Solved Examples on Linear Systems (All)

Samuel Dominic Chukwuemeka (SamDom For Peace) 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 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 a 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: 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 as applicable.
Use any method you prefer. However, indicate the method(s) used.
Show all work.
Check all solutions as applicable.
Interpret your solutions as applicable.

(1.) curriculum.gov.mt Two straight lines $L_1$ and $L_2$ have equations $y = 2x + 5$ and $4y + x = 11$ respectively.

(a.) Make y the subject of equation $L_2$

(b.) What are the gradient and the y-intercept of equation $L_2$?

(c.) $L_1$ and $L_2$ are drawn on a set of x and y axes.
The coordinates of their point of intersection are (p, q).
Find the value of p and the value of q without drawing their graphs.


$ L_1:\;\; y = 2x + 5 \\[3ex] L_2:\;\; 4y + x = 11 \\[3ex] (a) \\[3ex] 4y + x = 11 \\[3ex] 4y = 11 - x \\[3ex] y = \dfrac{11 - x}{4} \\[5ex] (b) \\[3ex] y = \dfrac{11}{4} - \dfrac{x}{4} \\[5ex] y = -\dfrac{1}{4}x + \dfrac{11}{4} \\[5ex] Compare\;\;to:\;\; y = mx + b \\[3ex] m = gradient = -\dfrac{1}{4} \\[5ex] b = y-intercept = \dfrac{11}{4} \\[5ex] (c) \\[3ex] Modified\;\;L_2:\;\; y = \dfrac{11 - x}{4} \\[5ex] L_1 = Modified\;\;L_2...because\;\;y = y \\[3ex] \implies \\[3ex] 2x + 5 = \dfrac{11 - x}{4} \\[5ex] 4(2x + 5) = 11 - x \\[3ex] 8x + 20 = 11 - x \\[3ex] 8x + x = 11 - 20 \\[3ex] 9x = -9 \\[3ex] x = -\dfrac{9}{9} \\[5ex] x = -1 \\[3ex] Substitute\;\; x = -1 \;\;into L_1 \\[3ex] y = 2(-1) + 5 \\[3ex] y = -2 + 5 \\[3ex] y = 3 \\[3ex] (x, y) = (p, q) = (-1, 3) $
(2.)



ACT Use the following information to answer Questions 3 — 5

Winter Fun Ski Resort sells only 2 types of tickets - adult and student.
On Monday, the resort sold 200 tickets, 1 ticket to each skier.
Of those tickets, 25 were sold to first-time skiers.
When Alyssa skis the resort's main run, her elevation, E feet, at any point on the run is modeled by the equation $E = \dfrac{300,000}{t + 100}$ where t is the number of seconds after she begins skiing at the start of the main run.


(3.) The resort collected a total of $6,000 in ticket sales on Monday.
The price of an adult ticket is $50 and the price of a student ticket is $25.
How many adult and student tickets were sold on Monday?

$ \;\;\;\;\;\;\;\;\;\; \underline{adult} \;\;\;\;\; \underline{student} \\[3ex] F.\;\;\;\;\;\;\;\; 40 \;\;\;\;\;\;\;\;\;\;\; 160 \\[3ex] G.\;\;\;\;\;\;\;\; 80 \;\;\;\;\;\;\;\;\;\;\; 120 \\[3ex] H.\;\;\;\;\;\;\; 100 \;\;\;\;\;\;\;\;\;\; 100 \\[3ex] J.\;\;\;\;\;\;\;\; 120 \;\;\;\;\;\;\;\;\;\; 80 \\[3ex] K.\;\;\;\;\;\;\; 160 \;\;\;\;\;\;\;\;\;\; 40 \\[3ex] $

Let:
the number of adult tickets = x
the number of student tickets = y

On Monday, the resort sold 200 tickets, 1 ticket to each skier.

$ x + y = 200 ...eqn.(1) \\[3ex] $ The resort collected a total of $6,000 in ticket sales on Monday.
The price of an adult ticket is $50 and the price of a student ticket is $25.

$ 50x + 25y = 6000 \\[3ex] 25(2x + y) = 25(240) \\[3ex] 2x + y = 240 ...eqn.(2) \\[3ex] \underline{Elimination\;\;Method} \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (2x - x) + (y - y) = (240 - 200) \\[3ex] x = 40 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(1) \\[3ex] y = 200 - x ...eqn.(3) \\[3ex] Substitute\;\; x = 40 \;\;in \;\;eqn.(3) \\[3ex] y = 200 - 40 \\[3ex] y = 160 \\[3ex] $ 40 adult tickets were sold.
160 student tickets were sold.
(4.) On Monday, the resort sold 1 ticket to each of the 8 members of the Herzog family.
Assume this family is a representative sample of all the skiers at the resort on Monday.
How many of the 8 members of the Herzog family are NOT first-time skiers?

$ A.\;\; 1 \\[3ex] B.\;\; 2 \\[3ex] C.\;\; 4 \\[3ex] D.\;\; 6 \\[3ex] E.\;\; 7 \\[3ex] $

On Monday, the resort sold 200 tickets, 1 ticket to each skier.
Of those tickets, 25 were sold to first-time skiers.
This implies that (200 − 25 = 175) tickets were sold to non first-time skiers.
Of 200 tickets, 175 to non first-time skiers
Of 8 tickets, how many would be non first-time?
Let the number of the 8 members of the Herzog family that are NOT first-time skiers = n

Proportional Reasoning Method
Sold Non First-Time Skiers
$200$ $175$
$8$ $n$

$ \dfrac{200}{8} = \dfrac{175}{n} \\[5ex] 200 * n = 8 * 175 \\[3ex] n = \dfrac{8 * 175}{200} \\[5ex] n = 7 \\[3ex] $ 7 members of the Herzog family are NOT first-time skiers
(5.) What is Alyssa's elevation, in feet, at the start of the main run?

$ F.\;\; 30 \\[3ex] G.\;\; 300 \\[3ex] H.\;\; 3,000 \\[3ex] J.\;\; 30,000 \\[3ex] K.\;\; 300,000 \\[3ex] $

At the start of the main run, the number of seconds after she begins skiing at the start of the main run is zero because she is yet to begin skiing
This implies that t = 0

$ E = \dfrac{300,000}{t + 100} \\[5ex] = \dfrac{300,000}{0 + 100} \\[5ex] = \dfrac{300,000}{100} \\[5ex] = 3000 \\[3ex] $ Alyssa's elevation at the start of the main run is 3,000 feet.
(6.)




ACT Use the following information to answer questions 7 — 9

A large theater complex surveyed 5,000 adults.
The results of the survey are shown in the tables below.

Age groups Number
21 - 30
31 - 40
41 - 50
51 or older 		2,750
1,225
625
400

Moviegoer category Number
Very often
Often
Sometimes
Rarely 		830
1,650
2,320
200

Tickets are $9.50 for all regular showings and $7.00 for matinees.


(7.) Based on the survey results, what was the average number of moviegoers for each of the 4 categories?

$ A.\;\; 610 \\[3ex] B.\;\; 1,060 \\[3ex] C.\;\; 1,240 \\[3ex] D.\;\; 1,250 \\[3ex] E.\;\; 1,985 \\[3ex] $

$ Average\;\;number\;\;of\;\;moviegoers \\[3ex] = \dfrac{830 + 1650 + 2320 + 200}{4} \\[5ex] = \dfrac{5000}{4} \\[5ex] = 1,250 \\[3ex] $ The average number of moviegoers for each of the 4 categories is 1,250 moviegoers.
(8.) Suppose all the adults surveyed happened to attend 1 movie each in one particular week.
The total amount spent on tickets by those surveyed in that week was $44,000.00
How many adults attended matinees that week?

$ F.\;\; 500 \\[3ex] G.\;\; 1,400 \\[3ex] H.\;\; 2,500 \\[3ex] J.\;\; 3,600 \\[3ex] K.\;\; 4,500 \\[3ex] $

Let the adults that attended:
regular showings = x
matinees = y

A large theater complex surveyed 5,000 adults.

$ x + y = 5000 ...eqn.(1) \\[3ex] $ Tickets are $9.50 for all regular showings and $7.00 for matinees.
Suppose all the adults surveyed happened to attend 1 movie each in one particular week.
The total amount spent on tickets by those surveyed in that week was $44,000.00

$ 9.5x + 7y = 44000 \\[3ex] 95x + 70y = 440000 \\[3ex] 19x + 14y = 88000 ...eqn.(2) \\[3ex] \underline{Elimination\;\;Method} \\[3ex] 19 * eqn.(1) - eqn.(2) \implies \\[3ex] 19(x + y) - (19x + 14y) = 19(5000) - 88000 \\[3ex] 19x + 19y - 19x - 14y = 95000 - 88000 \\[3ex] 5y = 7000 \\[3ex] y = \dfrac{7000}{5} \\[5ex] y = 1400 \\[3ex] $ 1400 adults attended matinees that week.
(9.) One of the following circle graphs represents the proportion by age group of the adults surveyed.
Which one?

Number 9


We have about a minute to solve this question.
So, we shall find the sectorial angle for the first age group and then use the process of elimination to identify our answer
If necessary, we shall find the sectorial angle for the second age group and also eliminate options
We shall repeat this process until we get our answer

$ Sectorial\;\;\angle\;\;for\;\;each\;\;age\;\;group = \dfrac{frequency\;\;of\;\;the\;\;age\;\;group}{\Sigma f} * 100 \\[5ex] \underline{21-30\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{2750}{5000} * 100 \\[5ex] = 55\% \\[3ex] $ Eliminate Options B., D., and E.
Let us calculate the sectorial angle for the second age group

$ \underline{31-40\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{1225}{5000} * 100 \\[5ex] = 24.5\% \\[3ex] $ Eliminate Option C.
Option A. is the correct answer.

For those who just want to complete the rest of the age groups

$ \underline{41-50\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{625}{5000} * 100 \\[5ex] = 12.5\% \\[3ex] \underline{51\;\;or\;\;Older\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{400}{5000} * 100 \\[5ex] = 8\% \\[5ex] \underline{Check} \\[3ex] \Sigma Sectorial\;\; \angle = 55 + 24.5 + 12.5 + 8 = 100\% $
(10.)

(11.) curriculum.gov.mt Solve these simultaneous equations. $$ 2x - y = 3 \\[3ex] xy - 2x = 12 $$


$ 2x - y = 3 ...eqn.(1) \\[3ex] xy - 2x = 12 ...eqn.(2) \\[3ex] From\;\;eqn.(1) \\[3ex] 2x - 3 = y \\[3ex] y = 2x - 3 ...eqn.(3) \\[3ex] Substitute\;\;(2x - 3)\;\;for\;\;y\;\;in\;\;eqn.(2) \\[3ex] x(2x - 3) - 2x = 12 \\[3ex] 2x^2 - 3x - 2x = 12 \\[3ex] 2x^2 - 5x - 12 = 0 \\[3ex] (2x^2)(-12) = -24x^2 \\[3ex] Factors\;\;are\;\;-8x\;\;and\;\;3x \\[3ex] \implies \\[3ex] 2x^2 - 8x + 3x - 12 = 0 \\[3ex] 2x(x - 4) + 3(x - 4) = 0 \\[3ex] (x - 4)(2x + 3) = 0 \\[3ex] x - 4 = 0 \;\;\;OR\;\;\; 2x + 3 = 0 \\[3ex] x = 4 \;\;\;OR\;\;\; 2x = -3 \\[3ex] x = 4 \;\;\;OR\;\;\; x = -\dfrac{3}{2} \\[5ex] Substitute\;\;each\;\;value\;\;of\;\;x\;\;in\;\;eqn.(3) \\[3ex] When\;\;x = 4 \\[3ex] y = 2(4) - 3 \\[3ex] y = 8 - 3 \\[3ex] y = 5 \\[3ex] (x, y) = (4, 5) \\[3ex] When\;\; x = -\dfrac{3}{2} \\[5ex] y = 2\left(-\dfrac{3}{2}\right) - 3 \\[5ex] y = -3 - 3 \\[3ex] y = -6 \\[3ex] (x, y) = \left(-\dfrac{3}{2}, -6\right) \\[5ex] $ Check
$ (x, y) = (4, 5) \\[3ex] (x, y) = \left(-\dfrac{3}{2}, -6\right) $
LHS RHS
$ 2x - y \\[3ex] 2(4) - 5 \\[3ex] 8 - 5 \\[3ex] 3 $
$ xy - 2x \\[3ex] 4(5) - 2(4) \\[3ex] 20 - 8 \\[3ex] 12 $ 		$3$
$12$
$ 2x - y \\[3ex] 2\left(-\dfrac{3}{2}\right) - (-6) \\[5ex] -3 + 6 \\[3ex] 3 $
$ xy - 2x \\[3ex] -\dfrac{3}{2}(-6) - 2\left(-\dfrac{3}{2}\right) \\[5ex] 9 + 3 \\[3ex] 12 $ 		$3$
$12$
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(17.) Solve the linear system using the Row Reduction method.
Check your solution accordingly.

$ x + y - z = 4 \\[3ex] 3x - 4y + 6z = -11 \\[3ex] x + 3y - 3z = 10 \\[3ex] $

$ x + y - z = 4 \\[3ex] 3x - 4y + 6z = -11 \\[3ex] x + 3y - 3z = 10 \\[3ex] \underline{Row\;\;Reduction\;\;Method} \\[3ex] \left[ \begin{array}{ccc|c} 1 & 1 & -1 & 4 \\[3ex] 3 & -4 & 6 & -11 \\[3ex] 1 & 3 & -3 & 10 \end{array} \right] \begin{matrix} \underrightarrow{-3R_1 + R_2} \\ \underrightarrow{-R_1 + R_3} \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & 4 \\[3ex] 0 & -7 & 9 & -23 \\[3ex] 0 & 2 & -2 & 6 \end{array} \right] \begin{matrix} \underrightarrow{R_2 + 7R_1} \\ \underrightarrow{2R_2 + 7R_3} \end{matrix} \left[ \begin{array}{ccc|c} 7 & 0 & 2 & 5 \\[3ex] 0 & -7 & 9 & -23 \\[3ex] 0 & 0 & 4 & -4 \end{array} \right] \begin{matrix} \underrightarrow{-R_3 + 2R_1} \\ \underrightarrow{-9R_3 + 4R_2} \end{matrix} \left[ \begin{array}{ccc|c} 14 & 0 & 0 & 14 \\[3ex] 0 & -28 & 0 & -56 \\[3ex] 0 & 0 & 4 & -4 \end{array} \right] \begin{matrix} \underrightarrow{\dfrac{R_1}{14}} \\ \underrightarrow{-\dfrac{R_2}{28}} \\ \underrightarrow{\dfrac{R_3}{4}} \end{matrix} \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\[3ex] 0 & 1 & 0 & 2 \\[3ex] 0 & 0 & 1 & -1 \end{array} \right] \\[5ex] \implies \\[3ex] x = 1 \\[3ex] y = 2 \\[3ex] z = -1 \\[3ex] $ Check
$x = 1,\;\;y = 2,\;\;z = -1$
LHS RHS
$ x + y - z \\[3ex] 1 + 2 - (-1) \\[3ex] 3 + 1 \\[3ex] 4 $ $4$
$ 3x - 4y + 6z \\[3ex] 3(1) - 4(2) + 6(-1) \\[3ex] 3 - 8 - 6 \\[3ex] -11 $ $-11$
$ x + 3y - 3z \\[3ex] 1 + 3(2) - 3(-1) \\[3ex] 1 + 6 + 3 \\[3ex] 10 $ $10$
(18.) Solve the linear system using the Row Reduction method.
Check your solution accordingly.

$ x + 2y - z = -3 \\[3ex] 4x - 4y + z = -13 \\[3ex] -4x + 2y - 3z = 10 \\[3ex] $

$ x + 2y - z = -3 \\[3ex] 4x - 4y + z = -13 \\[3ex] -4x + 2y - 3z = 10 \\[3ex] \underline{Row\;\;Reduction\;\;Method} \\[3ex] \left[ \begin{array}{ccc|c} 1 & 2 & -1 & -3 \\[3ex] 4 & -4 & 1 & -13 \\[3ex] -4 & 2 & -3 & 10 \end{array} \right] \begin{matrix} \underrightarrow{-4R_1 + R_2} \\ \underrightarrow{4R_1 + R_3} \end{matrix} \left[ \begin{array}{ccc|c} 1 & 2 & -1 & -3 \\[3ex] 0 & -12 & 5 & -1 \\[3ex] 0 & 10 & -7 & -2 \end{array} \right] \begin{matrix} \underrightarrow{R_2 + 6R_1} \\ \underrightarrow{5R_2 + 6R_3} \end{matrix} \left[ \begin{array}{ccc|c} 6 & 0 & -1 & -19 \\[3ex] 0 & -12 & 5 & -1 \\[3ex] 0 & 0 & -17 & -17 \end{array} \right] \begin{matrix} \underrightarrow{-R_3 + 17R_1} \\ \underrightarrow{5R_3 + 17R_2} \end{matrix} \left[ \begin{array}{ccc|c} 102 & 0 & 0 & -306 \\[3ex] 0 & -204 & 0 & -102 \\[3ex] 0 & 0 & -17 & -17 \end{array} \right] \begin{matrix} \underrightarrow{\dfrac{R_1}{102}} \\ \underrightarrow{-\dfrac{R_2}{204}} \\ \underrightarrow{-\dfrac{R_3}{17}} \end{matrix} \left[ \begin{array}{ccc|c} 1 & 0 & 0 & -3 \\[3ex] 0 & 1 & 0 & \dfrac{1}{2} \\[5ex] 0 & 0 & 1 & 1 \end{array} \right] \\[5ex] \implies \\[3ex] x = -3 \\[3ex] y = \dfrac{1}{2} \\[5ex] z = 1 \\[3ex] $ Check
$x = -3,\;\;y = \dfrac{1}{2},\;\;z = 1$
LHS RHS
$ x + 2y - z \\[3ex] -3 + 2\left(\dfrac{1}{2}\right) - 1 \\[5ex] -3 + 1 - 1 \\[3ex] -3 $ $-3$
$ 4x - 4y + z \\[3ex] 4(-3) - 4\left(\dfrac{1}{2}\right) + 1 \\[5ex] -12 - 2 + 1 \\[3ex] -13 $ $-13$
$ -4x + 2y - 3z \\[3ex] -4(-3) + 2\left(\dfrac{1}{2}\right) - 3(1) \\[3ex] 12 + 1 - 3 \\[3ex] 10 $ $10$
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