18. Algebraic Fractions

Question 1

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{35 x^2 y^4}{7x^3 y}

Select the correct answer from the list below.

A: \dfrac{5y^5}{x}

B: \dfrac{7y^3}{x}

C: 5xy^3

D: \dfrac{5y^3}{x}

CORRECT ANSWER:   D: \dfrac{5y^3}{x}

WORKED SOLUTION:

We see that top and bottom have common factors of x^2 and y, so they can cancel. Also \dfrac{35}{7} = 5. This leaves us with

\dfrac{5y^3}{x}

This can be no further simplified, and so we are done.

Question 2

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{x^2 + 3x}{4x + 12}

Select the correct answer from the list below.

A: 4x

B: x + 3

C: \dfrac{x + 3}{4}

D: \dfrac{x}{4}

CORRECT ANSWER:    D: \dfrac{x}{4}

WORKED SOLUTION:

Factorising the numerator and denominator of our algebraic fraction, we see that

\dfrac{x^2 + 3x}{4x + 12} = \dfrac{x(x + 3)}{4(x + 3)}

We see that top and bottom have a common factor (x + 3), so it can cancel. This leaves us with

\dfrac{x}{4}

This can be no further simplified, and so we are done.

Question 3

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{g^2 - g - 6}{g - 3}

Select the correct answer from the list below.

A: g + 2

B: g - 3

C: \dfrac{1}{g - 3}

D: g^2 - 3

CORRECT ANSWER:   A: g + 2

WORKED SOLUTION:

We should try to factorise the numerator in order to maybe find a common factor in the fraction which we can cancel. So, considering that -3 \times 2 = -6 and -3 + 2 = -1, we get that

g^2 - g - 6 = (g - 3)(g + 2)

Thus, our algebraic fraction becomes

\dfrac{(g - 3)(g + 2)}{g - 3}

We see that both top and bottom have a factor of g - 3, and so cancelling this factor we get

\dfrac{g + 2}{1} = g + 2

This cannot be simplified any further, and so it is the final answer.

Question 4

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{(a - 3)^2 - a^2}{-6a^2 + 9a}

Select the correct answer from the list below.

A: \dfrac{6}{a}

B: \dfrac{a}{6}

C: a

D: \dfrac{1}{a}

CORRECT ANSWER:   D: \dfrac{1}{a}

WORKED SOLUTION:

Firstly, we expand the numerator. See

(a - 3)^2 - a^2 = a^2 - 3a - 3a + 9 - a^2 = -6a + 9

Then, factorising a out of the denominator means our fraction becomes

\dfrac{-6a + 9}{a(-6a + 9)}

We can cancel the common factor of (6a - 9), which leaves us with

\dfrac{1}{a}

This cannot be simplified further so is the final answer.

Question 5

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{z^4 - 4z^3}{z^2 - 16}

Select the correct answer from the list below.

A: \dfrac{z^3}{z + 4}

B: \dfrac{z^2}{z - 4}

C: \dfrac{z - 4}{z + 4}

D: \dfrac{z - 4}{z}

CORRECT ANSWER:  A: \dfrac{z^3}{z + 4}

WORKED SOLUTION:

There is nothing to be expanded, so we look to factorise. Firstly, we can factorise a z^3 out of the numerator to get

z^4 - 4z^3 = z^3(z - 4)

Furthermore, we can factorise the denominator as it is a case of the difference of two squares. Noting that \sqrt{16} = 4, we get

z^2 - 16 = (z - 4)(z + 4)

So, our fraction becomes

\dfrac{z^3(z - 4)}{(z - 4)(z + 4)}

We can cancel the common factor of (z - 4), leaving us with

\dfrac{z^3}{z + 4}

This cannot be simplified further and so is the final answer.

Question 6

LEVEL 6

Simplify the following algebraic fraction:

\dfrac{m^2 - 2m - 8}{(m - 4)(m + 1) - 6}

Select the correct answer from the list below.

A: \dfrac{m + 4}{m + 1}

B: \dfrac{m - 4}{m - 5}

C: \dfrac{m + 2}{m - 2}

D: \dfrac{m - 2}{m - 4}

CORRECT ANSWER:    B: \dfrac{m - 4}{m - 5}

WORKED SOLUTION:

Firstly, we must expand and simplify the denominator. See:

(m - 4)(m + 1) - 6 = m^2 - 4m + m - 4 - 6 = m^2 - 3m - 10

Next, we can look at factorising. We’ve shown the denominator is a quadratic, so if we notice that 2 \times -5 = -10 and 2 + -5 = -3, we can see

m^2 - 3m - 10 = (m - 5)(m + 2)

Now, we wish to factorise the numerator (also a quadratic) similarly. So, observing that 2 \times -4 = -8 and 2 + (-4) = -2, we get that

m^2 - 2m - 8 = (m - 4)(m + 2)

Therefore, our algebraic fraction becomes

\dfrac{(m - 4)(m + 2)}{(m - 5)(m + 2)}

We can cancel the common factor of (m + 2), which gives us

\dfrac{m - 4}{m - 5}

This cannot be simplified further and so is the final answer.

Question 7

LEVEL 8

Write \dfrac{2}{x+2} + \dfrac{3}{2x+1} as a single fraction.

Select the correct answer from the list below.

A: \dfrac{7x+7}{(2x+1)(x+2)}

B:  \dfrac{14}{(2x+1)(x+2)}

C: \dfrac{3x+3}{(2x+1)(x+2)}

D: \dfrac{14x+14}{(2x+1)(x+2)}

CORRECT ANSWER:   A: \dfrac{7x+7}{(2x+1)(x+2)}

WORKED SOLUTION:

We need to multiply each fraction by the denominator of the other fraction.

\bigg(\dfrac{2}{x+2}\times {\dfrac{2x+1}{2x+1}}\bigg) + \bigg(\dfrac{3}{2x+1}\times {\dfrac{x+2}{x+2}}\bigg) = \dfrac{2{(2x+1)}}{(x+2){(2x+1)}} + \dfrac{3{(x+2)}}{(2x+1){(x+2)}}

Multiply out the numerators.

\dfrac{2{(2x+1)}}{(x+2){(2x+1)}} + \dfrac{3{(x+2)}}{(2x+1){(x+2)}} = \dfrac{4x + 1}{(x+2)(2x+1)} + \dfrac{3x+6}{(2x+1)(x+2)}

Add the fractions

\dfrac{4x + 1}{(x+2)(2x+1)} + \dfrac{3x+6}{(2x+1)(x+2)} =  \dfrac{7x+7}{(2x+1)(x+2)}

Question 8

LEVEL 8

Simplify the following:

\dfrac{(2x + 3)}{(x-2)} \div \dfrac{4x}{(x-2)}

Select the correct answer from the list below.

A: \dfrac{4}{2x+3}

B:  \dfrac{4x}{2x+3}

C: \dfrac{2x+3}{4}

D: \dfrac{2x+3}{4x}

CORRECT ANSWER:    D: \dfrac{2x+3}{4x}

WORKED SOLUTION:

Flip the second fraction upside down and change the \div to a \times

\dfrac{(2x + 3)}{(x-2)} \div \dfrac{4x}{(x-2)} = \dfrac{(2x + 3)}{(x-2)} \times \dfrac{(x-2)}{4x}

Then the (x-2) bracket can cancel, so we get

Question 9

LEVEL 6

Express the following as a single fraction:

\dfrac{m}{2} - \dfrac{m}{3}

Select the correct answer from the list below.

A: \dfrac{m}{6}

B: 0

C: \dfrac{5m}{6}

D: - \dfrac{m^2}{6}

CORRECT ANSWER:   A: \dfrac{m}{6}

WORKED SOLUTION:

Make the denominators of both fractions the same. So we multiply the first fraction by 3 on the top and bottom and multiply the second fraction by 2 on the top and bottom, to get 6 on the bottom of both fractions.

Takeaway the numerators of the fractions and the denominators of the fractions, separately. We get

Question 10

LEVEL 6

Express the following as a single fraction:

\dfrac{x}{20} + \dfrac{2x}{5}

Select the correct answer from the list below.

A: \dfrac{3x}{25}

B: \dfrac{9x}{100}

C: \dfrac{2x^2}{100}

D: \dfrac{9x}{20}

CORRECT ANSWER:    D: \dfrac{9x}{20}

WORKED SOLUTION:

Make the denominators of both fractions the same. So we multiply the second fraction by 4 on the top and bottom to get 20 on the bottom of both fractions.

Add the numerators of the fractions together and the denominators of the fractions together, separately. We get

Question 11

LEVEL 6

Express the following as a single fraction:

\dfrac{2a}{3} + \dfrac{3}{2}

Select the correct answer from the list below.

A: \dfrac{4a-9}{6}

B: \dfrac{4a+9}{6}

C: 2a-3

D: \dfrac{2a-3}{6}

CORRECT ANSWER:    A: \dfrac{4a-9}{6}

WORKED SOLUTION:

Make the denominators of both fractions the same. So we multiply the first fraction by 2 on the top and bottom and the second fraction by 3 on the top and bottom to get 6 on the bottom of both fractions.

Subtract the numerators of the fractions from one another and the denominators of the fractions from one another, separately. We get

Question 12

LEVEL 6

Express the following as a single fraction:

\dfrac{n}{4} \times \dfrac{3n}{5}

Select the correct answer from the list below.

A: \dfrac{3n^2}{20}

B: \dfrac{18n}{20}

C: \dfrac{4n}{9}

D: \dfrac{3n}{20}

CORRECT ANSWER:    A: \dfrac{3n^2}{20}

WORKED SOLUTION:

Multiply the numerators of the fractions together and the denominators of the fractions together, separately. We get