Solved Examples: Rational Equations

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 $

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.

Checking your solution checks whether your answers are correct or incorrect.

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

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] $

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} $

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.

$ -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 $

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} $

$ -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 $

$ -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 $