# Algebraic Fractions

Did you know that the word "algebra" originates from the Arabic word translated as "al-jabr" which literally means fixing parts of pieces of a whole such as fixing fractured bones in early times.

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Table of contents

Anyways, you won't be learning Arabic rather you would be understanding ways to simplify, multiply, add and fractorise algebraic fractions herein.

## Algebraic fractions definition

Before defining the algebraic fractions, we recall first the definition of the algebraic expressions.

Algebraic expressions are expressions containing variables and constants.

$a+b$, $2x-1$, $\frac{y}{5}$, $uv-3b$ are some examples of algebraic expressions.

We are now ready to define algebraic fractions.

Algebraic fractions are fractions whose numerator and denominator are algebraic expressions.

In other words, numerical fractions are of the form,

$\frac{2}{5},\frac{5}{6},\frac{1}{3}$

Meanwhile, rather than just numbers in either the numerator or denominator, algebraic expressions are present in the numerator and/or the denominator. And hence, algebraic fractions are of the form,

$\frac{2x}{7},\frac{3}{2y},\frac{x+1}{2x},\frac{{x}^{2}+2}{3y-4}$.

## Simplifying algebraic fractions

Simplifying algebraic fractions is reducing algebraic fractions to the smallest terms. Here our ability to divide accurately is crucial. We apply similar principles used in simplifying numerical fractions in carrying the simplification of algebraic fractions.

And this can be done while following one of the two following methods,

• Dividing by the greatest common divisor (GCD).
• Dividing with simpler common factors continuously.

We shall explain how the first method is carried out briefly in subsequent examples, however, we start first with an example of how to find GCDs between algebraic pairs.

Find the GCD between the algebraic pair 24x2y6 and 6x3y4

Solution

Here, you wish to find the highest expression which is a factor of both algebraic expressions. The best approach is to split them into their components separately.

For 24x2y6, you have 24, x2 and y6, and for 6x3y4, you have 6, x3, and y4.

Now compare similar components and find their GCD in pairs.

24 and 6: The GCD between 24 and 6 is 6.

x2 and x3: When dealing with exponents, the expression with a lower power is the GCD, this means that x2 is the GCD.

y6 and y4: When dealing with exponents, the expression with a lower power is the GCD, this means that y4 is the GCD.

The next step is to multiply all your GCDs from the comparisons, to get

$6×{x}^{2}×{y}^{4}=6{x}^{2}{y}^{4}$

Hence the GCD between the algebraic pair: 24x2y6 and 6x3y4 is 6x2y4

Now you know how to find the GCD among algebraic pairs, you should apply it in the simplification of algebraic fractions henceforth.

Simplify the following,

$\frac{15{t}^{3}{s}^{4}}{20t{s}^{5}}$

Solution

Step 1.

Find the GCD between the numerator and denominator. Applying the same reasoning explained in the previous example, the GCD between 15t3s4 and 20ts5 is 5ts4.

Step 2.

Divide the numerator and denominator by the GCD, to get

$\frac{15{t}^{3}{s}^{4}÷5t{s}^{4}}{20t{s}^{5}÷5t{s}^{4}}=\frac{\left(15÷5\right)\left({t}^{3}÷t\right)\left({s}^{4}÷{s}^{4}\right)}{\left(20÷5\right)\left(t÷t\right)\left({s}^{5}÷{s}^{4}\right)}=\frac{3{t}^{2}}{4s}$

Now, let's consider using the second method which involves continuous division. In order to make this method easier, you may choose to express the algebraic expression as a product of its factors.

Simplify the following algebraic fractions

a. $\frac{56{x}^{3}{y}^{2}}{42{x}^{2}{y}^{3}}$

b. $\frac{10{a}^{4}{b}^{2}}{2ab}$

Solution

a.

$\frac{56{x}^{3}{y}^{2}}{42{x}^{2}{y}^{3}}$

We divide directly through or we re-express the expression as a product of its factors,

$\frac{56{x}^{3}{y}^{2}}{42{x}^{2}{y}^{3}}=\frac{2×2×2×7×x×x×x×y×y}{2×3×7×x×x×y×y×y}\phantom{\rule{0ex}{0ex}}=\frac{\overline{)2}×2×2×\overline{)7}×\overline{)x}×\overline{)x}×x×\overline{)y}×\overline{)y}}{\overline{)2}×3×\overline{)7}×\overline{)x}×\overline{)x}×\overline{)y}×\overline{)y}×y}\phantom{\rule{0ex}{0ex}}=\frac{2×2×x}{3×y}\phantom{\rule{0ex}{0ex}}=\frac{4x}{3y}$

b.

$\frac{10{a}^{4}{b}^{2}}{2ab}$

We divide through to get,

$\frac{10{a}^{4}{b}^{2}}{2ab}=\frac{\overline{){10}^{5}×}{a}^{4}{b}^{2}}{\overline{){2}^{1}×}ab}=\frac{5{a}^{4}{b}^{2}}{ab}=\frac{5\overline{){a}^{4}}{b}^{2}}{\overline{)a}b}=\frac{5{a}^{3}{b}^{2}}{b}=\frac{5{a}^{3}\overline{){b}^{2}}}{\overline{)b}}=5{a}^{3}b$

## Factorising algebraic fractions

The knowledge of factorization is essential in solving problems related to algebraic fractions. As a matter of fact, let us see how much you remember factorization in the example below.

Factorize $4x-2xy$

Solution

When factorizing, express each algebraic expression as a product of its factors (prime factors for numbers).

Afterward, we compare and bring out common factors as seen in the illustration below,

Figure 1: describing how common factors are derived from algebraic fraction, StudySmarter Originals

Our factor is 2x. Taking 2x as a common factor from the expression 4x, we are left with 2. In fact,

$\frac{\overline{){}^{2}4}\overline{)x}}{\overline{)2}\overline{)x}}=2$

And taking 2x as a common factor from the expression -2xy, we are left with -y. In fact,

$\frac{-\overline{)2x}y}{\overline{)2x}}=-y$

Thus when $4x-2xy$ is factorized we get,

$2x\left(2-y\right)$.

Factorize $2{y}^{2}+4x-2xy-4y$

Solution

Step 1.

Bring like terms together to get,

$4x-2xy-4y+2{y}^{2}$

Step 2.

Use brackets to separate similar terms for ease in factorization, to get

$\left(4x-2xy\right)+\left(-4y+2{y}^{2}\right)$

Step 3.

Bring out common factors from what you have in the brackets.

The common factor between 4x and -2xy is 2x, thus we have

$4x-2xy=2x\left(2-y\right)$

The common factor between $-4y$and $2{y}^{2}$is 2y, thus we have

$-4y+2{y}^{2}=2y\left(-2+y\right)$

Thus, the algebraic expression $2{y}^{2}+4x-2xy-4y$ can be rewritten as,

$2{y}^{2}+4x-2xy-4y=2x\left(2-y\right)+2y\left(y-2\right)$

Note that the factorization is not complete yet. The expressions (2-y) and (y-2) differ by the minus sign. Thus, rewriting we get,

$2{y}^{2}+4x-2xy-4y=2x\left(2-y\right)-2y\left(2-y\right)$

Now we have the same expression in the brackets, we take (2-y) as a common factor between these expressions to get,

$2{y}^{2}+4x-2xy-4y=2x\left(2-y\right)-2y\left(2-y\right)=\left(2-y\right)\left(2x-2y\right)$

We notice that factor (2x-2y) has a common factor of 2, thus we have

$2x-2y=2\left(x-y\right)$

And hence, the complete factorized form for the initial algebraic expression is

$2{y}^{2}+4x-2xy-4y=\left(2-y\right)2\left(x-y\right)=2\left(2-y\right)\left(x-y\right)$

In order to verify that the factorized form is correct, we expand it and we wee that we obtain the same initial algebraic expression.

We are now ready to deepen with the following examples.

Simplify the following,

a. $\frac{x+1}{{x}^{2}-1}$

b. $\frac{15y+12}{25{y}^{2}-16}$

c. $\frac{{p}^{2}-4}{{p}^{2}-4p+4}$

Solution

a.$\frac{x+1}{{x}^{2}-1}$

Step 1.

We first notice that the denominator can be factorized to,

${x}^{2}-1=\left(x-1\right)\left(x+1\right)$

Step 2.

Now, we substitute the factorized expression into the main expression to get,

$\frac{x+1}{{x}^{2}-1}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}=\frac{{\overline{)x+1}}^{1}}{\left(x-1\right)\overline{){\left(x+1\right)}^{1}}}=\frac{1}{x-1}$

while ensuring the $x+1\ne 0$ that is $x\ne -1$.b. $\frac{15y+12}{25{y}^{2}-16}$

Step 1.

Both the numerator and denominator of the algebraic fraction can be factorized to give,

$15y+12=3\left(5y+4\right)\phantom{\rule{0ex}{0ex}}25{y}^{2}-16=\left(5y-4\right)\left(5y+4\right)$

Step 2.

Now, substitute the factorized expressions into the algebraic fraction to get,

$\frac{15y+12}{25{y}^{2}-16}=\frac{3\left(5y+4\right)}{\left(5y-4\right)\left(5y+4\right)}=\frac{3\left(\overline{)5y+4{\right)}^{1}}}{\left(5y-4\right){\overline{)\left(5y+4\right)}}^{1}}=\frac{3}{5y-4}$

while ensuring that $5y+4\ne 0$ that is $y\ne \frac{-4}{5}$.

c.$\frac{{p}^{2}-4}{{p}^{2}-4p+4}$

Step 1.

Both the numerator and denominator of the algebraic fraction can be factorized to give,

${p}^{2}-4=\left(p-2\right)\left(p+2\right)\phantom{\rule{0ex}{0ex}}{p}^{2}-4p+4=\left(p-2\right)\left(p-2\right)$

Step 2.

Now, substitute the factorized expressions into the algebraic fraction to get,

$\frac{{p}^{2}-4}{{p}^{2}-4p+4}=\frac{{\overline{)\left(p-2\right)}}^{1}\left(p+2\right)}{{\overline{)\left(p-2\right)}}^{1}\left(p-2\right)}=\frac{p+2}{p-2}$while ensuring that $p-2\ne 0$that is $p\ne 2$.

## Adding algebraic fractions

Just like numerical fractions, algebraic fractions can be added by having to find the lowest common denominator (LCD) between algebraic fractions under such operations.

Add the following algebraic fractions,

a. $\frac{1}{y}+\frac{1}{{y}^{2}}$

b. $\frac{2}{x+1}+\frac{3}{x-2}$

c. $\frac{1}{{p}^{2}-5p+6}+\frac{p}{p-3}$

Solution

a.$\frac{1}{y}+\frac{1}{{y}^{2}}$

Step 1.

We find the lowest common denominator (LCD) between the two denominators. The LCD between y and y2 is y2.

Next, we multiply the numerator and the denominator of $\frac{1}{y}$ by the LCD ${y}^{2}$, to get,

$\frac{1}{y}+\frac{1}{{y}^{2}}=\left(\frac{{y}^{2}}{{y}^{2}}×\frac{1}{y}\right)+\frac{1}{{y}^{2}}=\frac{y}{{y}^{2}}+\frac{1}{{y}^{2}}$

Step 2.

Now we have two fractions with the same denominator, we add only the numerators, to get,

$\frac{y}{{y}^{2}}+\frac{1}{{y}^{2}}=\frac{y+1}{{y}^{2}}$

b. $\frac{2}{x+1}+\frac{3}{x-2}$

Step 1.

We find the lowest common denominator (LCD) between the denominators. The LCD between x+1 and x-2 is their product (x+1)(x-2).

Step 2.

Next, we multiply the numerator and the denominator of the first fraction by (x-2), and the numerator and the denominator of the second fraction by (x+1) to get,

$\left(\frac{\left(x-2\right)}{\left(x-2\right)}×\frac{2}{x+1}\right)+\left(\frac{3}{x-2}×\frac{\left(x+1\right)}{\left(x+1\right)}\right)=\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}$

Since the two obtained fractions have the same denominator, we add directly the numerators while retaining the denominator, to get

$\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{2x-4}{\left(x+1\right)\left(x-2\right)}+\frac{3x+3}{\left(x+1\right)\left(x-2\right)}=\frac{2x-4+3x+3}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{2x+3x-4+3}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{5x-1}{\left(x+1\right)\left(x-2\right)}$

c.$\frac{1}{{p}^{2}-5p+6}+\frac{p}{p-3}$

Step 1.

Find the lowest common denominator (LCD). The LCD between p2-5p+6 and p-3 is p2-5p+6 because when p2-5p+6 is factorized as (p-2)(p-3). Replace p2-5p+6 with (p-2)(p-3) for ease in calculation. Now you have your LCD, multiply the numerator and denominator of the second fraction by $\left(p-2\right)$

$\frac{1}{\left(p-2\right)\left(p-3\right)}+\frac{p}{p-3}×\frac{\left(p-2\right)}{\left(p-2\right)}=\frac{1}{\left(p-2\right)\left(p-3\right)}+\frac{p\left(p-2\right)}{\left(p-2\right)\left(p-3\right)}$

Since they both have the same denominator, you can add directly while retaining the denominator. Therefore;

$\frac{1}{\left(p-2\right)\left(p-3\right)}+\frac{p\left(p-2\right)}{\left(p-2\right)\left(p-3\right)}=\frac{1}{\left(p-2\right)\left(p-3\right)}+\frac{{p}^{2}-2p}{\left(p-2\right)\left(p-3\right)}=\frac{{p}^{2}-2p+1}{\left(p-2\right)\left(p-3\right)}\phantom{\rule{0ex}{0ex}}$

Note that p2-2p+1 when factorized would give (p-1)(p-1). Thus when substituted into the expression gives,

$\frac{{p}^{2}-2p+1}{\left(p-2\right)\left(p-3\right)}=\frac{\left(p-1\right)\left(p-1\right)}{\left(p-2\right)\left(p-3\right)}=\frac{{\left(p-1\right)}^{2}}{\left(p-2\right)\left(p-3\right)}$

### Subtracting algebraic fractions

Just as algebraic fractions can be added, they can also be subtracted. The subtraction of algebraic fractions is quite similar to the addition, the difference is in the application of the minus (-) sign.

Simplify the following,

a. $\frac{1}{y}-\frac{1}{{y}^{2}}$

b. $\frac{2}{x+1}-\frac{3}{x-2}$

c. $\frac{1}{{p}^{2}-5p+6}-\frac{p}{p-3}$

Solution

We refer to the pervious example for the calculation of the LCD between the fractions in question.

a. $\frac{1}{y}-\frac{1}{{y}^{2}}$

Step 1.

Find the lowest common denominator (LCD). The LCD between y and y2 is y2.

Step 2.

Multiply the numerator and the denominator of the first fraction by the y to get,

$\frac{1}{y}-\frac{1}{{y}^{2}}=\frac{y}{y}×\frac{1}{y}-\frac{1}{{y}^{2}}=\frac{y-1}{{y}^{2}}$

b.$\frac{2}{x+1}-\frac{3}{x-2}$

Step 1.

The LCD between the two denominators is (x+1)(x-2).

Step 2.

Multiplying the numerator and the denominator of the first fraction by (x-2), and the numerator and the denominator of the second fraction by (x+1), we have

$\left(\frac{\left(x-2\right)}{\left(x-2\right)}×\frac{2}{x+1}\right)-\left(\frac{3}{x-2}×\frac{\left(x+1\right)}{\left(x+1\right)}\right)=\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}$

Step 3.

Since the two fractions have the same denominator, we subtract directly while retaining the denominator, to get

$\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{2x-4}{\left(x+1\right)\left(x-2\right)}-\frac{3x+3}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{2x-4-\left(3x+3\right)}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{2x-4-3x-3\right)}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{2x-3x-4-3}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{-x-7}{\left(x+1\right)\left(x-2\right)}\phantom{\rule{0ex}{0ex}}$

c.$\frac{1}{{p}^{2}-5p+6}-\frac{p}{p-3}$

Step 1.

The LCD between the denominators of the two fractions is ${p}^{2}-5p+6$.

Step 2.

We multiply the numerator and the denominator of the second fraction by (p-2) to get

$\frac{1}{{p}^{2}-5p+6}-\frac{p}{p-3}=\frac{1}{\left(p-2\right)\left(p-3\right)}-\frac{p\left(p-2\right)}{\left(p-2\right)\left(p-3\right)}$

Since the two fractions have the same denominator, we subtract directly while retaining the denominator, to get

$\frac{1}{\left(p-2\right)\left(p-3\right)}-\frac{p\left(p-2\right)}{\left(p-2\right)\left(p-3\right)}=\frac{1}{\left(p-2\right)\left(p-3\right)}-\frac{{p}^{2}-2p}{\left(p-2\right)\left(p-3\right)}\phantom{\rule{0ex}{0ex}}=\frac{1-\left({p}^{2}-2p\right)}{\left(p-2\right)\left(p-3\right)}\phantom{\rule{0ex}{0ex}}=\frac{1+2p-{p}^{2}}{\left(p-2\right)\left(p-3\right)}\phantom{\rule{0ex}{0ex}}$

## Multiplying algebraic fractions

The product of algebraic fractions can also be calculated as other numerical fractions. We multiply the numerators together and the denominators together.

Multiply the following,

$\frac{3}{y}×\frac{2x}{7}$

Solution

Multiply through to get

$\frac{3}{y}×\frac{2x}{7}=\frac{3×2x}{y×7}=\frac{6x}{7y}$

Multiply the following,

a. $\frac{4p}{3qr}×\frac{5rk}{6{q}^{2}p}$

b. $\frac{x-y}{z}×\frac{{z}^{2}}{{x}^{2}-{y}^{2}}$

c. $\frac{{p}^{2}-1}{{p}^{2}-5p+6}×\frac{p-3}{p-1}$

Solution

a.$\frac{4p}{3qr}×\frac{5rk}{6{q}^{2}p}$

We can either multiply both the numerators and denominators before eventually dividing the common factors directly, thus we have

$\frac{4p}{3qr}×\frac{5rk}{6{q}^{2}p}=\frac{2×2×p}{3×q×r}×\frac{5×r×k}{2××3×q×q×p}=\frac{\overline{)2}×2×\overline{)p}}{3×q×\overline{)r}}×\frac{5×\overline{)r}×k}{\overline{)2}××3×q×q×\overline{)p}}\phantom{\rule{0ex}{0ex}}=\frac{2×5×k}{3×3×q×q×q}=\frac{10k}{9{q}^{3}}$

b. $\frac{x-y}{z}×\frac{{z}^{2}}{{x}^{2}-{y}^{2}}$

We factorize the denominator of the second fraction to get

${x}^{2}-{y}^{2}=\left(x-y\right)\left(x+y\right)$

Now we substitute the factorized expression into the main expression to get,

$\frac{x-y}{z}×\frac{{z}^{2}}{{x}^{2}-{y}^{2}}=\frac{x-y}{z}×\frac{z×z}{\left(x-y\right)\left(x+y\right)}=\frac{\overline{)x-y}}{\overline{)z}}×\frac{\overline{)z}×z}{\overline{)\left(x-y\right)}\left(x+y\right)}=\frac{z}{x+y}$

c.$\frac{{p}^{2}-1}{{p}^{2}-5p+6}×\frac{p-3}{p-1}$

We ensure to factorize algebraic expressions to ease the multiplication,

${p}^{2}-1=\left(p-1\right)\left(p+1\right)\phantom{\rule{0ex}{0ex}}{p}^{2}-5p+6=\left(p-2\right)\left(p-3\right)$

Next, we substitute the factorized expressions into the main expression to get,

$\frac{{p}^{2}-1}{{p}^{2}-5p+6}=\frac{\left(p-1\right)\left(p+1\right)}{\left(p-2\right)\left(p-3\right)}×\frac{p-3}{p-1}=\frac{\overline{)\left(p-1\right)}\left(p+1\right)}{\left(p-2\right)\overline{)\left(p-3\right)}}×\frac{\overline{)p-3}}{\overline{)p-1}}=\frac{p+1}{p-2}$

### Dividing algebraic fractions

The quotient of two algebraic fractions can be calculated by multiplying the first fraction by the reciprocal of the second fraction.

That is,

Simplify

$\frac{3y}{x}÷\frac{5u}{w}$

Solution

$\frac{3y}{x}÷\frac{5u}{w}$

We write the reciprocal of the second algebraic fraction and change the division sign to multiplication sign to get

$\frac{3y}{x}÷\frac{5u}{w}=\frac{3y}{x}×\frac{w}{5u}=\frac{3y×w}{x×5u}=\frac{3wy}{5ux}$

Simplify the following,

a. $\frac{r{t}^{2}}{15}÷\frac{{r}^{2}}{5t}$

b. $\frac{{x}^{2}-9}{{x}^{2}+3x+2}÷\frac{{x}^{2}+6x+9}{{x}^{2}+8x+7}$

Solution

a.$\frac{r{t}^{2}}{15}÷\frac{{r}^{2}}{5t}$

We rearrange the expression by changing the division to multiplication sign and using the reciprocal of the algebraic fraction at the right hand, to get

$\frac{r{t}^{2}}{15}÷\frac{{r}^{2}}{5t}=\frac{r{t}^{2}}{15}×\frac{5t}{{r}^{2}}=\frac{r×t×t}{5×3}×\frac{5×t}{r×r}=\frac{\overline{)r}×t×t}{\overline{)5}×3}×\frac{\overline{)5}×t}{\overline{)r}×r}=\frac{{t}^{3}}{3r}$

b.$\frac{{x}^{2}-9}{{x}^{2}+3x+2}÷\frac{{x}^{2}+6x+9}{{x}^{2}+8x+7}$

We rearrange first the expression by changing the division to multiplication sign and using the reciprocal of the algebraic fraction at the right hand to get

$\frac{{x}^{2}-9}{{x}^{2}+3x+2}×\frac{{x}^{2}+8x+7}{{x}^{2}+6x+9}$

Next, factorize the quadratic expressions

${x}^{2}-9=\left(x-3\right)\left(x+3\right)\phantom{\rule{0ex}{0ex}}{x}^{2}+3x+2=\left(x+2\right)\left(x+1\right)\phantom{\rule{0ex}{0ex}}{x}^{2}+8x+7=\left(x+7\right)\left(x+1\right)\phantom{\rule{0ex}{0ex}}{x}^{2}+6x+9=\left(x+3\right)\left(x+3\right)$

Now we substitute the factorized expressions into the algebraic fractions and simplify to get

$\frac{\left(x-3\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}×\frac{\left(x+7\right)\left(x+1\right)}{\left(x+3\right)\left(x+3\right)}=\frac{\left(x-3\right)\overline{)\left(x+3\right)}}{\left(x+2\right)\overline{)\left(x+1\right)}}×\frac{\left(x+7\right)\overline{)\left(x+1\right)}}{\overline{)\left(x+3\right)}\left(x+3\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(x-3\right)\left(x+7\right)}{\left(x+2\right)\left(x+3\right)}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}+4x-21}{{x}^{2}+5x+6}$

While ensuring that $x+1\ne 0$ that is $x\ne -1$and $x+3\ne 0$ that is $x\ne -3$.

## Algebraic fractions examples

Although we have seen several examples on algebraic fractions, you shall see further applications of algebraic fractions in word problems hereafter.

When a certain number is reduced by 8, it is equivalent to the sum of one-third the number and half of it. Find the number.

Solution

Let us call the unknown number w. Now we can create an equation in terms of w.

The first part says the number is reduced by 8, it means,

$w-8$

This second part says the sum of one-third the number and half of it, this means

$\frac{w}{3}+\frac{w}{2}$

Now we are told the first part is equivalent to the second part. Thus, we have

$w-8=\frac{w}{3}+\frac{w}{2}$

Next, we find the Lowest Common Denominator (LCD) and multiply the whole equation by it. The LCD between 3 and 2 is 6. Therefore, we have

$6\left(w-8\right)=6\left(\frac{w}{3}\right)+6\left(\frac{w}{2}\right)\phantom{\rule{0ex}{0ex}}6w-48=2w+3w\phantom{\rule{0ex}{0ex}}6w-48=5w$

We bring like terms together and solve further,

$6w-48=5w\phantom{\rule{0ex}{0ex}}6w-5w=48\phantom{\rule{0ex}{0ex}}w=48$

When the square of a positive number is added to $\frac{3}{4}$, the result is 1. Find the integer.

Solution

Let us call the unknown integer z.

We are told the denominator is added to the square of the integer. Thus,

$\frac{3}{4}+{z}^{2}$

Next, we are told that when this is done, our result is 1. Therefore,

$\frac{3}{4}+{z}^{2}=1$

${z}^{2}=1-\frac{3}{4}=\frac{4}{4}-\frac{3}{4}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}$

Thus, $z=±\frac{1}{2}$

Recall that the question specifies that it is a positive number, thus $z=\frac{1}{2}$.

## Algebraic fractions - Key takeaways

• Algebraic fractions are fractions which bear algebraic expressions.
• Simplifying algebraic fractions means reducing algebraic fractions to the smallest terms.
• The knowledge of factorization is essential in solving problems related with algebraic fractions.
• Algebraic fractions can be added as well as subtracted.
• Algebraic fractions can be multiplied and divided.

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##### Frequently Asked Questions about Algebraic Fractions

What is an example of an algebraic fraction?

An example of an algebraic fraction appears in the form a/3, (3x+1)/2, 3/2y etc.

How do you solve algebraic fractions step by step?

In order to solve algebraic fractions step by step, you need to find the lowest common denominator and be able to simplify algebraic expressions.

What is the meaning of algebraic fraction?

Algebraic fractions are fractions which bear algebraic expressions.

How do you solve algebraic fractions with variables?

In order to solve algebraic fractions with variables, you must be able to factorize and simplify the common factors.

How to add algebraic fractions?

Algebraic fractions are added like normal fractions, you just need to know the lowest common denominator before summing.

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