If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

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 Dominic 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 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.

Pre-requisites:
(2.) Distributive Property
(3.) 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.) JAMB If the numbers $M, N, Q$ are in the ratio $5:4:3$, find the value of $\dfrac{2N - Q}{M}$

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

$\rightarrow M = 5, N = 4, Q = 3 \\[3ex] \dfrac{2N - Q}{M} \\[5ex] = \dfrac{2(4) - 3}{5} \\[5ex] = \dfrac{8 - 3}{5} \\[5ex] = \dfrac{5}{5} \\[5ex] = 1$
(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: $-\dfrac{9}{2}(2x - 28y)$

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

(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.) CSEC Express as a single fraction: $\dfrac{3p}{2} + \dfrac{q}{p}$

$\dfrac{3p}{2} + \dfrac{q}{p} \\[5ex] \dfrac{3p(p)}{2p} + \dfrac{q(2)}{2p} \\[5ex] \dfrac{3p^2}{2p} + \dfrac{2q}{2p} \\[5ex] \dfrac{3p^2 + 2q}{2p}$
(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.) ACT For all real values of p and r, which of the following expressions is equivalent to $p(2 - r) + 8(p - r)$?

$F.\;\; 6p - 9r \\[3ex] G.\;\; 6p - 8r \\[3ex] H.\;\; 6p - 8r - pr \\[3ex] J.\;\; 10p - 2r \\[3ex] K.\;\; 10p - 8r - pr \\[3ex]$

$p(2 - r) + 8(p - r) \\[3ex] 2p - pr + 8p - 8r \\[3ex] 10p - 8r - pr$
(25.) NZQA Find the value of $3x^2 + 8 + 2y^2 + x^2 - 2$ when $x = 2$ and $y = -3$

$3x^2 + 8 + 2y^2 + x^2 - 2 \\[3ex] x = 2 \\[3ex] y = -3 \\[3ex] = 3(2)^2 + 8 + 2(-3)^2 + 2^2 - 2 \\[3ex] = 3(4) + 8 + 2(9) + 4 - 2 \\[3ex] = 12 + 8 + 18 + 4 - 2 \\[3ex] = 40$
(26.) Simplify the expression: $\sqrt{75x^8} + \sqrt{48x^8}$

$\sqrt{75x^8} + \sqrt{48x^8} \\[3ex] One\;\;at\;\;a\;\;time \\[3ex] \sqrt{75x^8} \\[3ex] = \sqrt{75} * \sqrt{x^8} \\[3ex] = \sqrt{25 * 3} * (x^8)^{\dfrac{1}{2}} \\[3ex] = \sqrt{25} * \sqrt{3} * x^4 \\[3ex] = 5 * x^4 * \sqrt{3} \\[3ex] = 5x^4\sqrt{3} \\[5ex] \sqrt{48x^8} \\[3ex] = \sqrt{48} * \sqrt{x^8} \\[3ex] = \sqrt{16 * 3} * (x^8)^{\dfrac{1}{2}} \\[3ex] = \sqrt{16} * \sqrt{3} * x^4 \\[3ex] = 4 * x^4 * \sqrt{3} \\[3ex] = 4x^4\sqrt{3} \\[5ex] \therefore \sqrt{75x^8} + \sqrt{48x^8} \\[3ex] = 5x^4\sqrt{3} + 4x^4\sqrt{3} \\[3ex] = 9x^4\sqrt{3}$
(27.) ACT For every cent decrease in price of a gallon of gasoline, a station sells 250 more gallons per day.
The station normally sells 3,000 gallons of gasoline a day at $1.09 per gallon. Which of the following expressions represents the number of gallons of gasoline sold per day if the cost is reduced by x cents per gallon?$ F.\;\; 3,000(1.09 - x) \\[3ex] G.\;\; 1.09(250x + 3,000) \\[3ex] H.\;\; (1.09 - x)(250x + 3,000) \\[3ex] J.\;\; 1.09 - x \\[3ex] K.\;\; 3,000 + 250x \\[3ex] $Normal sell: 3000 gallons @$1.09 per gallon
For every cent decrease in price, 250 more gallons per day was sold.
Decrease of 1 cent: For $1.09 - 0.01 = \$1.08$,$3000 + 250 = 3250$gallons per day was sold Decrease of 2 cents: For$1.09 - 2(0.01) = \$1.07$, $3000 + 2(250) = 3500$ gallons per sold was sold

Decrease of 3 cents: For $1.09 - 3(0.01) = \$1.06$,$3000 + 3(250) = 3750$gallons per sold was sold Decrease of x cents: Similary; for$1.09 - x(0.01) = \$1.09 - 0.01x$, $3000 + x(250) = 3000 + 250x$ gallons per sold was sold

The number of gallons of gasoline sold per day if the cost is reduced by x cents per gallon is $3000 + 250x$
(28.) NZQA A picture is framed using four rectangular pieces of wood, as shown in the diagram below.
Find the area of the picture, in terms of $x$, giving your answer in the form $ax^2 + bx + c$

$Length \\[3ex] = x + (7x + 3) \\[3ex] = x + 7x + 3 \\[3ex] = 8x + 3 \\[3ex] Width \\[3ex] = (5x + 2) + x \\[3ex] = 5x + 2 + x \\[3ex] = 6x + 2 \\[3ex] Area = Length * Width \\[3ex] = (8x + 3)(6x + 2) \\[3ex] = 48x^2 + 16x + 18x + 6 \\[3ex] = 48x^2 + 34x + 6$
(29.) Simplify the algebraic linear expressions

$(a.)\;\; 9a + 10(6a - 1) \\[3ex] (b.)\;\; -9(6m - 3) + 6(1 + 4m) \\[3ex]$

$(a.) \\[3ex] 9a + 10(6a - 1) \\[3ex] 9a + 60a - 10 \\[3ex] 69a - 10 \\[3ex] (b.) \\[3ex] -9(6m - 3) + 6(1 + 4m) \\[3ex] -54m + 27 + 6 + 24m \\[3ex] -30m + 33$
(30.) CSEC Find the value of $2\pi \sqrt{\dfrac{l}{g}}$ where $\pi = 3.14,\; l = 0.625$ and $g = 10$

$\pi = 3.14 \\[3ex] l = 0.625 \\[3ex] g = 10 \\[3ex] 2\pi \sqrt{\dfrac{l}{g}} \\[5ex] = 2 * 3.14 * \sqrt{\dfrac{0.625}{10}} \\[5ex] = 6.28 * \sqrt{0.0625} \\[3ex] = 6.28 * 0.25 \\[3ex] = 1.57$
(31.) If $\dfrac{x}{y} = 2$ and $\dfrac{y}{z} = 3$, find the value of $\dfrac{x + y}{y + z}$

$\dfrac{x}{y} = 2 \\[5ex] \implies x = 2y \\[3ex] \dfrac{y}{z} = 3 \\[5ex] \implies y = 3z \\[5ex] \dfrac{x + y}{y + z} \\[5ex] = \dfrac{2y + y}{3z + z} \\[5ex] = \dfrac{3y}{4z} \\[5ex] = \dfrac{3}{4} * \dfrac{y}{z} \\[5ex] = \dfrac{3}{4} * 3 \\[5ex] = \dfrac{9}{4}$
(32.) SQA National 5 Maths Expand and simplify $(x + 5)(2x^2 - 7x - 3)$

$(x + 5)(2x^2 - 7x - 3) \\[3ex] x(2x^2) = 2x^3 \\[3ex] x(-7x) = -7x^2 \\[3ex] x(-3) = -3x \\[3ex] 5(2x^2) = 10x^2 \\[3ex] 5(-7x) = -35x \\[3ex] 5(-3) = -15 \\[3ex] \implies \\[3ex] = 2x^3 - 7x^2 + 10x^2 - 3x - 35x - 15 \\[3ex] = 2x^3 + 3x^2 - 38x - 15$
(33.)

(34.) ACT Which of the following is always equal to $a(5 - a) - 6(a + 4)$?

$A.\;\; -2a - 24 \\[3ex] B.\;\; -2a + 4 \\[3ex] C.\;\; -a^2 - a - 24 \\[3ex] D.\;\; -a^2 - a + 4 \\[3ex] E.\;\; -2a^3 - 24 \\[3ex]$

$a(5 - a) - 6(a + 4) \\[3ex] 5a - a^2 - 6a - 24 \\[3ex] -a^2 - a - 24$
(35.)

$(a.)\;\; \sqrt{512k^2} \\[3ex] (b.)\;\; \sqrt{216k^4} \\[3ex] (c.)\;\; \sqrt{512m^3} \\[3ex]$

$(a.) \\[3ex] \sqrt{512k^2} \\[3ex] = \sqrt{512 \cdot k^2} \\[3ex] = \sqrt{256 \cdot 2 \cdot k^2}\\[3ex] = \sqrt{256} \cdot \sqrt{2} \cdot \sqrt{k^2} \\[3ex] = 16 \cdot \sqrt{2} \cdot (k^2)^{\dfrac{1}{2}}...Law\;7...Exp \\[5ex] = 16 \cdot \sqrt{2} \cdot k^{2 \cdot \dfrac{1}{2}}...Law\;5...Exp \\[5ex] = 16 \cdot \sqrt{2} \cdot k \\[3ex] = 16k\sqrt{2} \\[5ex] (b.) \\[3ex] \sqrt{216k^4} \\[3ex] = \sqrt{216 \cdot k^4} \\[3ex] = \sqrt{36 \cdot 6 \cdot k^4}\\[3ex] = \sqrt{36} \cdot \sqrt{6} \cdot \sqrt{k^4} \\[3ex] = 6 \cdot \sqrt{6} \cdot (k^4)^{\dfrac{1}{2}}...Law\;7...Exp \\[5ex] = 6 \cdot \sqrt{6} \cdot k^{4 \cdot \dfrac{1}{2}}...Law\;5...Exp \\[5ex] = 6 \cdot \sqrt{6} \cdot k^2 \\[3ex] = 6k^2\sqrt{6} \\[5ex] (c.) \\[3ex] \sqrt{512m^3} \\[3ex] = \sqrt{512 \cdot m^3} \\[3ex] = \sqrt{256 \cdot 2 \cdot m^2 \cdot m^1}\\[3ex] = \sqrt{256} \cdot \sqrt{2} \cdot \sqrt{m^2} \cdot \sqrt{m} \\[3ex] = 16 \cdot \sqrt{2} \cdot (m^2)^{\dfrac{1}{2}} \cdot \sqrt{m} ...Law\;7...Exp \\[5ex] = 16 \cdot \sqrt{2} \cdot m^{2 \cdot \dfrac{1}{2}} \cdot \sqrt{m} ...Law\;5...Exp \\[5ex] = 16 \cdot \sqrt{2} \cdot m \cdot \sqrt{m} \\[3ex] = 16 \cdot m \cdot \sqrt{2} \cdot \sqrt{m} \\[3ex] = 16m \cdot \sqrt{2 \cdot m} \\[3ex] = 16m\sqrt{2m}$
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(40.) ACT If x is any number other than 4 and 5, then $\dfrac{(4 - x)(x - 5)}{(x - 4)(x - 5)} = ?$

$A.\;\; -20 \\[2ex] B.\;\; -1 \\[3ex] C.\;\; 0 \\[3ex] D.\;\; 1 \\[3ex] E.\;\; 20 \\[3ex]$

$\dfrac{(4 - x)(x - 5)}{(x - 4)(x - 5)} \\[5ex] = \dfrac{4 - x}{x - 4} \\[5ex] = \dfrac{-x + 4}{x - 4} \\[5ex] = \dfrac{-1(x - 4)}{x - 4} \\[5ex] = - 1$

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