Algebraic Fractions

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

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

$6\times x^2\times y^4=6x^2{y}^4$

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

Simplify the following,

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

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

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

Simplify the following algebraic fractions

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

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

Solution

a.

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

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

$$\begin{gathered} \frac{56x^3{y}^2}{42x^2{y}^3}=\frac{2\times 2\times 2\times 7\times x\times x\times x\times y\times y}{2\times 3\times 7\times x\times x\times y\times y\times y} \\ =\frac{\cancel{2}\times 2\times 2\times \cancel{7}\times \cancel{x}\times \cancel{x}\times x\times \cancel{y}\times \cancel{y}}{\cancel{2}\times 3\times \cancel{7}\times \cancel{x}\times \cancel{x}\times \cancel{y}\times \cancel{y}\times y} \\ =\frac{2\times 2\times x}{3\times y} \\ =\frac{4x}{3y} \end{gathered}$$

b.

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

We divide through to get,

$\frac{10a^4{b}^2}{2ab}=\frac{\cancel{{10}^5\times }{a}^4{b}^2}{\cancel{2^1\times }ab}=\frac{5a^4{b}^2}{ab}=\frac{5\cancel{a^4}{b}^2}{\cancel{a}b}=\frac{5a^3{b}^2}{b}=\frac{5a^3\cancel{b^2}}{\cancel{b}}=5a^{3}b$

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,

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

$\frac{\cancel{{}^{2}4}\cancel{x}}{\cancel{2}\cancel{x}}=2$

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

$\frac{-\cancel{2x}y}{\cancel{2x}}=-y$

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

$2x(2-y)$.

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

Solution

Step 1.

Bring like terms together to get,

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

Step 2.

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

$(4x-2xy)+(-4y+2y^2)$

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

$4x-2xy=2x(2-y)$

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

$-4y+2y^2=2y(-2+y)$

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

$2y^2+4x-2xy-4y=2x(2-y)+2y(y-2)$

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

$2y^2+4x-2xy-4y=2x(2-y)-2y(2-y)$

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

$2y^2+4x-2xy-4y=2x(2-y)-2y(2-y)=(2-y)(2x-2y)$

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

$2x-2y=2(x-y)$

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

$2y^2+4x-2xy-4y=(2-y)2(x-y)=2(2-y)(x-y)$

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

Simplify the following,

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

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

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,

$x^2-1=(x-1)(x+1)$

Step 2.

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

$\frac{x+1}{x^2-1}=\frac{x+1}{(x-1)(x+1)}=\frac{{\cancel{x+1}}^1}{(x-1)\cancel{{(x+1)}^1}}=\frac{1}{x-1}$

Step 1.

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

$$\begin{gathered} 15y+12=3(5y+4) \\ 25y^2-16=(5y-4)(5y+4) \end{gathered}$$

Step 2.

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

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

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,

$$\begin{gathered} p^2-4=(p-2)(p+2) \\ {p}^2-4p+4=(p-2)(p-2) \end{gathered}$$

Step 2.

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

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

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,

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

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,

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

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

$$\begin{gathered} \frac{2(x-2)}{(x+1)(x-2)}+\frac{3(x+1)}{(x+1)(x-2)}=\frac{2x-4}{(x+1)(x-2)}+\frac{3x+3}{(x+1)(x-2)}=\frac{2x-4+3x+3}{(x+1)(x-2)} \\ =\frac{2x+3x-4+3}{(x+1)(x-2)} \\ =\frac{5x-1}{(x+1)(x-2)} \end{gathered}$$

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 $(p-2)$

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

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

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

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

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

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,

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

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

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

Step 3.

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

$$\begin{gathered} \frac{2(x-2)}{(x+1)(x-2)}-\frac{3(x+1)}{(x+1)(x-2)}=\frac{2x-4}{(x+1)(x-2)}-\frac{3x+3}{(x+1)(x-2)} \\ =\frac{2x-4-(3x+3)}{(x+1)(x-2)} \\ =\frac{2x-4-3x-3)}{(x+1)(x-2)} \\ =\frac{2x-3x-4-3}{(x+1)(x-2)} \\ =\frac{-x-7}{(x+1)(x-2)} \end{gathered}$$

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

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

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

$$\begin{gathered} \frac{1}{(p-2)(p-3)}-\frac{p(p-2)}{(p-2)(p-3)}=\frac{1}{(p-2)(p-3)}-\frac{p^2-2p}{(p-2)(p-3)} \\ =\frac{1-(p^2-2p)}{(p-2)(p-3)} \\ =\frac{1+2p-p^2}{(p-2)(p-3)} \end{gathered}$$

Multiply the following,

$\frac{3}{y}\times \frac{2x}{7}$

Solution

Multiply through to get

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

Multiply the following,

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

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

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

Solution

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

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

$$\begin{gathered} \frac{4p}{3qr}\times \frac{5rk}{6q^{2}p}=\frac{2\times 2\times p}{3\times q\times r}\times \frac{5\times r\times k}{2\times \times 3\times q\times q\times p}=\frac{\cancel{2}\times 2\times \cancel{p}}{3\times q\times \cancel{r}}\times \frac{5\times \cancel{r}\times k}{\cancel{2}\times \times 3\times q\times q\times \cancel{p}} \\ =\frac{2\times 5\times k}{3\times 3\times q\times q\times q}=\frac{10k}{9q^3} \end{gathered}$$

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

We factorize the denominator of the second fraction to get

$x^2-y^2=(x-y)(x+y)$

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

$\frac{x-y}{z}\times \frac{z^2}{x^2-y^2}=\frac{x-y}{z}\times \frac{z\times z}{(x-y)(x+y)}=\frac{\cancel{x-y}}{\cancel{z}}\times \frac{\cancel{z}\times z}{\cancel{(x-y)}(x+y)}=\frac{z}{x+y}$

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

We ensure to factorize algebraic expressions to ease the multiplication,

$$\begin{gathered} p^2-1=(p-1)(p+1) \\ {p}^2-5p+6=(p-2)(p-3) \end{gathered}$$

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

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

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

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

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

$\frac{r{t}^2}{15}÷\frac{r^2}{5t}=\frac{r{t}^2}{15}\times \frac{5t}{r^2}=\frac{r\times t\times t}{5\times 3}\times \frac{5\times t}{r\times r}=\frac{\cancel{r}\times t\times t}{\cancel{5}\times 3}\times \frac{\cancel{5}\times t}{\cancel{r}\times r}=\frac{t^3}{3r}$

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

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

Next, factorize the quadratic expressions

$$\begin{gathered} x^2-9=(x-3)(x+3) \\ {x}^2+3x+2=(x+2)(x+1) \\ {x}^2+8x+7=(x+7)(x+1) \\ {x}^2+6x+9=(x+3)(x+3) \end{gathered}$$

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

$$\begin{gathered} \frac{(x-3)(x+3)}{(x+2)(x+1)}\times \frac{(x+7)(x+1)}{(x+3)(x+3)}=\frac{(x-3)\cancel{(x+3)}}{(x+2)\cancel{(x+1)}}\times \frac{(x+7)\cancel{(x+1)}}{\cancel{(x+3)}(x+3)} \\ =\frac{(x-3)(x+7)}{(x+2)(x+3)} \\ =\frac{x^2+4x-21}{x^2+5x+6} \end{gathered}$$

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

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

$$\begin{gathered} 6(w-8)=6\left(\frac{w}{3}\right)+6\left(\frac{w}{2}\right) \\ 6w-48=2w+3w \\ 6w-48=5w \end{gathered}$$

We bring like terms together and solve further,

$$\begin{gathered} 6w-48=5w \\ 6w-5w=48 \\ w=48 \end{gathered}$$

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,

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

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

Thus, $z=\pm \frac{1}{2}$

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