# Solved Examples: Rational Equations

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Unless otherwise specified;
For each rational equation:
Specify the values that are excluded from the domain.
Write the domain in interval notation.
Solve, using at least two methods.
Show all work.
Check all solutions.
Identify any extraneous root(s).

(1.) NZQA Solve the equation

$\dfrac{3}{x + 2} + \dfrac{5}{x - 4} = 2 \\[5ex]$

For both First and Second Methods

We need to find the LCD first.

$LCD = (x + 2)(x - 4) \\[3ex]$ This is the $LCD$ because it contains (a multiple of) the denominators.

First Method
Multiply all terms by the $LCD$ so we will not have to deal with fractions.

$LCD = (x + 2)(x - 4) \\[3ex] \dfrac{3}{x + 2} * (x + 2)(x - 4) + \dfrac{5}{x - 4} * (x + 2)(x - 4) = 2[(x + 2)(x - 4)] \\[5ex] 3(x - 4) + 5(x + 2) = 2(x^2 - 4x + 2x - 8) \\[3ex] 3x - 12 + 5x + 10 = 2(x^2 - 2x - 8) \\[3ex] 8x - 2 = 2x^2 - 4x - 16 \\[3ex] 0 = 2x^2 - 4x - 16 - 8x + 2 \\[3ex] 0 = 2x^2 - 12x - 14 \\[3ex] 2x^2 - 12x - 14 = 0 \\[3ex] Simplify:\;\;Divide\;\;all\;\;terms\;\;by\;\;2 \\[3ex] x^2 - 6x - 7 = 0 \\[3ex] (x + 1)(x - 7) = 0 \\[3ex] x + 1 = 0 \;\;OR\;\; x - 7 = 0 \\[3ex] x = -1 \;\;OR\;\; x = 7 \\[3ex]$ Second Method
Solve as fractions. To do so, all of them must have the same denominator: the $LCD$

$LCD = (x + 2)(x - 4) \\[3ex] \dfrac{3}{x + 2} = \dfrac{3(x - 4)}{(x + 2)(x - 4)} \\[5ex] \dfrac{5}{x - 4} = \dfrac{5(x + 2)}{(x + 2)(x - 4)} \\[5ex] 2 = \dfrac{2(x + 2)(x - 4)}{(x + 2)(x - 4)} \\[5ex]$ The denominators are the same.
We equate the numerators.

$3(x - 4) + 5(x + 2) = 2(x + 2)(x - 4) \\[3ex] 3x - 12 + 5x + 10 = 2(x^2 - 2x - 8) \\[3ex] 8x - 2 = 2x^2 - 4x - 16 \\[3ex] Swap \\[3ex] 2x^2 - 4x - 16 = 8x - 2 \\[3ex] 2x^2 - 4x - 8x - 16 + 2 = 0 \\[3ex] 2x^2 - 12x - 14 = 0 \\[3ex] Simplify:\;\;Divide\;\;all\;\;terms\;\;by\;\;2 \\[3ex] x^2 - 6x - 7 = 0 \\[3ex] (x + 1)(x - 7) = 0 \\[3ex] x + 1 = 0 \;\;OR\;\; x - 7 = 0 \\[3ex] x = -1 \;\;OR\;\; x = 7 \\[3ex]$ Check
Let us check the equation for both roots.
 $\underline{LHS} \\[3ex] \dfrac{3}{x + 2} + \dfrac{5}{x - 4} \\[5ex] x = -1 \\[3ex] \dfrac{3}{-1 + 2} + \dfrac{5}{-1 - 4} \\[5ex] = \dfrac{3}{1} + \dfrac{5}{-5} \\[5ex] = 3 + -1 \\[3ex] = 3 - 1 \\[3ex] = 2 \\[3ex]$ $\color{black}{x = -1}$ is a solution $x = 7 \\[3ex] \dfrac{3}{7 + 2} + \dfrac{5}{7 - 4} \\[5ex] = \dfrac{3}{9} + \dfrac{5}{3} \\[5ex] = \dfrac{3}{9} + \dfrac{15}{9} \\[5ex] = \dfrac{3 + 15}{9} \\[5ex] = \dfrac{18}{9} \\[5ex] = 2 \\[3ex]$ $\color{black}{x = 7}$ is a solution $\underline{RHS} \\[3ex] 2 \\[3ex]$ $2$
(2.) WASSCE Using completing the squares method, solve, correct to $2$ decimal places,

$\dfrac{x - 2}{4} = \dfrac{x + 2}{2x}$

Third method: Cross Multiply method

WASSCE is a timed exam. So, this method is recommended because it is the fastest method.

$\dfrac{x - 2}{4} = \dfrac{x + 2}{2x} \\[5ex] 2x(x - 2) = 4(x + 2) \\[3ex] 2x^2 - 4x = 4x + 8 \\[3ex] 2x^2 - 4x - 4x - 8 = 0 \\[3ex] 2x^2 - 8x - 8 = 0 \\[3ex] Factor\:\: by\:\: GCF \\[3ex] 2(x^2 - 4x - 4) = 0 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] x^2 - 4x - 4 = 0 \\[3ex]$ Specifically, we are asked to use the Completing the Squares method

$x^2 - 4x = 4 \\[3ex] Coefficient\:\: of\:\: x = -4 \\[3ex] Half\:\: of\:\: it = \dfrac{1}{2} * -4 = -2 \\[5ex] Square\:\: it = (-2)^2 \\[3ex] \implies x^2 - 4x + (-2)^2 = 4 + (-2)^2 \\[3ex] (x - 2)^2 = 4 + 4 \\[3ex] (x - 2)^2 = 8 \\[3ex] x - 2 = \pm \sqrt{8} \\[3ex] x = 2 \pm \sqrt{8} \\[3ex]$ We are required to round the final answer to $2$ decimal places.

In that regard, please round all intermediate calculations to "at least" $5$ decimal places ($3$ places more)

$x = 2 \pm 2.828427125 \\[3ex] x = 2 + 2.828427125 \:\:OR\:\: x = 2 - 2.828427125 \\[3ex] x = 4.828427125 \:\:OR\:\: x = -0.8284271247 \\[3ex] x \approx 4.83 \:\:OR\:\: x \approx -0.83 \\[3ex]$ First Method
Multiply all terms by the $LCD$ so you will not have to deal with fractions.

$LCD = 4x \\[3ex] 4x * \dfrac{x - 2}{4} = 4x * \dfrac{x + 2}{2x} \\[5ex] x(x - 2) = 2(x + 2) \\[3ex] x^2 - 2x = 2x + 4 \\[3ex] x^2 - 2x - 2x - 4 = 0 \\[3ex] x^2 - 4x - 4 = 0 \\[3ex]$ Specifically, we are asked to use the Completing the Squares method

$x^2 - 4x = 4 \\[3ex] Coefficient\:\: of\:\: x = -4 \\[3ex] Half\:\: of\:\: it = \dfrac{1}{2} * -4 = -2 \\[5ex] Square\:\: it = (-2)^2 \\[3ex] \implies x^2 - 4x + (-2)^2 = 4 + (-2)^2 \\[3ex] (x - 2)^2 = 4 + 4 \\[3ex] (x - 2)^2 = 8 \\[3ex] x - 2 = \pm \sqrt{8} \\[3ex] x = 2 \pm \sqrt{8} \\[3ex]$ We are required to round the final answer to $2$ decimal places.

In that regard, please round all intermediate calculations to "at least" $5$ decimal places ($3$ places more)

$x = 2 \pm 2.828427125 \\[3ex] x = 2 + 2.828427125 \:\:OR\:\: x = 2 - 2.828427125 \\[3ex] x = 4.828427125 \:\:OR\:\: x = -0.8284271247 \\[3ex] x \approx 4.83 \:\:OR\:\: x \approx -0.83 \\[3ex]$ Second Method
Solve as fractions. To do so, all of them must have the same denominator - the $LCD$

$LCD = 4x \\[3ex] \dfrac{x - 2}{4} = \dfrac{x(x - 2)}{4x} \\[5ex] \dfrac{x + 2}{2x} = \dfrac{2(x + 2)}{4x} \\[5ex]$ The denominators are the same.
We can now equate the numerators.

$x(x - 2) = 2(x + 2) \\[3ex] x^2 - 2x = 2x + 4 \\[3ex] x^2 - 2x - 2x - 4 = 0 \\[3ex] x^2 - 4x - 4 = 0 \\[3ex]$ Specifically, we are asked to use the Completing the Squares method

$x^2 - 4x = 4 \\[3ex] Coefficient\:\: of\:\: x = -4 \\[3ex] Half\:\: of\:\: it = \dfrac{1}{2} * -4 = -2 \\[5ex] Square\:\: it = (-2)^2 \\[3ex] \implies x^2 - 4x + (-2)^2 = 4 + (-2)^2 \\[3ex] (x - 2)^2 = 4 + 4 \\[3ex] (x - 2)^2 = 8 \\[3ex] x - 2 = \pm \sqrt{8} \\[3ex] x = 2 \pm \sqrt{8} \\[3ex]$ We are required to round the final answer to $2$ decimal places.

In that regard, please round all intermediate calculations to "at least" $5$ decimal places ($3$ places more)

$x = 2 \pm 2.828427125 \\[3ex] x = 2 + 2.828427125 \:\:OR\:\: x = 2 - 2.828427125 \\[3ex] x = 4.828427125 \:\:OR\:\: x = -0.8284271247 \\[3ex] x \approx 4.83 \:\:OR\:\: x \approx -0.83 \\[3ex]$ We were not asked to check.

However, it is better to check if you have time.

Check
We shall check with the radicals "as is"

You may decide to check with the decimals if you prefer.

$\sqrt{8} = \sqrt{4 * 2} = \sqrt{4} * \sqrt{2} = 2 * \sqrt{2} = 2\sqrt{2}$
 $\underline{LHS} \\[3ex] \dfrac{x - 2}{4} \\[5ex] x = 2 + \sqrt{8} \\[3ex] x = 2 + 2\sqrt{2} \\[3ex] \dfrac{2 + 2\sqrt{2} - 2}{4} \\[5ex] = \dfrac{2\sqrt{2}}{4} \\[5ex] = \dfrac{\sqrt{2}}{2} \\[5ex]$ $\color{black}{x = 2 + 2\sqrt{2}}$ is a solution $\dfrac{x - 2}{4} \\[5ex] x = 2 - \sqrt{8} \\[3ex] x = 2 - 2\sqrt{2} \\[3ex] \dfrac{2 - 2\sqrt{2} - 2}{4} \\[5ex] = \dfrac{-2\sqrt{2}}{4} \\[5ex] = \dfrac{-\sqrt{2}}{2} \\[5ex]$ $\color{black}{x = 2 - 2\sqrt{2}}$ is a solution $\underline{RHS} \\[3ex] \dfrac{x + 2}{2x} \\[5ex] x = 2 + \sqrt{8} \\[3ex] x = 2 + 2\sqrt{2} \\[3ex] \dfrac{2 + 2\sqrt{2} + 2}{2(2 + 2\sqrt{2})} \\[5ex] = \dfrac{4 + 2\sqrt{2}}{2 * 2(1 + \sqrt{2})} \\[5ex] = \dfrac{2(2 + \sqrt{2})}{4(1 + \sqrt{2})} \\[5ex] = \dfrac{2 + \sqrt{2}}{2(1 + \sqrt{2})} \\[5ex] = \dfrac{2 + \sqrt{2}}{2(1 + \sqrt{2})} * \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}} \\[5ex] = \dfrac{(2 + \sqrt{2})(1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})} \\[5ex] = \dfrac{2 - 2\sqrt{2} + \sqrt{2} - (\sqrt{2})^2}{2[(1)^2 - (\sqrt{2})^2]} \\[5ex] = \dfrac{2 - \sqrt{2} - 2}{2[(1)^2 - (\sqrt{2})^2]} \\[5ex] = \dfrac{-\sqrt{2}}{2(1 - 2)} \\[5ex] = \dfrac{-\sqrt{2}}{2(-1)} \\[5ex] = \dfrac{-\sqrt{2}}{-2} \\[5ex] = \dfrac{\sqrt{2}}{2} \\[5ex]$ $\color{black}{x = 2 + 2\sqrt{2}}$ is a solution $\dfrac{x + 2}{2x} \\[5ex] x = 2 - \sqrt{8} \\[3ex] x = 2 - 2\sqrt{2} \\[3ex] \dfrac{2 - 2\sqrt{2} + 2}{2(2 - 2\sqrt{2})} \\[5ex] = \dfrac{4 - 2\sqrt{2}}{2 * 2(1 - \sqrt{2})} \\[5ex] = \dfrac{2(2 - \sqrt{2})}{4(1 - \sqrt{2})} \\[5ex] = \dfrac{2 - \sqrt{2}}{2(1 - \sqrt{2})} \\[5ex] = \dfrac{2 - \sqrt{2}}{2(1 - \sqrt{2})} * \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[5ex] = \dfrac{(2 - \sqrt{2})(1 + \sqrt{2})}{2(1 - \sqrt{2})(1 + \sqrt{2})} \\[5ex] = \dfrac{2 + 2\sqrt{2} - \sqrt{2} - (\sqrt{2})^2}{2[(1)^2 - (\sqrt{2})^2]} \\[5ex] = \dfrac{2 + \sqrt{2} - 2}{2[(1)^2 - (\sqrt{2})^2]} \\[5ex] = \dfrac{\sqrt{2}}{2(1 - 2)} \\[5ex] = \dfrac{\sqrt{2}}{2(-1)} \\[5ex] = \dfrac{\sqrt{2}}{-2} \\[5ex] = -\dfrac{\sqrt{2}}{2} \\[5ex]$ $\color{black}{x = 2 - 2\sqrt{2}}$ is a solution
(3.) $\dfrac{p}{p + 3} + \dfrac{3}{p - 1} = \dfrac{13p - 1}{p^2 + 2p - 3}$

For both First and Second Methods

We need to find the LCD first.

Let us simplify: $p^2 + 2p - 3$

$p + 3 = p + 3 \\[3ex] p - 1 = p - 1 \\[3ex] p^2 + 2p - 3 \\[3ex] = (p + 3)(p - 1) \\[3ex] LCD = (p + 3)(p - 1) \\[3ex]$ This is the $LCD$ because it contains (a multiple of) all the denominators.

To find the values excluded from the domain, we shall look at each denominator
This is because the denominator cannot be $0$.
Any denominator that is $0$ makes the rational equation undefined.

$For\:\: p + 3 \\[3ex] p + 3 = 0 \\[3ex] p = -3 \\[3ex] For\:\: p - 1 \\[3ex] p - 1 = 0 \\[3ex] p = 1 \\[3ex] For\:\: p^2 + 2p - 3 \\[3ex] Has\:\: been\:\: taken\:\: care\:\: of \\[3ex]$ Values excluded from the domain = $-3, 1$
$D = (-\infty, -3) \cup (-3, 1) \cup (1, \infty) \\[3ex]$ First Method
Multiply all terms by the $LCD$ so we will not have to deal with fractions.

$LCD = (p + 3)(p - 1) \\[3ex] (p + 3)(p - 1) * \dfrac{p}{p + 3} + (p + 3)(p - 1) * \dfrac{3}{p - 1} = (p + 3)(p - 1) * \dfrac{13p - 1}{p^2 + 2p - 3} \\[5ex] (p + 3)(p - 1) * \dfrac{p}{p + 3} + (p + 3)(p - 1) * \dfrac{3}{p - 1} = (p + 3)(p - 1) * \dfrac{13p - 1}{(p + 3)(p - 1)} \\[5ex] p(p - 1) + 3(p + 3) = 13p - 1 \\[3ex] p^2 - p + 3p + 9 = 13p - 1 \\[3ex] p^2 + 2p + 9 - 13p + 1 = 0 \\[3ex] p^2 - 11p + 10 = 0 \\[3ex] (p - 10)(p - 1) = 0 \\[3ex] p - 10 = 0 \:\:OR\:\: p - 1 = 0 \\[3ex] p = 10 \:\:OR\:\: p = 1 \\[3ex]$ Second Method
Solve as fractions. To do so, all of them must have the same denominator: the $LCD$

$LCD = (p + 3)(p - 1) \\[3ex] \dfrac{p}{p + 3} = \dfrac{p(p - 1)}{(p + 3)(p - 1)} \\[5ex] \dfrac{3}{p - 1} = \dfrac{3(p + 3)}{(p + 3)(p - 1)} \\[5ex] \dfrac{13p - 1}{p^2 + 2p - 3} = \dfrac{13p - 1}{(p + 3)(p - 1)} \\[5ex]$ The denominators are the same.
We can now equate the numerators.

$p(p - 1) + 3(p + 3) = 13p - 1 \\[3ex] p^2 - p + 3p + 9 = 13p - 1 \\[3ex] p^2 + 2p + 9 - 13p + 1 = 0 \\[3ex] p^2 - 11p + 10 = 0 \\[3ex] (p - 10)(p - 1) = 0 \\[3ex] p - 10 = 0 \:\:OR\:\: p - 1 = 0 \\[3ex] p = 10 \:\:OR\:\: p = 1 \\[3ex]$ Check
We have to check the equation for both roots.
 $\underline{LHS} \\[3ex] p = 10 \\[3ex] \dfrac{p}{p + 3} + \dfrac{3}{p - 1} \\[5ex] \dfrac{10}{10 + 3} + \dfrac{3}{10 - 1} \\[5ex] = \dfrac{10}{13} + \dfrac{3}{9} \\[5ex] = \dfrac{10}{13} + \dfrac{1}{3} \\[5ex] = \dfrac{30}{39} + \dfrac{13}{39} \\[5ex] = \dfrac{30 + 13}{39} \\[5ex] = \dfrac{43}{39} \\[5ex]$ $\color{black}{p = 10}$ is a solution $p = 1 \\[3ex] \dfrac{p}{p + 3} + \dfrac{3}{p - 1} \\[5ex] \dfrac{1}{1 + 3} + \dfrac{3}{1 - 1} \\[5ex] = \dfrac{1}{4} + \dfrac{3}{0} \\[5ex] = \dfrac{1}{4} + undefined \\[5ex]$ $p = 1$ is an extraneous root $\underline{RHS} \\[3ex] p = 10 \\[3ex] \dfrac{13p - 1}{p^2 + 2p - 3} \\[5ex] = \dfrac{13(10) - 1}{10^2 + 2(10) - 3} \\[5ex] = \dfrac{130 - 1}{100 + 20 - 3} \\[5ex] = \dfrac{129}{117} \\[5ex] = \dfrac{43}{39} \\[5ex]$ $p = 1 \\[3ex] \dfrac{13p - 1}{p^2 + 2p - 3} \\[5ex] = \dfrac{13(1) - 1}{1^2 + 2(1) - 3} \\[5ex] = \dfrac{13 - 1}{1 + 2 - 3} \\[5ex] = \dfrac{12}{0} \\[5ex] = undefined \\[5ex]$
(4.) NSC Solve for $x$:

$x - \dfrac{2}{x} = 5 \\[5ex]$

For both First and Second Methods

We need to find the LCD first.

$LCD = x \\[3ex]$ First Method
Multiply all terms by the $LCD$ so we will not have to deal with fractions.

$LCD = x \\[3ex] x * x - \dfrac{2}{x} * x = 5 * x \\[5ex] x^2 - 2 = 5x \\[3ex] x^2 - 5x - 2 = 0 \\[3ex] a = 1, \;\; b = -5, \;\; c = -2 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} \\[5ex] x = \dfrac{5 \pm \sqrt{25 + 8}}{2} \\[5ex] x = \dfrac{5 \pm \sqrt{33}}{2} \\[5ex]$ Second Method
Solve as fractions. To do so, all of them must have the same denominator: the $LCD$

$LCD = x \\[3ex] x = \dfrac{x^2}{x} \\[5ex] \dfrac{2}{x} = \dfrac{2}{x} \\[5ex] 5 = \dfrac{5x}{x} \\[5ex]$ The denominators are the same.
We equate the numerators.

$x^2 - 2 = 5x \\[3ex] x^2 - 5x - 2 = 0 \\[3ex] a = 1, \;\; b = -5, \;\; c = -2 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} \\[5ex] x = \dfrac{5 \pm \sqrt{25 + 8}}{2} \\[5ex] x = \dfrac{5 \pm \sqrt{33}}{2}$
(5.) $\dfrac{4}{p + 8} + \dfrac{5}{p^2 - 8p + 64} = \dfrac{768}{p^3 + 512}$

For both First and Second Methods

We need to find the LCD first.

$p + 8 = p + 8 \\[3ex] p^2 - 8p + 64 = p^2 - 8p + 64 \\[3ex] p^3 + 512 \\[3ex] = p^3 + 8^3 \\[3ex] x^3 + y^3 = (x + y)(x^2 - xy + y^2)...Sum\:\: of\:\: Two\:\: Cubes \\[3ex] = (p + 8)(p^2 - 8p + 8^2) \\[3ex] = (p + 8)(p^2 - 8p + 64) \\[3ex] LCD = (p + 8)(p^2 - 8p + 64) \\[3ex]$ This is the $LCD$ because it contains (a multiple of) all the denominators.

To find the values excluded from the domain, we shall look at each denominator
This is because the denominator cannot be $0$.
Any denominator that is $0$ makes the rational equation undefined.
$For\:\: p + 8 \\[3ex] p + 8 = 0 \\[3ex] p = -8 \\[3ex] For\:\: p^2 - 8p + 64 \\[3ex] p^2 - 8p + 64 = 0 \\[3ex] Solutions\:\: are\:\: not\:\: real(are\:\: complex) \\[3ex] For\:\: p^3 + 512 \\[3ex] Has\:\: been\:\: taken\:\: care\:\: of \\[3ex]$ Value excluded from the domain = $8$
$D = (-\infty, -8) \cup (-8, \infty) \\[3ex]$ First Method
Multiply all terms by the $LCD$ so you will not have to deal with fractions.

$LCD = (p + 8)(p^2 - 8p + 64) \\[3ex] (p + 8)(p^2 - 8p + 64) * \dfrac{4}{p + 8} + (p + 8)(p^2 - 8p + 64) * \dfrac{5}{p^2 - 8p + 64} = (p + 8)(p^2 - 8p + 64) * \dfrac{768}{p^3 + 512} \\[5ex] (p + 8)(p^2 - 8p + 64) * \dfrac{4}{p + 8} + (p + 8)(p^2 - 8p + 64) * \dfrac{5}{p^2 - 8p + 64} = (p + 8)(p^2 - 8p + 64) * \dfrac{768}{(p + 8)(p^2 - 8p + 64)} \\[5ex] (p^2 - 8p + 64)(4) + (p + 8)(5) = 768 \\[3ex] 4(p^2 - 8p + 64) + 5(p + 8) = 768 \\[3ex] 4p^2 - 32p + 256 + 5p + 40 = 768 \\[3ex] 4p^2 - 32p + 5p + 256 + 40 - 768 = 0 \\[3ex] 4p^2 - 27p - 472 = 0 \\[3ex] 4p^2 + 32p - 59p - 472 = 0 \\[3ex] 4p^2 + 32p = 4p(p + 8) \\[3ex] -59p - 472 = -59(p + 8) \\[3ex] (p + 8)(4p - 59) = 0 \\[3ex] p + 8 = 0 \:\:OR\:\: 4p - 59 = 0 \\[3ex] p = -8 \:\:OR\:\: 4p = 59 \\[3ex] p = -8 \:\:OR\:\: p = \dfrac{59}{4} \\[5ex]$ We already know that $p = -8$ is an extraneous root because it is excluded from the domain.
However, we shall still check it.
Student: Why do we need to check it since we already it will not work?
Teacher: Just to be "thoroughly" sure. Besides, I want to teach you the importance of always checking your work. 😊

Second Method
Solve as fractions. To do so, all of them must have the same denominator - the $LCD$

$LCD = (p + 8)(p^2 - 8p + 64) \\[3ex] \dfrac{4}{p + 8} = \dfrac{4(p^2 - 8p + 64)}{(p + 8)(p^2 - 8p + 64)} = \dfrac{4p^2 - 32p + 256}{(p + 8)(p^2 - 8p + 64)} \\[5ex] \dfrac{5}{p^2 - 8p + 64} = \dfrac{5(p + 8)}{(p + 8)(p^2 - 8p + 64)} = \dfrac{5p + 40}{(p + 8)(p^2 - 8p + 64)} \\[5ex] \dfrac{768}{p^3 + 512} = \dfrac{768}{(p + 8)(p^2 - 8p + 64)}$
The denominators are the same.
We can now equate the numerators.

$(4p^2 - 32p + 256) + (5p + 40) = 768 \\[3ex] 4p^2 - 32p + 256 + 5p + 40 = 768 \\[3ex] 4p^2 - 32p + 5p + 256 + 40 - 768 = 0 \\[3ex] 4p^2 - 27p - 472 = 0 \\[3ex] 4p^2 + 32p - 59p - 472 = 0 \\[3ex] 4p^2 + 32p = 4p(p + 8) \\[3ex] -59p - 472 = -59(p + 8) \\[3ex] (p + 8)(4p - 59) = 0 \\[3ex] p + 8 = 0 \:\:OR\:\: 4p - 59 = 0 \\[3ex] p = -8 \:\:OR\:\: 4p = 59 \\[3ex] p = -8 \:\:OR\:\: p = \dfrac{59}{4} \\[5ex]$ Check
We have to check the equation for both roots.
 $\underline{LHS} \\[3ex] p = -8 \\[3ex] \dfrac{4}{p + 8} + \dfrac{5}{p^2 - 8p + 64} \\[5ex] = \dfrac{4}{-8 + 8} + \dfrac{5}{(-8)^2 - 8(-8) + 64} \\[5ex] = \dfrac{4}{0} + \dfrac{5}{64 + 64 + 64} \\[5ex] = undefined \\[3ex]$ $p = -8$ is an extraneous root. $p = \dfrac{59}{4} \\[5ex] \dfrac{4}{p + 8} + \dfrac{5}{p^2 - 8p + 64} \\[5ex] = \dfrac{4}{\left(\dfrac{59}{4}\right) + 8} + \dfrac{5}{\left(\dfrac{59}{4}\right)^2 - 8 * \left(\dfrac{59}{4}\right) + 64} \\[7ex] = \dfrac{4}{\left(\dfrac{59}{4}\right) + 8} + \dfrac{5}{\left(\dfrac{59}{4}\right)^2 - 8 * \left(\dfrac{59}{4}\right) + 64} \\[7ex] = \dfrac{4}{\left(\dfrac{59}{4}\right) + \left(\dfrac{32}{4}\right)} + \dfrac{5}{\left(\dfrac{3481}{16}\right) - \left(\dfrac{472}{4}\right) + 64} \\[7ex] = \dfrac{4}{\left(\dfrac{59 + 32}{4}\right)} + \dfrac{5}{\left(\dfrac{3481}{16} - \dfrac{1888}{16} + \dfrac{1024}{16}\right)} \\[7ex] = \dfrac{4}{\left(\dfrac{91}{4}\right)} + \dfrac{5}{\left(\dfrac{3481 - 1888 + 1024}{16}\right)} \\[7ex] = \dfrac{4}{\left(\dfrac{91}{4}\right)} + \dfrac{5}{\left(\dfrac{2617}{16}\right)} \\[7ex] = \left(4 \div \dfrac{91}{4}\right) + \left(5 \div \dfrac{2617}{16}\right) \\[5ex] = \left(4 * \dfrac{4}{91}\right) + \left(5 * \dfrac{16}{2617}\right) \\[5ex] = \dfrac{16}{91} + \dfrac{80}{2617} \\[5ex] = \dfrac{41872}{238147} + \dfrac{7280}{238147} \\[5ex] = \dfrac{41872 + 7280}{238147} \\[5ex] = \dfrac{49152}{238147} \\[5ex]$ $\color{black}{p = \dfrac{59}{4}}$ is the solution. $\underline{RHS} \\[3ex] p = -8 \\[3ex] \dfrac{768}{p^3 + 512} \\[5ex] = \dfrac{768}{(-8)^3 + 512} \\[5ex] = \dfrac{768}{-512 + 512} \\[5ex] = \dfrac{768}{0} \\[2ex] = undefined \\[3ex]$ $p = -8$ is an extraneous root. $p = \dfrac{59}{4} \\[5ex] \dfrac{768}{\left(\dfrac{59}{4}\right)^3 + 512} \\[5ex] = 768 \div \left[\left(\dfrac{59}{4}\right)^3 + 512\right] \\[5ex] = 768 \div \left[\left(\dfrac{205379}{64}\right) + 512\right] \\[5ex] = 768 \div \left(\dfrac{205379}{64} + \dfrac{32768}{64}\right) \\[5ex] = 768 \div \left(\dfrac{205379 + 32768}{64}\right) \\[5ex] = 768 \div \left(\dfrac{238147}{64}\right) \\[5ex] = 768 * \left(\dfrac{64}{238147}\right) \\[5ex] = \dfrac{768 * 64}{238147} \\[5ex] = \dfrac{49152}{238147} \\[5ex]$ $\color{black}{p = \dfrac{59}{4}}$ is the solution.
(6.)

$-2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex]$ Check
Check for both values.
 $\underline{LHS} \\[3ex] -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 0 \\[3ex] -2\left|3 - \dfrac{0}{3}\right| + 1 \\[5ex] -2|3 - 0| + 1 \\[5ex] -2|3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 0$ is a solution $-2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 18 \\[3ex] -2\left|3 - \dfrac{18}{3}\right| + 1 \\[5ex] -2|3 - 6| + 1 \\[5ex] -2|-3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 18$ is a solution $\underline{RHS} \\[3ex] -5$
(7.) $\dfrac{1}{2x + 8} - \dfrac{2}{x^2 - 16} = \dfrac{8}{x - 4}$

For both First and Second Methods

We need to find the LCD first.

Let us simplify: $x^2 - 16$

$2x + 8 = 2(x + 4) \\[3ex] x^2 - 16 \\[3ex] = x^2 - 4^2 \\[3ex] x^2 - y^2 = (x + y)(x - y)...Difference\:\: of\:\: Two\:\: Squares \\[3ex] = (x + 4)(x - 4) \\[3ex] x - 4 = x - 4 \\[3ex] LCD = 2(x + 4)(x - 4) \\[3ex]$ This is the $LCD$ because it contains (a multiple of) all the denominators.

To find the values excluded from the domain, we shall look at each denominator
This is because the denominator cannot be $0$.
Any denominator that is $0$ makes the rational equation undefined.

$For\:\: 2x + 8 \\[3ex] 2x + 8 = 0 \\[3ex] 2x = -8 \\[3ex] x = -\dfrac{8}{2} = -4 \\[5ex] For\:\: x^2 - 16 \\[3ex] x^2 - 16 = 0 \\[3ex] (x + 4)(x - 4) = 0 \\[3ex] x + 4 = 0 \:\:OR\:\: x - 4 = 0 \\[3ex] x = -4 \:\:OR\:\: x = 4 \\[3ex] x - 4 = 0 \\[3ex] x = 4 \\[3ex]$ Values excluded from the domain = $-4, 4$
$D = (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \\[3ex]$ First Method
Multiply all terms by the $LCD$ so you will not have to deal with fractions.

$LCD = 2(x + 4)(x - 4) \\[3ex] 2(x + 4)(x - 4) * \dfrac{1}{2x + 8} - 2(x + 4)(x - 4) * \dfrac{2}{x^2 - 16} = 2(x + 4)(x - 4) * \dfrac{8}{x - 4} \\[5ex] 2(x + 4)(x - 4) * \dfrac{1}{2(x + 4)} - 2(x + 4)(x - 4) * \dfrac{2}{(x + 4)(x - 4)} = 2(x + 4)(x - 4) * \dfrac{8}{x - 4} \\[5ex] 1(x - 4) - 2(2) = 2(8)(x + 4) \\[3ex] x - 4 - 4 = 16(x + 4) \\[3ex] x - 8 = 16x + 64 \\[3ex] -8 - 64 = 16x - x \\[3ex] -72 = 15x \\[3ex] 15x = -72 \\[3ex] x = -\dfrac{72}{15} \\[5ex] x = -\dfrac{24}{5} \\[5ex]$ Second Method
Solve as fractions. To do so, all of them must have the same denominator: the $LCD$

$LCD = 2(x + 4)(x - 4) \\[3ex] \dfrac{1}{2x + 8} = \dfrac{1}{2(x + 4)} = \dfrac{1(x - 4)}{2(x + 4)(x - 4)} = \dfrac{x - 4}{2(x + 4)(x - 4)} \\[5ex] \dfrac{2}{x^2 - 16} = \dfrac{2}{(x + 4)(x - 4)} = \dfrac{2(2)}{(2(x + 4)(x - 4)} = \dfrac{4}{2(x + 4)(x - 4)} \\[5ex] \dfrac{8}{x - 4} = \dfrac{8(2)(x + 4)}{2(x + 4)(x - 4)} = \dfrac{16(x + 4)}{2(x + 4)(x - 4)} = \dfrac{16x + 64}{2(x + 4)(x - 4)} \\[5ex]$ The denominators are the same.
We can now equate the numerators.

$(x - 4) - 4 = 16x + 64 \\[3ex] x - 4 - 4 = 16x + 64 \\[3ex] x - 8 = 16x + 64 \\[3ex] -8 - 64 = 16x - x \\[3ex] -72 = 15x \\[3ex] 15x = -72 \\[3ex] x = -\dfrac{72}{15} \\[5ex] x = -\dfrac{24}{5} \\[5ex]$ Check
We have to check the equation for both roots.
 $\underline{LHS} \\[3ex] x = -\dfrac{24}{5} \\[5ex] \dfrac{1}{2x + 8} - \dfrac{2}{x^2 - 16} \\[5ex] 2x = 2\left(-\dfrac{24}{5}\right) = -\dfrac{48}{5} \\[5ex] 2x + 8 = -\dfrac{48}{5} + \dfrac{40}{5} = \dfrac{-48 + 40}{5} = -\dfrac{8}{5} \\[5ex] \dfrac{1}{2x + 8} = 1 \div (2x + 8) \\[3ex] \dfrac{1}{2x + 8} = 1 \div -\dfrac{8}{5} \\[5ex] \dfrac{1}{2x + 8} = 1 * -\dfrac{5}{8} = -\dfrac{5}{8} \\[5ex] x^2 = \left(-\dfrac{24}{5}\right)^2 = \dfrac{(-24)^2}{5^2} = \dfrac{576}{25} \\[5ex] x^2 - 16 = \dfrac{576}{25} - \dfrac{400}{25} = \dfrac{576 - 400}{25} = \dfrac{176}{25} \\[5ex] \dfrac{2}{x^2 - 16} = 2 \div (x^2 - 16) \\[3ex] \dfrac{2}{x^2 - 16} = 2 \div \dfrac{176}{25} \\[5ex] \dfrac{2}{x^2 - 16} = 2 * \dfrac{25}{176} \\[5ex] \dfrac{2}{x^2 - 16} = \dfrac{25}{88} \\[5ex] \implies -\dfrac{5}{8} - \dfrac{25}{88} \\[5ex] = -\dfrac{55}{88} - \dfrac{25}{88} \\[5ex] = \dfrac{-55 - 25}{88} \\[5ex] = -\dfrac{80}{88} \\[5ex] = -\dfrac{10}{11}$ $\underline{RHS} \\[3ex] x = -\dfrac{24}{5} \\[5ex] \dfrac{8}{x - 4} \\[5ex] x - 4 = -\dfrac{24}{5} - 4 \\[5ex] x - 4 = -\dfrac{24}{5} - \dfrac{20}{5} \\[5ex] x - 4 = \dfrac{-24 - 20}{5} \\[5ex] x - 4 = -\dfrac{44}{5} \\[5ex] \dfrac{8}{x - 4} = 8 \div (x - 4) \\[3ex] \implies 8 \div -\dfrac{44}{5} \\[5ex] = 8 * -\dfrac{5}{44} \\[5ex] = 2 * -\dfrac{5}{11} \\[5ex] = -\dfrac{10}{11}$
(8.)

$-2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex]$ Check
Check for both values.
 $\underline{LHS} \\[3ex] -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 0 \\[3ex] -2\left|3 - \dfrac{0}{3}\right| + 1 \\[5ex] -2|3 - 0| + 1 \\[5ex] -2|3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 0$ is a solution $-2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 18 \\[3ex] -2\left|3 - \dfrac{18}{3}\right| + 1 \\[5ex] -2|3 - 6| + 1 \\[5ex] -2|-3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 18$ is a solution $\underline{RHS} \\[3ex] -5$
(9.)

(10.)

$-2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex]$ Check
Check for both values.
 $\underline{LHS} \\[3ex] -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 0 \\[3ex] -2\left|3 - \dfrac{0}{3}\right| + 1 \\[5ex] -2|3 - 0| + 1 \\[5ex] -2|3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 0$ is a solution $-2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 18 \\[3ex] -2\left|3 - \dfrac{18}{3}\right| + 1 \\[5ex] -2|3 - 6| + 1 \\[5ex] -2|-3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex]$ $p = 18$ is a solution $\underline{RHS} \\[3ex] -5$