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

Translate Three By Three Linear Systems

Calculators: Calculators

Prerequisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals, and Percents
(3.) Translate from English to Math

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.

Translate these word problems to Math.
Use appropriate variables.
Define your variables.
Do not solve unless you wish.

The names of the towns/cities/villages used in the questions are actual names.
But, I chose those names to make you smile/laugh 😊 while learning Math.

(1.) Godwin Constructions is building one-bedroom, two-bedroom, and three-bedroom units in their new apartment complex.
The number of two-bedroom and three-bedroom units is equal to the number of one-bedroom units.
The number of one-bedroom units is three times the number of three-bedroom units.
A total of $144$ units is to be built.
Find the number of one-bedroom, two-bedroom, and three-bedroom units to be built.

Let the number of one-bedroom units be c
The number of two-bedroom units be d and
The number of three-bedroom units be e
The number of two-bedroom and three-bedroom units is equal to the number of one-bedroom units.
$d + e = c ...eqn.(1)$
The number of one-bedroom units is three times the number of three-bedroom units.
$c = 3e ...eqn.(2)$
A total of 144 units is to be built.
$c + d + e = 144 ...eqn.(3)$

The three equations are:

$ d + e = c ...eqn.(1) \\[3ex] c = 3e ...eqn.(2) \\[3ex] c + d + e = 144 ...eqn.(3) $

(2.) Countries A, B, and C won a total of 126 medals in an international sports competition.
Country B won 8 more medals than Country C.
Country A won 34 more medals than the total amount won by the other two.
How many medals did each country win?

Let the number of medals won by Country A be A
The number of medals won by Country B be B and
The number of medals won by Country C be C
Countries A, B, and C won a total of 126 medals
$A + B + C = 126 ...eqn.(1)$
Country B won 8 more medals than Country C.
$B = 8 + C ...eqn.(2)$
Country A won 34 more medals than the total amount won by the other two.
$A = 34 + (B + C) ...eqn.(3)$

The three equations are:

$ A + B + C = 126 ...eqn.(1) \\[3ex] B = 8 + C ...eqn.(2) \\[3ex] A = 34 + B + C ...eqn.(3) $

(3.) Peter, James, and John can paint the exterior of a home in $8$ hours.
James and John have a painted a similar house together in $16$ hours.
One day, all three worked on this same kind of house for $6$ hours, after which John left.
Peter and James required $3$ more hours to finish.
Assume no gain or loss in efficiency, how long should it take each person to complete such a job alone?

Let the time (number of hours) required by Peter to do it alone be c
The time (number of hours) required by James to do it alone be d and
The time (number of hours) required by John to do it alone be e
Peter, James, and John can paint the exterior of a home
The exterior of a home is 1 job
Hourly rates (job per hour) for each of them:
Hourly rate for Peter = $\dfrac{1}{c}$

Hourly rate for James = $\dfrac{1}{d}$

Hourly rate for John = $\dfrac{1}{e}$

Hourly rate (job per hour) * number of hours = job
Peter, James, and John can paint the exterior of a home in 8 hours.
$8\left(\dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{e}\right) = 1 ...eqn.(1)$

James and John have a painted a similar house together in 16 hours.
$16\left(\dfrac{1}{d} + \dfrac{1}{e}\right) = 1 ...eqn.(2)$

One day, all three worked on this same kind of house for $6$ hours, after which John left.
Peter and James required $3$ more hours to finish.
This means that Peter, James, and John worked for 6 hours, and then Peter and James worked for 3 more hours to complete 1 job

$ 6\left(\dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{e}\right) + 3\left(\dfrac{1}{c} + \dfrac{1}{d}\right) = 1 ...eqn.(3) \\[5ex] \dfrac{6}{c} + \dfrac{6}{d} + \dfrac{6}{e} + \dfrac{3}{c} + \dfrac{3}{d} = 1 \\[5ex] \dfrac{9}{c} + \dfrac{9}{d} + \dfrac{6}{e} = 1 \\[5ex] $ The three equations are:

$ \dfrac{8}{c} + \dfrac{8}{d} + \dfrac{8}{e} = 1 ...eqn.(1) \\[5ex] \dfrac{16}{d} + \dfrac{16}{e} = 1 ...eqn.(2) \\[5ex] \dfrac{9}{c} + \dfrac{9}{d} + \dfrac{6}{e} = 1 ...eqn.(3) $

(4.) Zephaniah invested in stocks, bonds, and loans.
He had an annual return of 25% in stocks, 20% in bonds, and 30% in loans which amounted to $3000
His investment in loans was ten times his investment in stocks.
The interest earned from his investment in loans was equal to the dividends he received from his investment in bonds.
How much money did he invest in stocks? bonds? loans?

Annual Return is the same as Annual Interest
Investment is the same as Principal
Interest is the same as Dividends, and is also the same as Returns
Simple Interest Formula
Interest = Principal * Rate * Time
$I = P * r * t$
Let the investment in stocks be c
The investment in bonds be d and
The investment in loans be e

Stocks
Principal = $c$
Interest rate, r = 25% = $0.25$
Time, t = 1 year (annual means yearly)
Interest, I = $c * 0.25 * 1$ = $0.25c$

Bonds
Principal = $d$
Interest rate, r = 20% = $0.2$
Time, t = 1 year
Interest, I = $d * 0.2 * 1$ = $0.2d$

Loans
Principal = $e$
Interest rate, r = 30% = $0.3$
Time, t = 1 year
Interest, I = $e * 0.3 * 1$ = $0.3e$
He had an annual return of 25% in stocks, 20% in bonds, and 30% in loans which amounted to $$3000$.
$0.25c + 0.2d + 0.3e = 3000 ...eqn.(1)$
His investment in loans was ten times his investment in stocks.
$e = 10c ...eqn.(2)$
The interest earned from his investment in loans was equal to the dividends he received from his investment in bonds.
$0.3e = 0.2d ...eqn.(3)$

The three equations are:

$ 0.25c + 0.2d + 0.3e = 3000 ...eqn.(1) \\[3ex] e = 10c ...eqn.(2) \\[3ex] 0.3e = 0.2d ...eqn.(3) $

(5.) A Broadway theater in the town of Happyland, Oklahoma has $400$ seats, divided into orchestra, main, and balcony seating.
Orchestra seats sell for $60; main seats for $35, and balcony seats for $30.
If all the seats are sold, the gross revenue to the theater is $15,400.
If all the main and balcony seats are sold, but only half the orchestra seats are sold; the gross revenue is $13,000.
How many are there of each kind of seat?

Let the number of orchestra seats be c
The number of main seats be d and
The number of balcony seats be e
Two things are important here: Number and Cost

Number
A Broadway theater in the town of Happyland, Oklahoma has 400 seats, divided into orchestra, main, and balcony seating.
$c + d + e = 400 ...eqn.(1)$

Cost
Cost of $c$ orchestra seats @ $$60$ per seat = $60c$
Cost of $d$ main seats @ $$35$ per seat = $35d$
Cost of $e$ orchestra seats @ $$30$ per seat = $30e$
If all the seats are sold, the gross revenue to the theater is $15,400

$60c + 35d + 30e = 15400 ...eqn.(2)$
If all the main and balcony seats are sold, but only half the orchestra seats are sold; the gross revenue is $13,000
$\dfrac{60}{2}c + 35d + 30e = 13000 ...eqn.(3)$

The three equations are:

$ c + d + e = 400 ...eqn.(1) \\[3ex] 60c + 35d + 30e = 15400 ...eqn.(2) \\[3ex] 30c + 35d + 30e = 13000 ...eqn.(3) $

(6.) The measure of the largest angle of a triangle is twice the measure of the smallest angle.
The sum of the smallest angle and the largest angle is five times the other angle.
Calculate the measure of each angle.

Let the smallest angle be c
The other angle (middle angle) be d and
The largest angle be e
From our knowledge of Geometry, the sum of the angles of a triangle is 180°
$c + d + e = 180 ...eqn.(1)$
The measure of the largest angle of a triangle is twice the measure of the smallest angle.
$e = 2c ...eqn.(2)$
The sum of the smallest angle and the largest angle is five times the other angle.
$c + e = 5d ...eqn.(3)$

The three equations are:

$ c + d + e = 180 ...eqn.(1) \\[3ex] e = 2c ...eqn.(2) \\[3ex] c + e = 5d ...eqn.(3) $

(7.) A dietitian wishes a patient to have a meal that has 82 grams of protein, 76.5 grams of carbohydrates, and 690 milligrams of calcium on Mondays.
The hospital food service tells the dietitian that the Monday dinners consists of chicken, corn, and milk.
Each serving of chicken has 40 grams of protein, 15 grams of carbohydrates, and 100 milligrams of calcium.
Each serving of corn has 3 grams of protein, 16 grams of carbohydrates, and 10 milligrams of calcium.
Each glass of milk has 8 grams of protein, 11 grams of carbohydrates, and 260 milligrams of calcium.
How many servings of each food should the dietitian provide for the patient?

Three factors are important here: Proteins, Carbohydrates, and Calcium
Let the servings of chicken = c
The servings of corn be d and
The servings of the glass of milk be e

Proteins
Chicken: $c$ servings @ 40g proteins / one serving = $40c$
Corn: $d$ potatoes @ 3g proteins / one serving = $3d$
Glass of Milk: $e$ servings @ 8g proteins / glass = $8e$
From the question, 82g of proteins are needed
$40c + 3d + 8e = 82 ...eqn.(1)$

Carbohydrates
Chicken: $c$ servings @ 15g carbohydrates / one serving = $15c$
Corn: $d$ potatoes @ 16g carbohydrates / one serving = $16d$
Glass of Milk: $e$ servings @ 11g carbohydrates / glass = $11e$
From the question, 76.5g of carbohydrates are needed
$15c + 16d + 11e = 76.5 ...eqn.(2)$

Calcium
Chicken: $c$ servings @ 100mg calcium / one serving = $100c$
Corn: $d$ potatoes @ 10mg calcium / one serving = $10d$
Glass of Milk: $e$ servings @ 260mg calcium / glass = $260e$
From the question, 690mg of calcium are needed
$100c + 10d + 260e = 690 ...eqn.(3)$

The three equations are:

$ 40c + 3d + 8e = 82 ...eqn.(1) \\[3ex] 15c + 16d + 11e = 76.5 ...eqn.(2) \\[3ex] 100c + 10d + 260e = 690 ...eqn.(3) $

(8.) The perimeter of a triangle is 70 inches.
The longest side is 4 inches less than the sum of the other two sides.
Three times the shortest side is 9 inches more than the longest side.
Calculate the length of the three sides.

Let the length of the shortest side be c
The length of the other side (medium side) be d and
The length of the longest side be e
The perimeter of a triangle is 70 inches.
$c + d + e = 70 ...eqn.(1)$
The longest side is 4 inches less than the sum of the other two sides.
$e = (c + d) - 4 ...eqn.(2)$
Three times the shortest side is 9 inches more than the longest side.
$3 * c = 9 + e ...eqn.(3)$

The three equations are:

$ c + d + e = 70 ...eqn.(1) \\[3ex] e = c + d - 4 ...eqn.(2) \\[3ex] 3c = 9 + e ...eqn.(3) $

(9.) A dietitian works in a hospital under the guidance of a physician.
Suppose that for a particular patient, a physician prescribes a meal to have 750 calories, 50 grams of protein, and 120 milligrams of Vitamin C.
The dietitian decides to prepare the meal using steak, baked potatoes, and sprouts.
Each 3-ounce serving of steak contains 300 calories, 20 grams of protein, and no Vitamin C.
One baked potato contains 100 calories, 5 grams of protein, and 20 milligrams of Vitamin C.
One 156-gram serving contains 50 calories, 5 grams of protein, and 100 milligrams of Vitamin C.
How many servings of each food are required to satisfy the physician's requirements?

Three factors are important here: Calorie counts, Proteins, and Vitamin C
Let the amount/servings of steak = $c$
The amount of baked potatoes be $d$ and
The amount/servings of sprouts be $e$

Calories
Steak: $c$ servings @ 300 calories / one 3-oz = $300c$
Potato: $d$ potatoes @ 100 calories / one = $100d$
Sprout: $e$ servings @ 50 calories / one 156-g = $50e$
From the question, 750 calories are needed
$300c + 100d + 50e = 750 ...eqn.(1)$

Proteins
Steak: $c$ servings @ 20g proteins / one 3-oz = $20c$
Potato: $d$ potatoes @ 5g proteins / one = $5d$
Sprout: $e$ servings @ 5g proteins / one 156-g = $5e$
From the question, 50g of proteins are needed
$20c + 5d + 5e = 50 ...eqn.(2)$

Vitamin C
Steak: $c$ servings @ 0 mg Vitamin C / one 3-oz = $0c = 0$
Potato: $d$ potatoes @ 20 mg Vitamin C / one = $20d$
Sprout: $e$ servings @ 100g Vitamin C / one 156-g = $100e$
From the question, 120mg of Vitamin C are needed
$20d + 100e = 120 ...eqn.(3)$

The three equations are:

$ 300c + 100d + 50e = 750 ...eqn.(1) \\[3ex] 20c + 5d + 5e = 50 ...eqn.(2) \\[3ex] 20d + 100e = 120 ...eqn.(3) $

(10.) Obadiah Auto Sales has allocated $$1.98$ million to buy sedans, hybrids, and convertibles.
Sedans cost $16,500 each; hybrids cost $18,000 each, and convertibles cost $21,000 each.
Obadiah wants to purchase twice as many sedans as hybrids.
The total number of cars to be purchased is 95.
Determine the number of cars of each type to be purchased.
Assume that the entire budget will be used.

Let the number of convertibles be c
The number of hybrids be d and
The number of sedans be e
Two factors are important here - Cost and Number

Cost
$c$ convertibles @ $$21000$ each = $21000c$
$d$ hybrids @ $18000 each = $18000d$
$e$ sedans @ $16500 each = $16500e$
Sedans cost $16,500 each; hybrids cost $18,000 each, and convertibles cost $21,000 each.
$21000c + 18000d + 16500e = 1980000 ...eqn.(1)$

Number
Obadiah wants to purchase twice as many sedans as hybrids.
$e = 2d ...eqn.(2)$
The total number of cars to be purchased is $95$.
$c + d + e = 95 ...eqn.(3)$

The three equations are:

$ 21000c + 18000d + 16500e = 1980000 ...eqn.(1) \\[3ex] e = 2d ...eqn.(2) \\[3ex] c + d + e = 95 ...eqn.(3) $

(11.) Peace Investment Company has $$125,000$ earmarked for investment in stocks.
To arrive at an acceptable overall level of risk, the management of the company has classified the stocks into three categories: high-risk, medium-risk, and low-risk.
The management estimates that high-risk stocks will have a return rate of 25% per year; medium-risk stocks, 15% per year; and low-risk stocks, 5% per year.
The members have decided that the investment in low-risk stocks should be equal to three times the sum of the investments in the stocks of the other two categories.
Determine how much the company should invest in each type of stock if the investment goal is to have a return of $12,000 per year on the total investment.
Assume that all money available for investment is invested.

Annual Return is the same as Annual Interest
Investment is the same as Principal
Interest is the same as Dividends, and is also the same as Returns

Simple Interest Formula
Interest = Principal * Rate * Time
$I = P * r * t$
Let the investment in high-risk stocks be $c$
The investment in medium-risk stocks be $d$ and
The investment in low-risk stocks be $e$

High-risk Stocks
Principal = $c$
Interest rate, r = 25% = $0.25$
Time, t = 1 year
Interest, I = $c * 0.25 * 1$ = $0.25c$

Medium-risk Stocks
Principal = $d$
Interest rate, r = 15% = $0.15$
Time, t = 1 year
Interest, I = $d * 0.15 * 1$ = $0.15d$

Low-risk Stocks
Principal = $e$
Interest rate, r = 5% = $0.05$
Time, t = 1 year
Interest, I = $e * 0.05 * 1$ = $0.05e$
Peace Investment Company has $125,000 earmarked for investment in stocks.
$c + d + e = 125000 ...eqn.(1)$
The members have decided that the investment in low-risk stocks should be equal to three times the sum of the investments in the stocks of the other two categories.
$e = 3(c + d) ...eqn.(2)$
... if the investment goal is to have a return of $12,000 per year on the total investment.
$0.25c + 0.15d + 0.05e = 12000 ...eqn.(3)$

The three equations are:

$ c + d + e = 125000 ...eqn.(1) \\[3ex] e = 3c + 3d ...eqn.(2) \\[3ex] 0.25c + 0.15d + 0.05e = 12000 ...eqn.(3) $

(12.) Sarah receives $143 per year in simple interest from three investments.
A part is invested at 2%.
Another part is invested at 3%.
The third part is invested at 4%.
There is $500 more invested at 3% than at 2%.
The amount invested at 4% is three times the amount invested at 3%.
Calculate the amount invested at each rate.

Let the part invested at 2% be c. Let us call it 1st Investment.
The part invested at 3% be d. Let us call it 2nd Investment.
The part invested at 4% be e. Let us call it 3rd Investment.
Investment is the same as the Principal

Simple Interest Formula
Interest = Principal * Rate * Time
$I = P * r * t$

1st Investment
Principal = $c$
Interest rate, r = 2% = $0.02$
Time, t = 1 year
Interest, I = $c * 0.02 * 1$ = $0.02c$

2nd Investment
Principal = $d$
Interest rate, r = 3% = $0.03$
Time, t = 1 year
Interest, I = $d * 0.03 * 1$ = $0.03d$

3rd Investment
Principal = $e$
Interest rate, r = 4% = $0.04$
Time, t = 1 year
Interest, I = $e * 0.04 * 1$ = $0.04e$
Sarah receives $143 per year in simple interest from three investments.
$0.02c + 0.03d + 0.04e = 143 ...eqn.(1)$
There is $500 more invested at 3% than at 2%.
In other words, the 2nd Investment is $$500$ more than the 1st Investment.
$d = 500 + c ...eqn.(2)$
The amount invested at 4% is three times the amount invested at 3%.
In other words, the 3rd Investment is three times the 2nd Investment.
$e = 3 * d ...eqn.(3)$

The three equations are:

$ 0.02c + 0.03d + 0.04e = 143 ...eqn.(1) \\[3ex] d = 500 + c ...eqn.(2) \\[3ex] e = 3d ...eqn.(3) $

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