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 - Translating Word Problems to Algebraic 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.

Translate each word problem from English to Math.
Use appropriate variables as applicable.
Do not solve.

(1.) ACT Jane and Margaret moved to Newcity at the same time several years ago and have lived there ever since.
Jane has lived there $\dfrac{1}{2}$ of her life, while Margaret has lived there $\dfrac{3}{5}$ of her life.
If $j$ represents Jane's present age, which of the following expressions represents Margaret's present age?

$ F.\:\: \dfrac{3j}{10} \\[5ex] G.\:\: \dfrac{j}{2} \\[5ex] H.\:\: \dfrac{5j}{6} \\[5ex] J.\:\: \dfrac{6j}{5} \\[5ex] K.\:\: 2j \\[3ex] $

(2.) CSEC Write the following statement as an algebraic expression.
The sum of a number and its multiplicative inverse is five times the number.


Let the number be $p$
Multiplicative inverse of $p = \dfrac{1}{p}$

$ p + \dfrac{1}{p} = 5p $
(3.) ACT Which of the following mathematical expressions is equivalent to the verbal expression "A number, $x$, squared is $39$ more than the product of $10$ and $x$"?

$ F\:\: 2x = 39 + 10x \\[3ex] G.\:\: 2x = 39x + 10x \\[3ex] H.\:\: x^2 = 39 - 10x \\[3ex] J.\:\: x^2 = 39 + x^{10} \\[3ex] K.\:\: x^2 = 39 + 10x \\[3ex] $

A number, $x$ means $x$

squared means $x^2$

is means $=$

product of $10$ and $x$ means $10 * x = 10x$

$39$ more than the product of $10$ and $x$ means $39 + 10x$

A number, $x$, squared is $39$ more than the product of $10$ and $x$ means $x^2 = 39 + 10x$
(4.) Nahum bought $y$ children's admission tickets for $\$2$ each and $x$ adult's admission tickets for $\$7$ each.
Write an algebraic expression for the total amount spent by Nahum.


$y$ children's admission tickets for $\$2$ each = $y * 2 = 2y$

$x$ adult's admission tickets for $\$7$ each = $x * 7 = 7x$

$ Total = 2y + 7x $
(5.)


$ \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.) Samson's debt is seven less than half of David's debt.
If $d$ represents David's debt, write an expression for Samson's debt.


$ David's\:\:debt = d \\[3ex] Half\:\:of\:\:David's\:\:debt = \dfrac{1}{2} * d = \dfrac{d}{2} \\[5ex] Seven\:\:less\:\:than\:\: \dfrac{d}{2} = \dfrac{d}{2} - 7 \\[5ex] Samson's\:\:debt = \dfrac{d}{2} - 7 $
(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.) $\sqrt{2p + 6} - \sqrt{7 - 2p} = 1$


$\sqrt{2p + 6} - \sqrt{7 - 2p} = 1$

To find the domain;

$ 2p + 6 \ge 0 \\[3ex] 2p \ge 0 - 6 \\[3ex] 2p \ge -6 \\[3ex] p \ge -\dfrac{6}{2} \\[5ex] p \ge -3 \\[3ex] Also: \\[3ex] 7 - 2p \ge 0 \\[3ex] 7 \ge 2p \\[3ex] 2p \le 7 \\[3ex] p \le \dfrac{7}{2} \\[5ex] Combine\:\: both\:\: (who\:\: satisfies\:\: both) \\[3ex] p \ge -3 \\[3ex] -3 \le p \\[3ex] p \le \dfrac{7}{2} \\[5ex] -3 \le p \le \dfrac{7}{2} \\[5ex] $ The values excluded from the domain are all real numbers less than $-3$ and all real numbers greater than $\dfrac{7}{2}$

$ D = \left[-3, \dfrac{7}{2}\right] \\[5ex] \sqrt{2p + 6} - \sqrt{7 - 2p} = 1 \\[3ex] \sqrt{2p + 6} = 1 + \sqrt{7 - 2p} \\[3ex] Square\:\: both\:\: sides \\[3ex] \left(\sqrt{2p + 6}\right)^2 = \left(1 + \sqrt{7 - 2p}\right)^2 \\[3ex] 2p + 6 = (1 + \sqrt{7 - 2p})(1 + \sqrt{7 - 2p}) \\[3ex] (1 + \sqrt{7 - 2p})(1 + \sqrt{7 - 2p}) \\[3ex] (1)(1) = 1 \\[3ex] (1)(\sqrt{7 - 2p}) = \sqrt{7 - 2p} \\[3ex] \sqrt{7 - 2p}(1) = \sqrt{7 - 2p} \\[3ex] (\sqrt{7 - 2p})(\sqrt{7 - 2p}) = \left(\sqrt{7 - 2p}\right)^2 = 7 - 2p \\[3ex] \implies 2p + 6 = 1 + \sqrt{7 - 2p} + \sqrt{7 - 2p} + (7 - 2p) \\[3ex] 2p + 6 = 1 + 2\sqrt{7 - 2p} + 7 - 2p \\[3ex] 2p + 6 = 2\sqrt{7 - 2p} - 2p + 8 \\[3ex] 2\sqrt{7 - 2p} - 2p + 8 = 2p + 6 \\[3ex] 2\sqrt{7 - 2p} = 2p + 6 + 2p - 8 \\[3ex] 2\sqrt{7 - 2p} = 4p - 2 \\[3ex] 2\sqrt{7 - 2p} = 2(2p - 1) \\[3ex] Divide \:\:both\:\: sides\:\: 2 \\[3ex] \dfrac{2\sqrt{7 - 2p}}{2} = \dfrac{2(2p - 1)}{2} \\[5ex] \sqrt{7 - 2p} = 2p - 1 \\[3ex] Square\:\: both\:\: sides\:\: again \\[3ex] \left(\sqrt{7 - 2p}\right)^2 = (2p - 1)^2 \\[3ex] 7 - 2p = (2p - 1)(2p - 1) \\[3ex] 7 - 2p = 4p^2 - 2p - 2p + 1 \\[3ex] 7 - 2p = 4p^2 - 4p + 1 \\[3ex] 0 = 4p^2 - 4p + 1 - 7 + 2p \\[3ex] 0 = 4p^2 - 2p - 6 \\[3ex] 4p^2 - 2p - 6 = 0 \\[3ex] 2(2p^2 - p - 3) = 0 \\[3ex] Divide \:\:both\:\: sides\:\: 2 \\[3ex] \dfrac{2(2p^2 - p - 3)}{2} = \dfrac{0}{2} \\[3ex] 2p^2 - p - 3 = 0 \\[3ex] 2p^2 + 2p - 3p - 3 = 0 \\[3ex] 2p(p + 1) - 3(p + 1) = 0 \\[3ex] p + 1 = 0 \:\:OR\:\: 2p - 3 = 0 \\[3ex] p = -1 \:\:OR\:\: 2p = 3 \\[3ex] p = -1 \:\:OR\:\: p = \dfrac{3}{2} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] p = -1 \\[5ex] \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] \sqrt{2(-1) + 6} - \sqrt{7 - 2(-1)} \\[3ex] = \sqrt{-2 + 6} - \sqrt{7 + 2} \\[3ex] = \sqrt{4} - \sqrt{9} \\[3ex] = 2 - 3 \\[3ex] = -1 \\[3ex] $ $p = -1$ is NOT a solution.
$p = -1$ is an extraneous solution.


$ \sqrt{2p + 6} - \sqrt{7 - 2p} \\[3ex] p = \dfrac{3}{2} \\[5ex] \sqrt{2\left(\dfrac{3}{2}\right) + 6} - \sqrt{7 - 2\left(\dfrac{3}{2}\right)} \\[5ex] \sqrt{3 + 6} - \sqrt{7 - 3} \\[3ex] = \sqrt{9} - \sqrt{4} \\[3ex] = 3 - 2 \\[3ex] = 1 \\[3ex] $ $\color{black}{p = \dfrac{3}{2}}$ is a solution.
$ \underline{RHS} \\[3ex] 1 $
(10.) $\sqrt{p + 2} - 6 = 4$


$\sqrt{p + 2} - 6 = 4$

To find the domain;

$ p + 2 \ge 0 \\[3ex] p \ge -2 \\[3ex] $ The values excluded from the domain are all real numbers less than $-2$

$ D = [-2, \infty) \\[3ex] \sqrt{p + 2} - 6 = 4 \\[3ex] \sqrt{p + 2} = 4 + 6 \\[3ex] \sqrt{p + 2} = 10 \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{p + 2})^2 = 10^2 \\[3ex] p + 2 = 100 \\[3ex] p = 100 - 2 \\[3ex] p = 98 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] \sqrt{p + 2} - 6 \\[3ex] p = 98 \\[3ex] \sqrt{98 + 2} - 6 \\[3ex] = \sqrt{100} - 6 \\[3ex] = 10 - 6 \\[3ex] = 4 \\[3ex] $ $\color{black}{p = 98}$ is a solution $ \underline{RHS} \\[3ex] 4 $