brackets too soon as you may be able to simplify the fraction before solving the equation.

By multiplying each fraction by the denominator of the other fraction, we get

Multiplying the equation throughout by (x+2)(x-4) (the denominator of the fractional term), we get

Example 6: Equation including a quadratic

Here, we have a single fraction and so we do not need to find a common denominator.

Multiplying the equation throughout by x (the denominator of the fractional terms), we get

Here, we have two possible solutions for x so we can check both:

Common misconceptions

Multiplying the numerator by the denominator

Let us look at example 2 . When multiplying throughout by 10 to remove the denominator from each fraction, the numerator has also been multiplied by 10 . This means that the fractions have been multiplied by 100 , instead of 10 , leaving the next line of work to be incorrect.

Ignoring the denominators

When given an equation including an algebraic fraction, if the denominators are ignored the question will be answered incorrectly.

Not multiplying all terms by the denominator

When rearranging an equation, the denominator is moved to the other side of the equals sign, instead of each term being multiplied by it. You must remember to multiply throughout by any value, not just the opposite side of the equals sign.

Simplifying fractions incorrectly (1)

When adding two fractions, the denominator must be the same. A common misconception for adding two fractions is to add the numerators and the denominators together because this method is emphasised when looking at multiplying fractions.

Simplifying fractions incorrectly (2)

Seeing the same term on the numerator and denominator allow the misconception that they can both be cancelled.

Practice algebraic fractions questions

1. Solve the equation

$how to do algebraic fractions equations$

2. Solve the equation

3. Solve the equation

4. Solve the equation

5. Solve the equation

6. Solve the equation

Algebraic fractions GCSE questions

1. Arron earns £40 per hour. One day, he receives a bonus of £5 . He shares this day’s earnings equally with his brother and sister. If he gives away £190 , how many hours did Arron work that day?

2. Use the quadratic formula to solve the equation

2x(x+2)+5x(x+3)=4(x+3)(x+2) or equivalent

x=3 or x=-\frac{8}{3}

3. Two shapes given below have the same area. Calculate the value for x .

Area of Triangle = \frac{2(x+8)}{2}=x+8

Area of Square = \left(\frac{3}{\sqrt{x}}\right)^{2}=\frac{9}{x}

Conclusion: x=1 only

Learning checklist

You have now learned how to:

The next lessons are

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$how to do algebraic fractions equations$

Many techniques will simplify your work as you perform operations with algebraic fractions. As you review the examples, note the steps involved in each operation and any methods that will save you time.

Reducing algebraic fractions

To reduce an algebraic fraction to lowest terms, first factor the numerator and the denominator; then reduce , (or divide out) common factors.

$how to do algebraic fractions equations$

Warning: Do not reduce through an addition or subtraction sign as shown here.

$how to do algebraic fractions equations$

Multiplying algebraic fractions

To multiply algebraic fractions, first factor the numerators and denominators that are polynomials; then, reduce where possible. Multiply the remaining numerators together and denominators together. (If you've reduced properly, your answer will be in reduced form.)

$how to do algebraic fractions equations$

Dividing algebraic fractions

To divide algebraic fractions, invert the second fraction and multiply. Remember, you can reduce only after you invert.

$how to do algebraic fractions equations$

Adding or subtracting algebraic fractions

To add or subtract algebraic fractions having a common denominator, simply keep the denominator and combine (add or subtract) the numerators. Reduce if possible.

Perform the indicated operation.

$how to do algebraic fractions equations$

To add or subtract algebraic fractions having different denominators, first find a lowest common denominator (LCD), change each fraction to an equivalent fraction with the common denominator, and then combine each numerator. Reduce if possible.

$how to do algebraic fractions equations$

If there is a common variable factor with more than one exponent, use its greatest exponent.

$how to do algebraic fractions equations$

To find the lowest common denominator, it is often necessary to factor the denominators and proceed as follows.

$how to do algebraic fractions equations$

Occasionally, a problem will require reducing what appears to be the final result. A problem like that is found in the next example.

$how to do algebraic fractions equations$

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Solving Equations with Algebraic Fractions

Solving Equations with Algebraic Fractions

$how to do algebraic fractions equations$

Hello! Today we are going to take a look at how to solve equations that have algebraic fractions . But first, let’s review what algebraic fractions are. An algebraic fraction is any fraction that contains an algebraic expression. In other words, it’s a fraction that has a variable in it anywhere. For example,

\(\frac{5}{x}\), \(\frac{2x-9}{7}\), and \(\frac{3x}{2}\)

are all algebraic fractions.

So, now that we know what they are, let’s jump into a problem where we have to solve these. So we’re going to take a look at:

\(\frac{x}{8}=7\)

This is a nice and simple one. We know that in order to solve an equation, we need to isolate our variable. Here we can see that our variable, \(x\), is being divided by 8. So, to undo this division, we simply need to multiply both sides by 8.

\(8\cdot \frac{x}{8}=7\cdot 8\)

When we do that, our division is canceled out and we’re left with \(x\) on the left side, and \(7\times 8=56\), so:

Not too hard, right? Let’s try one that’s a little bit more complicated.

\(\frac{3x}{4}-\frac{2x}{12}=7\)

To solve this, I’m first going to simplify my fractions on the left side. To do this, I’ll need to convert \(\frac{3x}{4}\) to a fraction with a denominator of 12. I can do this by multiplying both the numerator and denominator by 3.

\(\frac{3x\cdot 3}{4\cdot 3}=\frac{9x}{12}\)

So that gives me the fraction, \(\frac{9x}{12}\). And then the rest of our equation is going to stay the same, so:

\(\frac{9x}{12}-\frac{2x}{12}=7\)

Since our denominators are the same, we can now subtract our numerators, so \(9x-2x=7x\).

\(\frac{7x}{12}=7\)

Another way we can write this is:

\(\frac{7}{12}\cdot x=7\)

So if we look at it this way, the way that we can get rid of this \(\frac{7}{12}\) over here is by multiplying by the reciprocal. So the reciprocal of \(\frac{7}{12}\), is \(\frac{12}{7}\), so we’ll multiply by that on both sides.

\(\frac{12}{7}\cdot \frac{7}{12}\cdot x=7\cdot \frac{12}{7}\)

When we do that, it cancels our fractions out over here, and we’re left with \(x\), which is what we want. And then if we do that over here, we can simplify first by doing \(7\div 7\), which gives us 1, and then do \(1\times 12\), which is 12.

So \(x=12\), which is our answer.

Let’s take a look at another problem.

\(\frac{4x}{5}=\frac{3x}{10}+9\)

This one we’re going to solve a little bit differently. For this problem, I’m going to solve for \(x\) by eliminating the fractions in our first step. To do this, we simply multiply the entire equation by the least common multiple of the denominators, of the fractions in the problem.

In this case, our denominators are 5 and 10, and the least common multiple of 5 and 10 is 10. So multiply the entire equation by 10.

\(10(\frac{4x}{5}=\frac{3x}{10}+9)\)

So that means we’re going to multiply each part of our equation by 10. So, \(10\cdot \frac{4x}{5}\). We can do \(10\div 5\) first, because remember it doesn’t matter if you multiply or divide first, multiplication and division can happen simultaneously. So for this, I’m going to divide \(10\div 5\) and get 2, and then multiply by \(4x\) to get \(8x\). If I multiply 10 by \(\frac{3x}{10}\), our 10s will cancel out and we’ll be left with \(3x\). And \(10\cdot 9=90\).

\(8x=3x+90\)

Now we can solve it like a normal algebra problem . So I’m going to subtract \(3x\) from both sides.

\(8x-3x=3x-3x+90\)

And then we divide both sides by 5.

\(\frac{5x}{5}=\frac{90}{5}\)

Either of these methods will help you get the right answer when solving equations with algebraic fractions, so feel free to use whichever one you feel most comfortable with.

Before we go, I want to show you one more problem you might come across.

\(\frac{x+17}{5}=21\)

So we have an expression in the numerator instead of just a variable, and maybe a number being multiplied by it, so it’s a little bit different than the ones we’ve looked at before. So, to solve problems like these, we first get rid of the fraction by multiplying by the denominator on both sides of the equation. So we’re going to multiply by 5 on both sides.

\(5\cdot \frac{x+17}{5}=21\cdot 5\)

That cancels out our denominator, so we no longer have a fraction, and we’re just left with the expression in the numerator: \(x+17\). And \(21\cdot 5=105\).

\(x+17=105\)

So now we’re going to solve it like normal. We’ll subtract 17 from both sides,

\(x+17-17=105-17\) \(x=88\)

And that’s it! I hope this video helped you better understand how to solve equations with algebraic fractions. Thanks for watching, and happy studying!

Practice Questions

Solve the equation: \(\frac{2x}{3}-\frac{4x}{9}=6\).

We can combine the fractions on the left side by converting \(\frac{2x}{3}\) to a fraction with a denominator of 9. To do so, multiply the first fraction by \(\frac{3}{3}\).

\(\frac{3}{3}\cdot\frac{2x}{3}-\frac{4x}{9}=6\) \(\frac{6x}{9}-\frac{4x}{9}=6\) \(\frac{2x}{9}=6\)

Now, multiply both sides by 9.

\(9\cdot\frac{2x}{9}=6\cdot9\) \(2x=54\)

Finally, divide both sides by 2.

\(\frac{2x}{2}=\frac{54}{2}\) \(x=27\)

Solve the equation: \(\frac{5x}{6}=\frac{11x}{15}+2\).

We can multiply both sides of the equation by the least common multiple of the denominators. The least common multiple of 6 and 15 is 30, so multiply the entire equation by 30

\(30\left(\frac{5}{6}x=\frac{11}{15}x+2\right)\) \(\frac{30\cdot5}{6}x=\frac{30\cdot11}{15}x+30\cdot2\) \(25x=22x+60\)

To get the variable terms on the left side, subtract 22x from both sides of the equation.

\(25x-22x=22x+60-22x\) \(3x=60\)

Now, divide both sides by 3.

\(\frac{3x}{3}=\frac{60}{3}\) \(x=20\)

Solve the equation: \(2x+79=5\).

We can get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 9.

\(9\cdot \frac{2x+7}{9}=5\cdot9\) \(2x+7=45\)

Now, subtract 7 from both sides.

\(2x+7-7=45-7\) \(2x=38\)

Then, divide both sides by 2.

\(\frac{2x}{2}=\frac{38}{2}\) \(x=19\)

The quotient of thirteen less than a number and 7 equals twice the number. What is the number?

Let x be the number you are trying to find. Since quotient means to divide, we can write the quotient of thirteen less than a number and 7 as \(\frac{x-13}{7}\). Twice the number can be written as \(2x\). Set these two expressions equal to each other to get the equation:

\(\frac{x-13}{7}=2x\)

To solve for x, first, get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 7.

\(7\cdot\frac{x-13}{7}=2x\cdot7\) \(x-13=14x\)

To get the variable terms on the left side, subtract \(14x\) from both sides of the equation.

\(x-13-14x=14x-14x\) \(-13x-13=0\)

Now, add the constant of 13 to both sides.

\(-13x-13+13=0+13\) \(-13x=13\)

Then, divide both sides by –13.

\(\frac{-13x}{-13}=\frac{13}{-13}\) \(x=-1\)

So, the number is –1.

You and a friend play on the same soccer team. Both of you decide to save money to buy a soccer goal to practice shooting. One-third of the money, in dollars, you and your friend have saved is $30 less than twice of what you have saved. If your friend has saved $25, how much money have you saved?

Let x be the amount of money you have saved. Since your friend has saved $25, we can write one-third of the money you and your friend have saved as \(\frac{x+25}{3}\). Thirty dollars less than twice of what you have saved can be written as \(2x-30\). Setting these two expressions equal to each other gives us the equation:

\(\frac{x+25}{3}=2x-30\)

To solve for x, first, get rid of the fraction by multiplying by the denominator on both sides of the equation. That is, multiply both sides of the equation by 3.

\(3\cdot\frac{x+25}{3}=3(2x-30)\) \(x+25=6x-90\)

To get the variable terms on the left side, subtract \(6x\) from both sides of the equation.

\(x+25-6x=6x-90-6x\) \(-5x+25=-90\)

Now, subtract 25 from both sides.

\(-5x+25-25=-90-25\) \(-5x=-115\)

Then, divide both sides by –5.

\(\frac{-5x}{-5}=\frac{115}{-5}\) \(x=23\)

So, you have saved $23.

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Solving Equations with Fractions

I know fractions are difficult, but with these easy step-by step instructions you'll be solving equations with fractions in no time.

Do you start to get nervous when you see fractions? Do you have to stop and review all the rules for adding, subtracting, multiplying and dividing fractions?

If so, you are just like almost every other math student out there! But... I am going to make your life so much easier when it comes to solving equations with fractions!

Our first step when solving these equations is to get rid of the fractions because they are not easy to work with!

Let see what happens with a typical two-step equation with the distributive property.

In this problem, we would typically distribute the 3/4 throughout the parenthesis and then solve. Let's see what happens:

Yuck! That just made this problem worse! Now we have two fractions to contend with and that means subtracting fractions and multiplying fractions.

So... let's stop here and say,

We DO NOT want to do this! DO NOT distribute fractions.

We are going to learn how to get rid of the fractions and make this much more simple!

So... what do we do? We are going to get rid of just the denominator in the fraction, so we will be left with the numerator, or just an integer!

I know, easier said than done! It's really not hard, but before I get into it, I want to go over one algebra definition.

We need to discuss the word term.

In Algebra, each term within an equation is separated by a plus (+) sign, minus (-) sign or an equals sign (=). Variable or quantities that are multiplied or divided are considered the same term.

Identifying terms within an equation

That last example is the most important to remember. If a quantity is in parentheses, it it considered one term!

Let's look at a few examples of how to solve these crazy looking problems!

Example 1 - Equations with Fractions

$Solving equations with fractions by eliminating the fractions$

Take a look at this example on video if you are feeling overwhelmed.

Hopefully you were able to follow that example. I know it's tough, but if you can get rid of the fraction, it will make these problems so much easier. Keep going, you'll get the hang of it!

In the next example, you will see two fractions. Since they have the same denominator, we will multiply by the denominator and get rid of both fractions.

Example 2 - Equations with Fractions with the Same Denominator

$how to do algebraic fractions equations$

Did you notice how multiplying by 2 (the denominator of both fractions) allowed us to get rid of the fractions? This is the best way to deal with equations that contain fractions.

In the next example, you will see what happens when you have 2 fractions that have different denominators.

We still want to get rid of the fractions all in one step. Therefore, we need to multiply all terms by the least common multiple. Remember how to find the LCM? If not, check out the LCM lesson here .

Example 3 - Equations with Two Fractions with Different Denominators

$Solving equations with two fractions that have different denominators$

Yes, the equations are getting harder, but if you take it step-by-step, you will arrive at the correct solution. Keep at it - I know you'll get it!

$how to do algebraic fractions equations$

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Course: Algebra 1 > Unit 2

Equation with variables on both sides: fractions

Equation with the variable in the denominator

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Video transcript

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Algebraic Fractions

Algebraic fractions are simply fractions with algebraic expressions on the top and/or bottom.

When adding or subtracting algebraic fractions, the first thing to do is to put them onto a common denominator (by cross multiplying).

The video below shows you how to calculate algebraic fractions.

e.g. 1 + 4 (x + 1) (x + 6)

= 1(x + 6) + 4(x + 1) (x + 1)(x + 6)

= x + 6 + 4x + 4 (x + 1)(x + 6)

= 5x + 10 (x + 1)(x + 6)

Solving equations

When solving equations containing algebraic fractions, first multiply both sides by a number/expression which removes the fractions.

Solve 10 - 2 = 1 (x + 3) x

multiply both sides by x(x + 3): ∴ 10x(x + 3) - 2x(x + 3) = x(x + 3) (x + 3) x

∴ 10x - 2(x + 3) = x 2 + 3x [after cancelling] ∴ 10x - 2x - 6 = x 2 + 3x ∴ x 2 - 5x + 6 = 0 ∴ (x - 3)(x - 2) = 0 ∴ either x = 3 or x = 2

$how to do algebraic fractions equations$

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MEDIAN Don Steward mathematics teaching: algebraic fractions
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Solving linear equations with two fractions
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Algebra
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Simplifying algebraic fractions
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VIDEO

Solving equations with fractions
Solving Equations with Fractions
Algebraic Fractions
Algebraic fraction
Equations with fractions (part 2)
solving equations

COMMENTS

Equations: Advanced Algebraic Fractions
Equations involving algebraic fractions, where there are variables (letters) on the denominators of the fractions.
08
This makes the equation much harder to solve because we must use the rules of fraction addition and fraction multiplication to isolate the
Solving Algebraic Fractions
Algebraic fractions are simply fractions with algebraic expressions either on the top, bottom or both. We treat them in the same way as we
Algebraic Fractions
How to solve equations including algebraic fractions · Convert each fraction so they all have a common denominator. · Multiply the equation throughout by the
Operations with Algebraic Fractions
To multiply algebraic fractions, first factor the numerators and denominators that are polynomials; then, reduce where possible. Multiply the remaining
Solving Equations with Algebraic Fractions (Video)
In this second video we look at an alternative method to solve linear equations with algebraic fractions in. If you are confident, this method is more
Solving Equations with Algebraic Fractions (PQ Video)
are all algebraic fractions. ... This is a nice and simple one. We know that in order to solve an equation, we need to isolate our variable. Here
Solving Equations With Fractions
Do you have to stop and review all the rules for adding, subtracting, multiplying and dividing fractions? If so, you are just like almost every other math
Equation with variables on both sides: fractions
The general rule for solving equations with fractions — whether it be only on one side or both — is to try to get rid of all of them. The most common way to
Algebraic Fractions
When adding or subtracting algebraic fractions, the first thing to do is to put them onto a common denominator (by cross multiplying). ... When solving equations