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

Solve all questions.
Check your solutions as applicable.
If there is no solution, please state so, and state your reason(s).
If there are many solutions, please state so, and state your reason(s).

Determine whether the ordered pairs are solutions of the equations.
(1.) $-7x - 3y = 15$
Ordered pair = $(-3, 2)$


$(-3, 2)$ implies that $x = -3, y = 2$
The ordered pair is a solution if the LHS = RHS when those values are substituted in the equation.

$ -7x - 3y = 15 \\[3ex] \underline{LHS} \\[3ex] -7x - 3y \\[3ex] Substitute\:\:the\:\:variables\:\:with\:\:those\:\:values \\[3ex] -7(-3) - 3(2) \\[3ex] 21 - 6 \\[3ex] 15 \\[3ex] \underline{RHS} \\[3ex] 15 \\[3ex] LHS = RHS \\[3ex] \therefore (-3, 2)\:\:is\:\:a\:\:solution. $
Determine whether the ordered pairs are solutions of the equations.
(2.) $9x - 8y = 5$
Ordered pair = $\left(\dfrac{5}{3}, \dfrac{3}{4}\right)$


$\left(\dfrac{5}{3}, \dfrac{3}{4}\right)$ implies that $x = \dfrac{5}{3}, y = \dfrac{3}{4}$
The ordered pair is a solution if the LHS = RHS when those values are substituted in the equation.

$ 9x - 8y = 5 \\[3ex] \underline{LHS} \\[3ex] 9x - 8y \\[3ex] Substitute\:\:the\:\:variables\:\:with\:\:those\:\:values \\[3ex] 9\left(\dfrac{5}{3}\right) - 8\left(\dfrac{3}{4}\right) \\[5ex] 3(5) - 2(3) \\[3ex] 15 - 6 \\[3ex] 9 \\[3ex] \underline{RHS} \\[3ex] 5 \\[3ex] LHS \ne RHS \\[3ex] \therefore \left(\dfrac{5}{3}, \dfrac{3}{4}\right)\:\:is\:\:not\:\:a\:\:solution. $
Determine whether the ordered pairs are solutions of the equations.
(3.) $7c + 5d = 1$
Ordered pair = $\left(-\dfrac{6}{7}, -\dfrac{2}{5}\right)$


$\left(-\dfrac{6}{7}, -\dfrac{2}{5}\right)$ implies that $c = -\dfrac{6}{7}, d = -\dfrac{2}{5}$
The ordered pair is a solution if the LHS = RHS when those values are substituted in the equation.

$ 7c + 5d = 1 \\[3ex] \underline{LHS} \\[3ex] 7c + 5d \\[3ex] Substitute\:\:the\:\:variables\:\:with\:\:those\:\:values \\[3ex] 7\left(-\dfrac{6}{7}\right) + 5\left(-\dfrac{2}{5}\right) \\[5ex] 1(-6) + 1(-2) \\[3ex] -6 - 2 \\[3ex] -8 \\[3ex] \underline{RHS} \\[3ex] 1 \\[3ex] LHS \ne RHS \\[3ex] \therefore \left(-\dfrac{6}{7}, -\dfrac{2}{5}\right)\:\:is\:\:not\:\:a\:\:solution. $
(4.) $3d = 9$


$ 3d = 9 \\[3ex] d = \dfrac{9}{3} \\[5ex] d = 3 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 3d \\[3ex] d = 3 \\[3ex] 3 * 3 \\[3ex] 9 $ $ \underline{RHS} \\[3ex] 9 $
(5.) $d + 7 = 9$


$ d + 7 = 9 \\[3ex] d = 9 - 7 \\[3ex] d = 2 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] d + 7 \\[3ex] d = 2 \\[3ex] 2 + 7 \\[3ex] 9 $ $ \underline{RHS} \\[3ex] 9 $
(6.) $d - 7 = 9$


$ d - 7 = 9 \\[3ex] d = 9 + 7 \\[3ex] d = 16 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] d - 7 \\[3ex] d = 16 \\[3ex] 16 - 7 \\[3ex] 9 $ $ \underline{RHS} \\[3ex] 9 $
(7.) ACT What is the solution set of the equation $x + 6 = 2(x + 3) - x$?

$ A.\:\: The\:\: empty\:\: set(no\:\: solution) \\[3ex] B.\:\: \{0\} \\[3ex] C.\:\: \{2\} \\[3ex] D.\:\: \{3\} \\[3ex] E.\:\: The\:\: set\:\: of\:\: all\:\: real\:\: numbers $


$ x + 6 = 2(x + 3) - x \\[3ex] x + 6 = 2x + 6 - x \\[3ex] x + 6 = x + 6 \\[3ex] x - x = 6 - 6 \\[3ex] 0 = 0 \\[3ex] $ There are many solutions
The solution is the set of all real numbers
(8.) ACT If $\dfrac{3}{5}x + 10 = 17$, then $x = ?$

$ F.\:\: -\dfrac{35}{3} \\[5ex] G.\:\: \dfrac{5}{3} \\[5ex] H.\:\: \dfrac{35}{3} \\[5ex] J.\:\: \dfrac{21}{5} \\[5ex] K.\:\: 45 \\[3ex] $

$ \dfrac{3}{5}x + 10 = 17 \\[5ex] \dfrac{3}{5} x = 17 - 10 \\[5ex] \dfrac{3x}{5} = 7 \\[5ex] \dfrac{3x}{5} = \dfrac{7}{1} \\[5ex] Cross\:\:Multiply \\[3ex] 3x(1) = 5(7) \\[3ex] 3x = 35 \\[3ex] x = \dfrac{35}{3} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] \dfrac{3}{5}x + 10 \\[5ex] x = \dfrac{35}{3} \\[5ex] 3 * 3 \\[3ex] \dfrac{3}{5} * \dfrac{35}{3} + 10 \\[5ex] 7 + 10 \\[3ex] 17 $ $ \underline{RHS} \\[3ex] 17 $
(9.) $\dfrac{d}{3} = 9$


$ \dfrac{d}{3} = 9 \\[5ex] LCD = 3 \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] 3 * \dfrac{d}{3} = 3 * 9 \\[5ex] d = 27 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] \dfrac{d}{3} \\[5ex] d = 27 \\[3ex] \dfrac{27}{3} \\[5ex] 9 $ $ \underline{RHS} \\[3ex] 9 $
(10.) $6y + 10 = 52$


$ 6y + 10 = 52 \\[3ex] 6y = 52 - 10 \\[3ex] 6y = 42 \\[3ex] y = 7 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 6y + 10 \\[3ex] y = 7 \\[3ex] 6(7) + 10 \\[3ex] 42 + 10 \\[3ex] 52 $ $ \underline{RHS} \\[3ex] 52 $
(11.) $\dfrac{2x}{5} - x = \dfrac{x}{10} - \dfrac{35}{2}$


$ \dfrac{2x}{5} - x = \dfrac{x}{10} - \dfrac{35}{2} \\[5ex] LCD = 10 \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] 10 * \dfrac{2x}{5} - 10 * x = 10 * \dfrac{x}{10} - 10 * \dfrac{35}{2} \\[5ex] 2(2x) - 10x = x - 5(35) \\[3ex] 4x - 10x - x = -175 \\[3ex] -7x = -175 \\[3ex] x = 25 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] \dfrac{2x}{5} - x \\[5ex] x = 25 \\[3ex] \dfrac{2 * 25}{5} - 25 \\[5ex] 10 - 25 \\[3ex] -15 $ $ \underline{RHS} \\[3ex] \dfrac{x}{10} - \dfrac{35}{2} \\[5ex] x = 25 \\[3ex] \dfrac{25}{10} - \dfrac{35}{2} \\[5ex] \dfrac{5}{2} - \dfrac{35}{2} \\[5ex] \dfrac{5 - 35}{2} \\[5ex] \dfrac{-30}{2} \\[5ex] -15 $
(12.) $\dfrac{2y}{5} - 1 = \dfrac{1}{4}$


$ \dfrac{2y}{5} - 1 = \dfrac{1}{4} \\[5ex] \dfrac{2y}{5} = \dfrac{1}{4} + 1 \\[5ex] \dfrac{2y}{5} = \dfrac{1}{4} + \dfrac{4}{4} \\[5ex] \dfrac{2y}{5} = \dfrac{1 + 4}{4} \\[5ex] \dfrac{2y}{5} = \dfrac{5}{4} \\[5ex] Cross\:\:Multiply \\[3ex] 2y * 4 = 5 * 5 \\[3ex] 8y = 25 \\[3ex] y = \dfrac{25}{8} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] \dfrac{2y}{5} - 1 \\[5ex] y = \dfrac{25}{8} \\[5ex] \dfrac{2 * \dfrac{25}{8}}{5} - 1 \\[5ex] \dfrac{50}{8} \div 5 - 1 \\[5ex] \dfrac{50}{8} * \dfrac{1}{5} - 1 \\[5ex] \dfrac{10}{8} - 1 \\[5ex] \dfrac{10}{8} - \dfrac{8}{8} \\[5ex] \dfrac{10 - 8}{8} \\[5ex] \dfrac{2}{8} \\[5ex] \dfrac{1}{4} $ $ \underline{RHS} \\[3ex] \dfrac{1}{4} $
(13.) $9m - 1 + 7m = 4m + 2 - 7m$


$ 9m - 1 + 7m = 4m + 2 - 7m \\[3ex] 9m + 7m - 1 = 4m - 7m + 2 \\[3ex] 16m - 1 = -3m + 2 \\[3ex] 16m + 3m = 2 + 1 \\[3ex] 19m = 3 \\[3ex] m = \dfrac{3}{19} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 9m - 1 + 7m \\[3ex] m = \dfrac{3}{19} \\[5ex] 9 * \dfrac{3}{19} - 1 + 7 * \dfrac{3}{19} \\[5ex] \dfrac{27}{19} - \dfrac{19}{19} + \dfrac{21}{19} \\[5ex] \dfrac{27 - 19 + 21}{19} \\[5ex] \dfrac{29}{19} $ $ \underline{RHS} \\[3ex] 4m + 2 - 7m \\[3ex] m = \dfrac{3}{19} \\[5ex] 4 * \dfrac{3}{19} + 2 - 7 * \dfrac{3}{19} \\[5ex] \dfrac{12}{19} + \dfrac{38}{19} - \dfrac{21}{19} \\[5ex] \dfrac{12 + 38 - 21}{19} \\[5ex] \dfrac{29}{19} $
(14.) $7 - p = 14$


$ 7 - p = 14 \\[3ex] 7 - 14 = p \\[3ex] -7 = p \\[3ex] p = -7 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 7 - p \\[3ex] p = -7 \\[3ex] 7 - (-7) \\[3ex] 7 + 7 \\[3ex] 14 $ $ \underline{RHS} \\[3ex] 14 $
(15.) $2p - [p - (3p - (6p + 9))] = 8p - 1$


$ 2p - [p - (3p - (6p + 9))] = 8p - 1 \\[3ex] 2p - [p - (3p - 6p - 9)] = 8p - 1 \\[3ex] 2p - [p - (-3p - 9)] = 8p - 1 \\[3ex] 2p - [p + 3p + 9] = 8p - 1 \\[3ex] 2p - [4p + 9] = 8p - 1 \\[3ex] 2p - 4p - 9 = 8p - 1 \\[3ex] -2p - 9 = 8p - 1 \\[3ex] -2p - 8p = -1 + 9 \\[3ex] -10p = 8 \\[3ex] p = \dfrac{8}{-10} \\[5ex] p = -\dfrac{8}{10} \\[5ex] p = -\dfrac{4}{5} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 2p - [p - (3p - (6p + 9))] \\[3ex] p = -\dfrac{4}{5} \\[5ex] 2 * -\dfrac{4}{5} - \left[-\dfrac{4}{5} - \left(3 * -\dfrac{4}{5} - \left(6 * -\dfrac{4}{5} + 9\right)\right)\right] \\[5ex] -\dfrac{8}{5} - \left[-\dfrac{4}{5} - \left(-\dfrac{12}{5} - \left(-\dfrac{24}{5} + 9\right)\right)\right] \\[5ex] -\dfrac{8}{5} - \left[-\dfrac{4}{5} - \left(-\dfrac{12}{5} + \dfrac{24}{5} - 9\right)\right] \\[5ex] -\dfrac{8}{5} - \left[-\dfrac{4}{5} - \left(-\dfrac{12}{5} + \dfrac{24}{5} - \dfrac{45}{5}\right)\right] \\[5ex] -\dfrac{8}{5} - \left[-\dfrac{4}{5} + \dfrac{12}{5} - \dfrac{24}{5} + \dfrac{45}{5}\right] \\[5ex] -\dfrac{8}{5} + \dfrac{4}{5} - \dfrac{12}{5} + \dfrac{24}{5} - \dfrac{45}{5} \\[5ex] \dfrac{-8 + 4 - 12 + 24 - 45}{5} \\[5ex] -\dfrac{37}{5} $ $ \underline{RHS} \\[3ex] 8p - 1 \\[3ex] p = -\dfrac{4}{5} \\[5ex] 8 * -\dfrac{4}{5} - 1 \\[5ex] -\dfrac{32}{5} - \dfrac{5}{5} \\[5ex] \dfrac{-32 - 5}{5} \\[5ex] -\dfrac{37}{5} $
(16.) ACT Given $\dfrac{3}{x} = 12$ and $\dfrac{x}{y} = 2$, what is the value of $y$?


$ \dfrac{3}{x} = 12 \\[5ex] Cross\:\:Multiply \\[3ex] \dfrac{3}{x} = 12 \\[5ex] 3 = 12x \\[3ex] 12x = 3 \\[3ex] x = \dfrac{3}{12} = \dfrac{1}{4} \\[5ex] \dfrac{x}{y} = 2 \\[5ex] x = 2y \\[3ex] 2y = x \\[3ex] y = \dfrac{x}{2} \\[5ex] y = x \div 2 \\[3ex] x = \dfrac{1}{4} \\[5ex] y = \dfrac{1}{4} \div 2 \\[5ex] y = \dfrac{1}{4} * \dfrac{1}{2} \\[5ex] y = \dfrac{1 * 1}{4 * 2} \\[5ex] y = \dfrac{1}{8} $
(17.) $5(p + 1) = 3 - 2(5p + 4)$


$ 5(p + 1) = 3 - 2(5p + 4) \\[3ex] Use\:\:Distributive\:\:Property \\[3ex] 5p + 5 = 3 - 10p - 8 \\[3ex] 5p + 10p = 3 - 8 - 5 \\[3ex] 15p = -10 \\[3ex] p = \dfrac{-10}{15} \\[5ex] p = -\dfrac{2}{3} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 5(p + 1) \\[3ex] p = -\dfrac{2}{3} \\[5ex] 5\left(-\dfrac{2}{3} + 1\right) \\[5ex] 5\left(-\dfrac{2}{3} + \dfrac{3}{3}\right) \\[5ex] 5\left(\dfrac{-2 + 3}{3}\right) \\[5ex] 5\left(\dfrac{1}{3}\right) \\[5ex] \dfrac{5}{3} $ $ \underline{RHS} \\[3ex] 3 - 2(5p + 4) \\[3ex] p = -\dfrac{2}{3} \\[5ex] 3 - 2\left(5\left(-\dfrac{2}{3}\right) + 4\right) \\[5ex] 3 - 2\left(-\dfrac{10}{3} + 4\right) \\[5ex] 3 - 2\left(-\dfrac{10}{3} + \dfrac{12}{3}\right) \\[5ex] 3 - 2\left(\dfrac{-10 + 12}{3}\right) \\[5ex] 3 - 2\left(\dfrac{2}{3}\right) \\[5ex] 3 - \dfrac{4}{3} \\[5ex] \dfrac{9}{3} - \dfrac{4}{3} \\[5ex] \dfrac{9 - 4}{3} \\[5ex] \dfrac{5}{3} $
(18.) $4k - 7 = 2k - 1$


$ 4k - 7 = 2k - 1 \\[3ex] 4k - 2k = -1 + 7 \\[3ex] 2k = 6 \\[3ex] k = \dfrac{6}{2} \\[5ex] k = 3 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 4k - 7 \\[3ex] k = 3 \\[3ex] 4(3) - 7 \\[3ex] 12 - 7 \\[3ex] 5 $ $ \underline{RHS} \\[3ex] 2k - 1 \\[3ex] k = 3 \\[3ex] 2(3) - 1 \\[3ex] 6 - 1 \\[3ex] 5 $
(19.) ACT Consider the equation $k = \dfrac{7}{5}j + 54$.

For what value of $j$ is the value of $k$ equal to $40$?


$ k = \dfrac{7}{5}j + 54 \\[5ex] k = 40 \\[3ex] 40 = \dfrac{7}{5}j + 54 \\[5ex] \dfrac{7}{5}j + 54 = 40 \\[5ex] \dfrac{7}{5}j = 40 - 54 \\[5ex] \dfrac{7}{5}j = -14 \\[5ex] 7j = -14(5) \\[5ex] 7j = -70 \\[3ex] j = -\dfrac{70}{7} \\[5ex] j = -10 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] k \\[3ex] 40 $ $ \underline{RHS} \\[3ex] \dfrac{7}{5}j + 54 \\[5ex] j = -10 \\[3ex] \dfrac{7}{5} * -10 + 54 \\[5ex] = 7(-2) + 54 \\[3ex] = -14 + 54 \\[3ex] = 40 $
(20.) $5g + 36 = g$


$ 5g + 36 = g \\[3ex] 5g - g = -36 \\[3ex] 4g = -36 \\[3ex] g = -\dfrac{36}{4} \\[5ex] g = -9 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 5g + 36 \\[3ex] g = -9 \\[3ex] 5(-9) + 36 \\[3ex] -45 + 36 \\[3ex] -9 $ $ \underline{RHS} \\[3ex] g \\[3ex] -9 $




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(21.) ACT What is the solution to the equation below?
$3(x - 4) - 2(x - 3) = 5(-x - 3) + 6$


$ 3(x - 4) - 2(x - 3) = 5(-x - 3) + 6 \\[3ex] Use\:\:Distributive\:\:Property \\[3ex] 3x - 12 - 2x + 6 = -5x - 15 + 6 \\[3ex] x - 6 = -5x - 9 \\[3ex] x + 5x = -9 + 6 \\[3ex] 6x = -3 \\[3ex] x = -\dfrac{3}{6} \\[5ex] x = -\dfrac{1}{2} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 3(x - 4) - 2(x - 3) \\[3ex] x = -\dfrac{1}{2} \\[5ex] -\dfrac{1}{2} - 4 = -\dfrac{1}{2} - \dfrac{8}{2} = \dfrac{-1 - 8}{2} = -\dfrac{9}{2} \\[5ex] -\dfrac{1}{2} - 3 = -\dfrac{1}{2} - \dfrac{6}{2} = \dfrac{-1 - 6}{2} = -\dfrac{7}{2} \\[5ex] = 3\left(-\dfrac{9}{2}\right) - 2\left(-\dfrac{7}{2}\right) \\[5ex] = -\dfrac{27}{2} + \dfrac{14}{2} \\[5ex] = \dfrac{-27 + 14}{2} \\[5ex] = -\dfrac{13}{2} $ $ \underline{RHS} \\[3ex] 5(-x - 3) + 6 \\[3ex] x = -\dfrac{1}{2} \\[5ex] -x = -1 * x = -1 * -\dfrac{1}{2} = \dfrac{1}{2} \\[5ex] -x - 3 = \dfrac{1}{2} - 3 = \dfrac{1}{2} - \dfrac{6}{2} = \dfrac{1 - 6}{2} = -\dfrac{5}{2} \\[5ex] = 5\left(-\dfrac{5}{2}\right) + 6 \\[5ex] = -\dfrac{25}{2} + \dfrac{12}{2} \\[5ex] = \dfrac{-25 + 12}{2} \\[5ex] = -\dfrac{13}{2} $
(22.) ACT For which of the equations below is its solution an integer?

$ I.\:\: 3n + 5 = 24 \\[3ex] II.\:\: 5n + 3 = 23 \\[3ex] III.\:\: 5(n + 3) = 25 \\[5ex] A.\:\: I\:\: only \\[3ex] B.\:\: II\:\: only \\[3ex] C.\:\: III\:\: only \\[3ex] D.\:\: I \:\:and\:\: II\:\: only \\[3ex] E.\:\: II \:\:and\:\: III\:\: only $


An integer is any whole number or its opposite

$ 3n + 5 = 24 \\[3ex] 3n = 24 - 5 \\[3ex] 3n = 19 \\[3ex] n = \dfrac{19}{3}...NO \\[5ex] 5n + 3 = 23 \\[3ex] 5n = 23 - 3 \\[3ex] 5n = 20 \\[3ex] n = \dfrac{20}{5} \\[5ex] n = 4 ...YES \\[3ex] 5(n + 3) = 25 \\[3ex] 5n + 15 = 25 \\[3ex] 5n = 25 - 15 \\[3ex] 5n = 10 \\[3ex] n = \dfrac{10}{5} \\[5ex] n = 2 ...YES \\[3ex] II \:\:and\:\: III\:\: only $
(23.) $0.3p + 0.3(20) = 0.1(p + 40)$


$ 0.3p + 0.3(20) = 0.1(p + 40) \\[3ex] 0.3p + 6 = 0.1p + 4 \\[3ex] 0.3p - 0.1p = 4 - 6 \\[3ex] 0.2p = -2 \\[3ex] p = \dfrac{-2}{0.2} \\[5ex] p = -10 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 0.3p + 0.3(20) \\[3ex] p = -10 \\[3ex] 0.3(-10) + 6 \\[3ex] -3 + 6 \\[3ex] 3 $ $ \underline{RHS} \\[3ex] 0.1(p + 40) \\[3ex] p = -10 \\[3ex] 0.1(-10 + 40) \\[3ex] 0.1(30) \\[3ex] 3 $
(24.) $5v - 1 = 23 - v$


$ 5v - 1 = 23 - v \\[3ex] 5v + v = 23 + 1 \\[3ex] 6v = 24 \\[3ex] v = \dfrac{24}{6} \\[5ex] v = 4 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 5v - 1 \\[3ex] v = 4 \\[3ex] 5(4) - 1 \\[3ex] 20 - 1 \\[3ex] 19 $ $ \underline{RHS} \\[3ex] 23 - v \\[3ex] v = 4 \\[3ex] 23 - 4 \\[3ex] 19 $
(25.) $2(4x + 1) = 8 - 3(x - 5)$


$ 2(4x + 1) = 8 - 3(x - 5) \\[3ex] 8x + 2 = 8 - 3x + 15 \\[3ex] 8x + 3x = 8 + 15 - 2 \\[3ex] 11x = 21 \\[3ex] x = \dfrac{21}{11} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 2(4x + 1) \\[3ex] x = \dfrac{21}{11} \\[5ex] 2\left(4 * \dfrac{21}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11}\ + \dfrac{11}{11}\right) \\[5ex] 2\left(\dfrac{84 + 11}{11}\right) \\[5ex] 2\left(\dfrac{95}{11}\right) \\[5ex] \dfrac{190}{11} $ $ \underline{RHS} \\[3ex] 8 - 3(x - 5) \\[3ex] x = \dfrac{21}{11} \\[5ex] 8 - 3\left(\dfrac{21}{11} - 5\right) \\[5ex] 8 - 3\left(\dfrac{21}{11} - \dfrac{55}{11}\right) \\[5ex] 8 - 3\left(\dfrac{21 - 55}{11}\right) \\[5ex] 8 - 3\left(-\dfrac{34}{11}\right) \\[5ex] 8 - -\dfrac{102}{11} \\[5ex] 8 + \dfrac{102}{11} \\[5ex] \dfrac{88}{11} + \dfrac{102}{11} \\[5ex] \dfrac{88 + 102}{11} \\[5ex] \dfrac{190}{11} $
(26.) $4 - m = 2m + 16$


$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 4 - m \\[3ex] m = -4 \\[3ex] 4 - (-4) \\[3ex] 4 + 4 \\[3ex] 8 $ $ \underline{RHS} \\[3ex] 2m + 16 \\[3ex] m = -4 \\[3ex] 2(-4) + 16 \\[3ex] -8 + 16 \\[3ex] 8 $
(27.) $4(3p + 2) - 7 = 3(p - 2)$


$ 4(3p + 2) - 7 = 3(p - 2) \\[3ex] 12p + 8 - 7 = 3p - 6 \\[3ex] 12p - 3p = -6 - 8 + 7 \\[3ex] 9p = -7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 4(3p + 2) - 7 \\[3ex] 4\left(3 * -\dfrac{7}{9} + 2\right) - 7 \\[5ex] 4\left(-\dfrac{21}{9} + \dfrac{18}{2}\right) - 7 \\[5ex] 4\left(\dfrac{-21 + 18}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{3}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{1}{3}\right) - 7 \\[5ex] -\dfrac{4}{3} - 7 \\[5ex] -\dfrac{4}{3} - \dfrac{21}{3} \\[5ex] \dfrac{-4 - 21}{3} \\[5ex] -\dfrac{25}{3} $ $ \underline{RHS} \\[3ex] 3(p - 2) \\[3ex] 3\left(-\dfrac{7}{9} - 2\right) \\[5ex] 3\left(-\dfrac{7}{9} - \dfrac{18}{9}\right) \\[5ex] 3\left(\dfrac{-7 - 18}{9}\right) \\[5ex] 3\left(-\dfrac{25}{9}\right) \\[5ex] -\dfrac{25}{3} $
(28.) ACT Given that $3x + 2 = 4$ and $2y + 6 = 5$, what is $x + y$?

$ F.\:\: -\dfrac{1}{2} \\[5ex] G.\:\: \dfrac{1}{6} \\[5ex] H.\:\: \dfrac{2}{3} \\[5ex] J.\:\: \dfrac{7}{6} \\[5ex] K.\:\: \dfrac{15}{2} \\[5ex] $

$ 3x + 2 = 4 \\[3ex] 3x = 4 - 2 \\[3ex] 3x = 2 \\[3ex] x = \dfrac{2}{3} \\[5ex] 2y + 6 = 5 \\[3ex] 2y = 5 - 6 \\[3ex] 2y = -1 \\[3ex] y = -\dfrac{1}{2} \\[5ex] x + y \\[3ex] = \dfrac{2}{3} + -\dfrac{1}{2} \\[5ex] = \dfrac{2}{3} - \dfrac{1}{2} \\[5ex] = \dfrac{4}{6} - \dfrac{3}{6} \\[5ex] = \dfrac{4 - 3}{6} \\[5ex] = \dfrac{1}{6} $
(29.) ACT Given $3x - 7 = 5x - 13$ is true, $x = ?$

$ A.\:\: -10 \\[3ex] B.\:\: -3 \\[3ex] C.\:\: -\dfrac{5}{2} \\[5ex] D.\:\: \dfrac{5}{2} \\[5ex] E.\:\: 3 \\[3ex] $

$ 3x - 7 = 5x - 13 \\[3ex] -7 + 13 = 5x - 3x \\[3ex] 6 = 2x \\[3ex] 2x = 6 \\[3ex] x = \dfrac{6}{2} \\[5ex] x = 3 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 3x - 7 \\[3ex] 3(3) - 7 \\[3ex] 9 - 7 \\[3ex] 2 $ $ \underline{RHS} \\[3ex] 5x - 13 \\[3ex] 5(3) - 13 \\[3ex] 15 - 13 \\[3ex] 2 $
(30.) JAMB If $-2$ is the solution of the equation $2x + 1 - 3c = 2c + 3x - 7$, find the value of $c$

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

$-2$ is a solution implies that $x = -2$

$ 2x + 1 - 3c = 2c + 3x - 7 \\[3ex] 2(-2) + 1 - 3c = 2c + 3(-2) - 7 \\[3ex] -4 + 1 - 3c = 2c - 6 - 7 \\[3ex] -3 - 3c = 2c - 13 \\[3ex] -3 + 13 = 2c + 3c \\[3ex] 10 = 5c \\[3ex] 5c = 10 \\[3ex] c = \dfrac{10}{5} \\[5ex] c = 2 $