Algebraic Fractions

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

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,

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 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 $x\ne -3$.