Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples - Expressions

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

Pre-requisites:
(1.) Distributive Property
(2.) Fractions

Solve each expression.
Indicate the method(s) used as applicable.
Use at least two methods (two or more methods) whenever applicable.
Show all work.

(1.) ACT Which of the following expressions is equivalent to $a(4 - a) - 5(a + 7)?$

$ A.\:\: -2a - 35 \\[3ex] B.\:\: -2a + 7 \\[3ex] C.\:\: -a^2 - a - 35 \\[3ex] D.\:\: -a^2 - a + 7 \\[3ex] E.\:\: -2a^3 - 35 \\[3ex] $

$ a(4 - a) - 5(a + 7) \\[3ex] \underline{Distributive\:\:Property} \\[3ex] 4a - a^2 - 5a - 35 \\[3ex] -a^2 - a - 35 $
(2.) ACT Which of the following expressions is equivalent to $(3x + 6)(2x - 1)?$

$ A.\:\: 15x - 6 \\[3ex] B.\:\: 15x - 1 \\[3ex] C.\:\: 6x^2 - 6 \\[3ex] D.\:\: 6x^2 + 9x - 6 \\[3ex] E.\:\: 6x^2 + 12x - 6 \\[3ex] $

$ (3x + 6)(2x - 1) \\[3ex] \underline{FOIL\:\:Method} \\[3ex] 3x(2x) = 6x^2 \\[3ex] 3x(-1) = -3x \\[3ex] 6(2x) = 12x \\[3ex] 6(-1) = -6 \\[3ex] \rightarrow 6x^2 - 3x + 12x - 6 \\[3ex] = 6x^2 + 9x - 6 $
(3.) ACT Given that $r = 4,\:\:b = 2,\:\:and\:\:g = -5$, $(r + b - g)(b + g) = ?$

$ A.\:\: -33 \\[3ex] B.\:\: -3 \\[3ex] C.\:\: 3 \\[3ex] D.\:\: 7 \\[3ex] E.\:\: 8 \\[3ex] $

$ r = 4 \\[3ex] b = 2 \\[3ex] g = -5 \\[3ex] r + b - g \\[3ex] = 4 + 2 - (-5) \\[3ex] = 6 + 5 \\[3ex] = 11 \\[3ex] b + g \\[3ex] = 2 + -5 \\[3ex] = 2 - 5 \\[3ex] = -3 \\[3ex] (r + b - g)(b + g) \\[3ex] = 11(-3) \\[3ex] = -33 $
(4.) ACT Given that $r = 6,\:\:b = 4,\:\:and\:\:g = -9$, $(r + b - g)(b + g) = ?$

$ F.\:\: -95 \\[3ex] G.\:\: -5 \\[3ex] H.\:\: 5 \\[3ex] J.\:\: 13 \\[3ex] K.\:\: 14 \\[3ex] $

$ r = 6 \\[3ex] b = 4 \\[3ex] g = -9 \\[3ex] r + b - g \\[3ex] = 6 + 4 - (-9) \\[3ex] = 6 + 4 + 9 \\[3ex] = 19 \\[3ex] b + g \\[3ex] = 4 + -9 \\[3ex] = 4 - 9 \\[3ex] = -5 \\[3ex] (r + b - g)(b + g) \\[3ex] = 19(-5) \\[3ex] = -95 $
(5.) CSEC Write as a single fraction:

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

$ \dfrac{2x + 3}{3} + \dfrac{x - 4}{4} \\[5ex] \dfrac{4(2x + 3)}{12} + \dfrac{3(x - 4)}{12} \\[5ex] \dfrac{8x + 12}{12} + \dfrac{3x - 12}{12} \\[5ex] \dfrac{8x + 12 + (3x - 12)}{12} \\[5ex] \dfrac{8x + 12 + 3x - 12}{12} \\[5ex] \dfrac{11x}{12} $
(6.) CSEC Express as a single fraction in its simplest form
$\dfrac{a}{3} + \dfrac{3a}{2}$


$ \dfrac{a}{3} + \dfrac{3a}{2} \\[5ex] \dfrac{2a}{6} + \dfrac{3(3a)}{6} \\[5ex] \dfrac{2a}{6} + \dfrac{9a}{6} \\[5ex] \dfrac{2a + 9a}{6} \\[5ex] \dfrac{11a}{6} $
(7.) WASSCE Express $\dfrac{2}{x + 3} - \dfrac{1}{x - 2}$ as a simple fraction.

$ A.\:\: \dfrac{x - 7}{x^2 + x - 6} \\[5ex] B.\:\: \dfrac{x - 1}{x^2 + x - 6} \\[5ex] C.\:\: \dfrac{x - 2}{x^2 + x - 6} \\[5ex] D.\:\: \dfrac{x + 7}{x^2 + x - 6} \\[5ex] $

$ \dfrac{2}{x + 3} - \dfrac{1}{x - 2} \\[5ex] \dfrac{2(x - 2)}{(x + 3)(x - 2)} - \dfrac{1(x + 3)}{(x + 3)(x - 2)} \\[5ex] \dfrac{2(x - 2) - 1(x + 3)}{(x + 3)(x - 2)} \\[5ex] \dfrac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6} \\[5ex] \dfrac{x - 7}{x^2 + x - 6} $
(8.) CSEC Express as a single fraction: $\dfrac{3p}{2} + \dfrac{q}{p}$


$ \dfrac{3p}{2} + \dfrac{q}{p} \\[5ex] \dfrac{3p(p)}{2p} + \dfrac{q(2)}{2p} \\[5ex] \dfrac{3p^2}{2p} + \dfrac{2q}{2p} \\[5ex] \dfrac{3p^2 + 2q}{2p} $
(9.) Simplify the expression: $\dfrac{1}{11}x + \dfrac{1}{5}(x + 7)$


$ \dfrac{1}{11}x + \dfrac{1}{5}(x + 7) \\[5ex] = \dfrac{1 * x}{11} + \dfrac{1 * x}{5} + \dfrac{1 * 7}{5} \\[5ex] = \dfrac{x}{11} + \dfrac{x}{5} + \dfrac{7}{5} \\[5ex] Collect\:\:like\:\:terms\:\:in\:\:x \\[3ex] LCD\:\:of\:\:11\:\:and\:\:5 = 55 \\[3ex] = \dfrac{5x}{55} + \dfrac{11x}{55} + \dfrac{7}{5} \\[5ex] = \dfrac{5x + 11x}{55} + \dfrac{7}{5} \\[5ex] = \dfrac{16x}{55} + \dfrac{7}{5} $
(10.) Simplify $3k + 4k - 2(10k + 8)$


$ 3k + 4k - 2(10k + 8) \\[3ex] = 7k - 20k - 16 \\[3ex] = -13k - 16 $
(11.) Simplify the expressions:

$ (a.)\:\: (7w + 2) + (-3w + 4) \\[3ex] (b.)\:\: (7w + 2) - (-3w + 4) \\[3ex] (c.)\:\: 3(7w + 2) + 5(-3w + 4) \\[3ex] (d.)\:\: -2(7w - 5) + 5(-2w + 1) \\[3ex] (e.)\:\: (7w + 2)(-3w + 4) \\[3ex] $

$ (a.) \\[3ex] (7w + 2) + (-3w + 4) \\[3ex] = 1(7w + 2) + 1(-3w + 4) \\[3ex] = 7w + 2 - 3w + 4 \\[3ex] = 4w + 6 \\[3ex] (b.) \\[3ex] (7w + 2) - (-3w + 4) \\[3ex] = 1(7w + 2) -1(-3w + 4) \\[3ex] = 7w + 2 + 3w - 4 \\[3ex] = 10w - 2 \\[3ex] (c.) \\[3ex] 3(7w + 2) + 5(-3w + 4) \\[3ex] = 21w + 6 - 15w + 20 \\[3ex] = 6w + 26 \\[3ex] (d.) \\[3ex] -2(7w - 5) + 5(-2w + 1) \\[3ex] = -14w + 10 - 10w + 5 \\[3ex] = -24w + 15 \\[3ex] (e.) \\[3ex] (7w + 2)(-3w + 4) \\[3ex] 7w(-3w) = -21w^2 \\[3ex] 7w(4) = 28w \\[3ex] 2(-3w) = -6w \\[3ex] 2(4) = 8 \\[3ex] = -21w^2 + 28w - 6w + 8 \\[3ex] = -21w^2 + 22w + 8 $
(12.) ACT Given that $(x + 2)$ and $(x - 1)$ are factors of the quadratic expression below, what are the values of $a$ and $b$?

$ x^2 + (a + 2)x + a + b \\[3ex] \:\:\:\:\:\:\:\:\:\:\underline{a}\:\:\:\:\:\:\underline{b} \\[3ex] F.\:\:-4\:\:\:\:\:\:5 \\[3ex] G.\:\:-3\:\:\:\:\:\:1 \\[3ex] H.\:\:-3\:\:\:\:\:\:5 \\[3ex] J.\:\:-1\:\:\:\:\:\:\:3 \\[3ex] K.\:-1\:\:\:\:-1 \\[3ex] $

$ Factors\:\:are\:\:(x + 2)\:\:and\:\:(x - 1) \\[3ex] Quadratic\:\:Expression = (x + 2)(x - 1) \\[3ex] x(x) = x^2 \\[3ex] x(-1) = -x \\[3ex] 2(x) = 2x \\[3ex] 2(-1) = -2 \\[3ex] \rightarrow x^2 - x + 2x - 2 \\[3ex] = x^2 + x - 2 \\[3ex] Quadratic\:\:Expression = x^2 + 1x - 2 \\[3ex] Compare:\:\: x^2 + (a + 2)x + a + b \\[3ex] \rightarrow a + 2 = 1...eqn.(1) \\[3ex] and\:\: a + b = -2...eqn.(2) \\[3ex] From\:\:eqn.(1);\:\: a = 1 - 2 \\[3ex] a = -1 \\[3ex] Substitute\:\:a = -1\:\:in\:\:eqn.(2) \\[3ex] -1 + b = -2 \\[3ex] b = -2 + 1 \\[3ex] b = -1 \\[3ex] \therefore a = -1, b = -1 $
(13.) JAMB If $y = \dfrac{x}{x - 3} + \dfrac{x}{x + 4}$, find $y$ when $x = -2$

$ (a.)\:\: -\dfrac{3}{5} \\[5ex] (b.)\:\: \dfrac{3}{5} \\[5ex] (c.)\:\: -\dfrac{7}{5} \\[5ex] (d.)\:\: \dfrac{7}{5} \\[5ex] $

$ y = \dfrac{x}{x - 3} + \dfrac{x}{x + 4} \\[5ex] x = -2 \\[3ex] y = \dfrac{-2}{-2 - 3} + \dfrac{-2}{-2 + 4} \\[5ex] = \dfrac{-2}{-5} + \dfrac{-2}{2} \\[5ex] = \dfrac{2}{5} + (-1) \\[5ex] = \dfrac{2}{5} - 1 \\[5ex] = \dfrac{2}{5} - \dfrac{5}{5} \\[5ex] = \dfrac{2 - 5}{5} \\[5ex] = -\dfrac{3}{5} $
(14.) ACT $(9m - 4n) - (2n + 5m)$ is equivalent to:

$ A.\:\: 4m - 6n \\[3ex] B.\:\: 4m - 2n \\[3ex] C.\:\: 5m + 3m \\[3ex] D.\:\: 7m - 9n \\[3ex] E.\:\: 7m + n \\[3ex] $

$ (9m - 4n) - (2n + 5m) \\[3ex] = 1(9m - 4n) - 1(2n + 5m) \\[3ex] = 9m - 4n - 2n - 5m \\[3ex] = 9m - 5m - 4n - 2n \\[3ex] = 4m - 6n $
(15.) CSEC Given that $m = -2$ and $n = 4$, calculate the value of $(2m + n)(2m - n)$


$ \underline{First\:\:Method} \\[3ex] (2m + n)(2m - n) \\[3ex] m = -2, n = 4 \\[3ex] = (2(-2) + 4)(2(-2) - 4) \\[3ex] = (-4 + 4)(-4 - 4) \\[3ex] = 0(-8) \\[3ex] = 0 \\[3ex] \underline{Second\:\:Method} \\[3ex] (2m + n)(2m - n) = (2m)^2 - n^2...Difference\:\:of\:\:Two\:\:Squares \\[3ex] m = -2, n = 4 \\[3ex] (2m)^2 = [2(-2)]^2 = (-4)^2 = -4(-4) = 16 \\[3ex] n^2 = 4^2 = 4(4) = 16 \\[3ex] \rightarrow (2m)^2 - n^2 = 16 - 16 = 0 $
(16.) Add: $4x + (-5)y + (-3)x + (-10)y$


$ 4x + (-5)y + (-3)x + (-10)y \\[3ex] Rearrange...write\:\:it\:\;well \\[3ex] 4x + y(-5) + x(-3) + y(-10) \\[3ex] = 4x - 5y - 3x - 10y \\[3ex] = 4x - 3x - 5y - 10y \\[3ex] = x - 15y $
(17.) Simplify completely: $4(3x - 2y) - (x + 7y)$


$ 4(3x - 2y) - (x + 7y) \\[3ex] = 4(3x - 2y) -1(x + 7y) \\[3ex] = 12x - 8y - x - 7y \\[3ex] = 12x - x - 8y - 7y \\[3ex] = 11x - 15y $
(18.) CSEC If $a = 2$, $b = -3$, and $c = 4$, evalaute:

$ (i)\:\: ab - bc \\[3ex] (ii)\:\: b(a - c)^2 \\[3ex] $

$ a = 2, b = -3, c = 4 \\[3ex] (i) \\[3ex] ab - bc \\[3ex] = 2(-3) - -3(4) \\[3ex] = -6 - -12 \\[3ex] = -6 + 12 \\[3ex] = 6 \\[3ex] (ii) \\[3ex] b(a - c)^2 \\[3ex] = b * (a - c)^2 \\[3ex] = -3 * (2 - 4)^2 \\[3ex] = -3 * (-2)^2 \\[3ex] (-2)^2 = -2(-2) = 4 \\[3ex] = -3 * 4 \\[3ex] = -12 $
(19.) Simplify completely: $[6 + (4x + 8)2x]3 + 5(x^2 + 10x + 12)$


$ [6 + (4x + 8)2x]3 + 5(x^2 + 10x + 12) \\[3ex] Rearrange...write\:\:it\:\;well \\[3ex] = 3[6 + 2x(4x + 8)] + 5(x^2 + 10x + 12) \\[3ex] = 3[6 + 8x^2 + 16x] + 5x^2 + 50x + 60 \\[3ex] = 18 + 24x^2 + 48x + 5x^2 + 50x + 60 \\[3ex] = 24x^2 + 5x^2 + 48x + 50x + 18 + 60 \\[3ex] = 29x^2 + 98x + 78 $
(20.) Simplify completely: $-\dfrac{9}{2}(2x - 28y)$


$ -\dfrac{9}{2}(2x - 28y) \\[5ex] -\dfrac{9}{2} * 2x = -9(x) = -9x \\[5ex] -\dfrac{9}{2} * -28y = 9(14y) = 126y \\[5ex] = -9x + 126y $




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(21.) CSEC Expand $(x + 3)^2(x - 4)$, writing your answer in descending powers of $x$


Writing your answer in descending powers of $x$ means that the exponents of $x$ in your answer has to be ordered from greatest to least

$ (x + 3)^2(x - 4) \\[3ex] = (x + 3)(x + 3)(x - 4) \\[3ex] \underline{First\:\:two\:\:Binomials} \\[3ex] (x + 3)(x + 3) \\[3ex] x(x) = x^2 \\[3ex] x(3) = 3x \\[3ex] 3(x) = 3x \\[3ex] 3(3) = 9 \\[3ex] = x^2 + 3x + 3x + 9 \\[3ex] = x^2 + 6x + 9 \\[3ex] \underline{Trinomial\:\:and\:\:Binomial} \\[3ex] (x^2 + 6x + 9)(x - 4) \\[3ex] x^2(x) = x^3 \\[3ex] x^2(-4) = -4x^2 \\[3ex] 6x(x) = 6x^2 \\[3ex] 6x(-4) = -24x \\[3ex] 9(x) = 9x \\[3ex] 9(-4) = -36 \\[3ex] = x^3 - 4x^2 + 6x^2 - 24x + 9x - 36 \\[3ex] = x^3 + 2x^2 - 15x - 36 $
(22.) JAMB If the numbers $M, N, Q$ are in the ratio $5:4:3$, find the value of $\dfrac{2N - Q}{M}$

$ A.\:\: 3 \\[3ex] B.\:\: 2 \\[3ex] C.\:\: 1 \\[3ex] D.\:\: 4 \\[3ex] $

$ \rightarrow M = 5, N = 4, Q = 3 \\[3ex] \dfrac{2N - Q}{M} \\[5ex] = \dfrac{2(4) - 3}{5} \\[5ex] = \dfrac{8 - 3}{5} \\[5ex] = \dfrac{5}{5} \\[5ex] = 1 $
(23.) JAMB If $\dfrac{y}{2} = x$, evaluate $\left(\dfrac{x^3}{y^3}+ \dfrac{1}{2}\right) \div \left(\dfrac{1}{2} - \dfrac{x^2}{y^2}\right)$

$ A.\:\: \dfrac{5}{8} \\[5ex] B.\:\: \dfrac{5}{2} \\[5ex] C.\:\: \dfrac{5}{4} \\[5ex] D.\:\: \dfrac{5}{16} \\[5ex] $

It is much better to work without fractions
So, let us make it easier and faster to solve

$ \dfrac{y}{2} = x \rightarrow y = 2x \\[5ex] \left(\dfrac{x^3}{y^3}+ \dfrac{1}{2}\right) \div \left(\dfrac{1}{2} - \dfrac{x^2}{y^2}\right) \\[5ex] \dfrac{x^3}{y^3} = \dfrac{x^3}{(2x)^3} = \dfrac{x^3}{8x^3} = \dfrac{1}{8} \\[5ex] \dfrac{x^3}{y^3}+ \dfrac{1}{2} = \dfrac{1}{8} + \dfrac{1}{2} = \dfrac{2}{16} + \dfrac{8}{16} = \dfrac{2 + 8}{16} = \dfrac{10}{16} = \dfrac{5}{8} \\[5ex] \dfrac{x^2}{y^2} = \dfrac{x^2}{(2x)^2} = \dfrac{x^2}{4x^2} = \dfrac{1}{4} \\[5ex] \dfrac{1}{2} - \dfrac{x^2}{y^2} = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{2}{4} - \dfrac{1}{4} = \dfrac{2 - 1}{4} = \dfrac{1}{4} \\[5ex] \dfrac{5}{8} \div \dfrac{1}{4} = \dfrac{5}{8} * \dfrac{4}{1} = \dfrac{5}{2} $
(24.)

$ r = 6 \\[3ex] b = 4 \\[3ex] g = -9 \\[3ex] r + b - g \\[3ex] = 6 + 4 - (-9) \\[3ex] = 6 + 4 + 9 \\[3ex] = 19 \\[3ex] b + g \\[3ex] = 4 + -9 \\[3ex] = 4 - 9 \\[3ex] = -5 \\[3ex] (r + b - g)(b + g) \\[3ex] = 19(-5) \\[3ex] = -95 $