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

Word Problems on 3 × 3 Linear Systems

Calculators: Calculators

Prerequisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals, and Percents
(3.) Expressions and Equations

Notable Notes About Word Problems
(1.) Word problems are written in standard British English.
(2.) Some word problems are very lengthy and some are unnecessary. Those ones are meant to discourage you from even trying.
(3.) Word problems in Mathematics demonstrate the real-world applications of mathematical concepts.
(4.) Embrace word problems. See it as writing from "English to Math".
Take time to:
(a.) Read to understand. Paraphrase and shorten long sentences as necessary.
(b.) Re-read and note/underline the vocabulary Math terms written in English.
(c.) Translate/Write important sentences one at a time.
(d.) Review what you wrote to ensure correctness.
(e.) Solve the math, and check your solution in the word problem.
Does that solution makes sense?
If it does, you may be correct.
If it does not, please re-do it. For example, if you were asked to calculate the length of something and you get a negative number, then you will need to re-do it.

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

For WASSCE Students: Unless specified otherwise:
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For GCSE 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.

Solve all questions.
Use appropriate variables.
Indicate the method(s) used.
Show all work.
Interpret your solutions.

(1.) Love Thy Neighbor Landscaping bought 5 tons of topsoil, 4 tons of mulch, and 5 tons of pea gravel for $2740.
The next week the firm bought 4 tons of topsoil, 3 tons of mulch, and 6 tons of pea gravel for $2544.
Pea gravel costs $16 less per ton than topsoil.
Find the cost per ton for each item.

Let the cost per ton of topsoil be c
The cost per ton of mulch be d and
The cost per ton of pea gravel be e
Love Thy Neighbor Landscaping bought $5$ tons of topsoil, $4$ tons of mulch, and $5$ tons of pea gravel for $\$2740$

$5c + 4d + 5e = 2740 ...eqn.(1)$

The next week the firm bought $4$ tons of topsoil, $3$ tons of mulch, and $6$ tons of pea gravel for $\$2544$

$4c + 3d + 6e = 2544 ...eqn.(2)$

Pea gravel costs $\$16$ less per ton than topsoil.

$e = c - 16 ...eqn.(3)$

The three equations are:

$ 5c + 4d + 5e = 2740 ...eqn.(1) \\[3ex] 4c + 3d + 6e = 2544 ...eqn.(2) \\[3ex] e = c - 16 ...eqn.(3) \\[3ex] \underline{Substitution\:\: Method} \\[3ex] Subst.\:\: eqn.(3)\:\: into\:\: eqn.(1) \\[3ex] 5c + 4d + 5(c - 16) = 2740 \\[3ex] 5c + 4d + 5c - 80 = 2740 \\[3ex] 10c + 4d = 2740 + 80 \\[3ex] 10c + 4d = 2820 ..eqn.(4) \\[3ex] Subst.\:\: eqn.(3)\:\: into\:\: eqn.(2) \\[3ex] 4c + 3d + 6(c - 16) = 2544 \\[3ex] 4c + 3d + 6c - 96 = 2544 \\[3ex] 10c + 3d = 2544 + 96 \\[3ex] 10c + 3d = 2640 ...eqn.(5) \\[3ex] \underline{Elimination\:\: Method} \\[3ex] To\:\: find\:\: d, \:\:eliminate\:\: c \\[3ex] eqn.(4) - eqn.(5) \\[3ex] \implies (10c + 4d) - (10c + 3d) = 2820 - 2640 \\[5ex] 10c + 4d - 10c - 3d = 180 \\[5ex] d = 180 ...eqn.(6) \\[3ex] To\:\: find\:\: c, \:\:eliminate\:\: d \\[3ex] \underline{Substitution\:\: Method} \\[3ex] Subst.\:\: eqn.(6)\:\: into\:\: eqn.(5) \\[3ex] 10c + 3(180) = 2640 \\[3ex] 10c + 540 = 2640 \\[3ex] 10c = 2640 - 540 \\[3ex] 10c = 2100 \\[3ex] c = \dfrac{2100}{10} \\[5ex] c = 210 ...eqn.(7) \\[3ex] Subst.\:\: eqn.(7)\:\: into\:\: eqn.(3) \\[3ex] e = 210 - 16 \\[3ex] e = 194 \\[3ex] $ This means that the cost per ton of topsoil is $\$210$
The cost per ton of mulch is $\$180$ and
The cost per ton of pea gravel is $\$194$

(4.) ACT What is the value of b in the solution to the system of equations below?

$ 3a - 2b = 21 \\[3ex] a + 3b = -4 \\[5ex] A.\:\: 17 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: 5 \\[3ex] D.\:\: -3 \\[3ex] E.\:\: -5 \\[3ex] $

$ 3a - 2b = 21 ...eqn.(1) \\[3ex] a + 3b = -4 ...eqn.(2) \\[3ex] From\;\;eqn.(2) \\[3ex] a = -4 - 3b \\[3ex] Substitute\;\;(-4 - 3b)\;\;for\;\;a\;\;in\;\;eqn.(1) \\[3ex] 3(-4 - 3b) - 2b = 21 \\[3ex] -12 - 9b - 2b = 21 \\[3ex] -12 - 21 = 9b + 2b \\[3ex] -33 = 11b \\[3ex] 11b = -33 \\[3ex] b = -\dfrac{33}{11} \\[5ex] b = -3 $

(5.) ACT Which of the following (x, y) pairs is the solution for the system of equations: x + 2y = 4 and −2x + y = 7?

$ A.\:\: (-2, 3) \\[3ex] B.\:\: (-1.2, 5) \\[3ex] C.\:\: (1, 1.5) \\[3ex] D.\:\: (2, 1) \\[3ex] E.\:\: (4, 0) \\[3ex] $

We can solve this question using at least two methods.

First Method: Test the Options
LHS = RHS for both equations

LHS	RHS
$ (-2, 3) \implies x = -2, \;\;\; y = 3 $ $ x + 2y \\[3ex] -2 + 2(3) \\[3ex] -2 + 6 \\[3ex] 4 \checkmark $ $ -2x + y \\[3ex] -2(-2) + 3 \\[3ex] 4 + 3 \\[3ex] 7 \checkmark $	4 7 Option A is YES
$ (-1.2, 5) \implies x = -1.2, \;\;\; y = 5 $ $ x + 2y \\[3ex] -1.2 + 2(5) \\[3ex] -1.2 + 10 \\[3ex] 8.8 \large\unicode{x2718} $ $ -2x + y \\[3ex] -2(-1.2) + 5 \\[3ex] 2.4 + 5 \\[3ex] 7.4 \large\unicode{x2718} $	4 7 Option B is NO
$ (1, 1.5) \implies x = 1, \;\;\; y = 1.5 $ $ x + 2y \\[3ex] 1 + 2(1.5) \\[3ex] 1 + 3 \\[3ex] 4 \checkmark $ $ -2x + y \\[3ex] -2(1) + 1.5 \\[3ex] -2 + 1.5 \\[3ex] -0.5 \large\unicode{x2718} $	4 7 Option C is NO
$ (2, 1) \implies x = 2, \;\;\; y = 1 $ $ x + 2y \\[3ex] 2 + 2(1) \\[3ex] 2 + 2 \\[3ex] 4 \checkmark $ $ -2x + y \\[3ex] -2(2) + 1 \\[3ex] -4 + 1 \\[3ex] -3 \large\unicode{x2718} $	4 7 Option D is NO
$ (4, 0) \implies x = 4, \;\;\; y = 0 $ $ x + 2y \\[3ex] 4 + 2(0) \\[3ex] 4 + 0 \\[3ex] 4 \checkmark $ $ -2x + y \\[3ex] -2(4) + 0 \\[3ex] -8 \large\unicode{x2718} $	4 7 Option E is NO

Second Method: Solution by Substitution

$ x + 2y = 4 ...eqn.(1) \\[3ex] -2x + y = 7 ...eqn.(2) \\[3ex] From\;\;eqn.(1) \\[3ex] x = 4 - 2y ...eqn.(3) \\[3ex] Substitute\;\;(4 - 2y)\;\;for\;\;x\;\;in\;\;eqn.(2) \\[3ex] -2(4 - 2y) + y = 7 \\[3ex] -8 + 4y + y = 7 \\[3ex] 5y = 7 + 8 \\[3ex] 5y = 15 \\[3ex] y = \dfrac{15}{5} \\[5ex] y = 3 \\[3ex] Substitute\;\;3\;\;for\;\;y\;\;in\;\;eqn.(3) \\[3ex] x = 4 - 2(3) \\[3ex] x = 4 - 6 \\[3ex] x = -2 \\[3ex] (x, y) = (-2, 3) $

(7.) ACT Ms. Johnson purchased 1000 prizes for the school carnival for $90.
Each prize costs either $0.05 or $0.25.
How many of the less expensive prizes did she buy?

$ F.\:\: 200 \\[3ex] G.\:\: 360 \\[3ex] H.\:\: 500 \\[3ex] J.\:\: 640 \\[3ex] K.\:\: 800 \\[3ex] $

Let the:
Less expensive prize be x
More expensive prize be y
Ms. Johnson purchased 1,000 prizes

$x + y = 1000 ...eqn.(1)$

Less expensive prize goes with $\$0.05$ and the More expensive prize goes with $\$0.25$
This is because $0.25 \gt 0.05$
Ms. Johnson purchased these prizes for $\$90$

$0.05x + 0.25y = 90 ...eqn.(2)$

The two equations are:

$ x + y = 1000 ...eqn.(1) \\[3ex] 0.05x + 0.25y = 90 ...eqn.(2) \\[3ex] \underline{Substitution\:\:Method} \\[3ex] From\:\:eqn.(1) \\[3ex] y = 1000 - x...eqn.(3) \\[3ex] Substitute\:\:eqn.(3)\:\:into\:\:eqn.(2) \\[3ex] 0.05x + 0.25(1000 - x) = 90 \\[3ex] 0.05x + 250 - 0.25x = 90 \\[3ex] -0.2x = 90 - 250 \\[3ex] -0.2x = -160 \\[3ex] x = \dfrac{-160}{-0.2} \\[5ex] x = 800 \\[3ex] $ Ms. Johnson purchased $800$ prizes of the $\$0.05$ prize

(10.) Two planes travel toward each other from cities that are about 1725km apart.
They travel at rates of 360km/hr and 330km/hr.
They started at the same time.
In how many hours will they meet?

The two planes started at the same time.
Hence, their times are equal.
Let the two planes be Plane A and Plane B
Let the speed of Plane A = 360 km/hr
and the speed of Plane B = 330 km/hr
Let the time be $t$

$ Speed = \dfrac{distance}{time} \\[5ex] \therefore distance = speed * time \\[3ex] $ Distance traveled by Plane A = $360 * t = 360t$
Distance traveled by Plane B = $330 * t = 330t$
The sum of the distance should be the total distance

$ 360t + 330t = 1725 \\[3ex] 690t = 1725 \\[3ex] t = \dfrac{1725}{690} \\[5ex] t = \dfrac{5}{2}\: hours $

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