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

Solutions on the Overview and Classification of Linear Systems

Samuel Dominic Chukwuemeka (SamDom For Peace) Calculators: Calculators
Prerequisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals, and Percents
(3.) Expressions and 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 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.
Check all solutions where applicable.
Show all work.

(1.) ACT For what value of a would the following system of equations have an infinite number of solutions? $$ 2x - 3y = 6 \\[3ex] 4x - 6y = 3a \\[3ex] $$ $ A.\;\; 2 \\[3ex] B.\;\; 4 \\[3ex] C.\;\; 6 \\[3ex] D.\;\; 12 \\[3ex] E.\;\; 18 \\[3ex] $

Infinite number of solutions implies that the linear system is consistent-dependent
This means that there is the same relationship for the respective variables and the constants when the linear system is complete and lined up accordingly.

$ \underline{Consistent-Dependent:\;\;Many\;\;Solutions} \\[3ex] x:\;\; \dfrac{4}{2} = 2 \\[5ex] x:\;\; 2 * \color{darkblue}{2} = 4 \\[3ex] y:\;\; -3 * \color{darkblue}{2} = -6 \\[3ex] same\;\;relationship:\;\;multiply\;\;by\;\;2 \\[3ex] \implies \\[3ex] 6 * \color{darkblue}{2} = 3a \\[3ex] 12 = 3a \\[3ex] 3a = 12 \\[3ex] a = \dfrac{12}{3} \\[5ex] a = 4 $
(2.) ACT What is the (x, y) solution, if one exists, to the system of equations y = 2x + 6 and 6x + 12 = 3y?

$ A.\:\: (-3, 0) \\[3ex] B.\:\: (-2, 0) \\[3ex] C.\:\: (0, 4) \\[3ex] D.\:\: (0, 6) \\[3ex] $ E. There is no solution to this system


We can solve this question using at least two methods.

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

LHS RHS
$ (-3, 0) \implies x = -3, \;\;\; y = 0 $
$ y \\[3ex] 0 \checkmark $
$ 6x + 12 \\[3ex] 6(-3) + 12 \\[3ex] -18 + 12 \\[3ex] -6 \large\unicode{x2718} $
$ 2x + 6 \\[3ex] 2(-3) + 6 \\[3ex] -6 + 6 \\[3ex] 0 \checkmark $
$ 3y \\[3ex] 3(0) \\[3ex] 0 \large\unicode{x2718} $
Option A is NO
$ (-2, 0) \implies x = -2, \;\;\; y = 0 $
$ y \\[3ex] 0 \large\unicode{x2718} $
$ 6x + 12 \\[3ex] 6(-2) + 12 \\[3ex] -12 + 12 \\[3ex] 0 \checkmark $
$ 2x + 6 \\[3ex] 2(-2) + 6 \\[3ex] -4 + 6 \\[3ex] 2 \large\unicode{x2718} $
$ 3y \\[3ex] 3(0) \\[3ex] 0 \checkmark $
Option B is NO
$ (0, 4) \implies x = 0, \;\;\; y = 4 $
$ y \\[3ex] 4 \large\unicode{x2718} $
$ 6x + 12 \\[3ex] 6(0) + 12 \\[3ex] 0 + 12 \\[3ex] 12 \checkmark $
$ 2x + 6 \\[3ex] 2(0) + 6 \\[3ex] 0 + 6 \\[3ex] 6 \large\unicode{x2718} $
$ 3y \\[3ex] 3(4) \\[3ex] 12 \checkmark $
Option C is NO
$ (0, 6) \implies x = 0, \;\;\; y = 6 $
$ y \\[3ex] 6 \checkmark $
$ 6x + 12 \\[3ex] 6(0) + 12 \\[3ex] 0 + 12 \\[3ex] 12 \large\unicode{x2718} $
$ 2x + 6 \\[3ex] 2(0) + 6 \\[3ex] 0 + 6 \\[3ex] 6 \checkmark $
$ 3y \\[3ex] 3(6) \\[3ex] 18 \large\unicode{x2718} $
Option D is NO

This implies that Option E is the correct answer
There is no solution to this system.

Second Method: Classify the System
Because this is an ACT question (it is timed), it is recommended you use this method especially if one of the options is that there is no solution to this system

$ y = 2x + 6 ...eqn.(1) \\[3ex] 6x + 12 = 3y ...eqn.(2) \\[3ex] \underline{Express\;\;in\;\;standard\;\;form} \\[3ex] y = 2x + 6 \\[3ex] 0 = 2x + 6 - y \\[3ex] -6 = 2x - y \\[3ex] 2x - y = -6...modified\;\;eqn.(1) \\[3ex] 6x + 12 = 3y \\[3ex] 6x - 3y = -12 ...modifies\;\;eqn.(2) \\[3ex] \underline{Check\;\;for\;\;relationship} \\[3ex] 2 * 3 = 6 \\[3ex] -1 * 3 = -3 \\[3ex] -6 * 3 \ne -12 ...Inconsistent\;\;system \\[3ex] No\;\;solution \\[3ex] $ There is no solution to this system.
(3.) (a.) Classify these linear systems.

$ (i.) \\[3ex] 3x = -y + 5 \\[3ex] -10 = -6x - 2y \\[5ex] (ii.) \\[3ex] -2y = -x + 8 \\[3ex] -3x = -5 - 6y \\[5ex] (iii) \\[3ex] x = 4 - y \\[3ex] 5 = -y + 5x \\[5ex] $ (b.) Give reasons for your answers.

(c.) If the system is consistent-independent, determine the solution.
If the system is consistent-dependent, determine the simplified general solution by expressing y in terms of x and according to the slope-intercept form: y = mx + b
If the system is inconsistent, indicate that there is no solution.


$ (i) \\[3ex] 3x = -y + 5 \\[3ex] -10 = -6x - 2y \\[5ex] Arrange\;\;in\;\;standard\;\;form \implies \\[3ex] 3x + y = 5 ...eqn.(1) \\[3ex] 6x + 2y = 10 ...eqn.(2) \\[3ex] \underline{Check\;\;for\;\;relationships} \\[3ex] 3 * 2 = 6 \\[3ex] 1 * 2 = 2 \\[3ex] 5 * 2 = 10 \\[3ex] Yes\;\;on\;\;the\;\;LHS \\[3ex] Yes\;\;on\;\;the\;\;RHS \\[3ex] The\;\;system\;\;is\;\;consistent-dependent \\[3ex] It\;\;has\;\;many\;\;solutions \\[3ex] From\;\;eqn.(1) \\[3ex] y = 5 - 3x \\[3ex] y = -3x + 5 ...General\;\;Solution \\[5ex] (ii.) \\[3ex] -2y = -x + 8 \\[3ex] -3x = -5 - 6y \\[5ex] Arrange\;\;in\;\;standard\;\;form \implies \\[3ex] x - 2y = 8 ...eqn.(1) \\[3ex] -3x + 6y = -5 ...eqn.(2) \\[3ex] \underline{Check\;\;for\;\;relationships} \\[3ex] 1 * -3 = -3 \\[3ex] -2 * -3 = 6 \\[3ex] 8 * -3 \ne -5 \\[3ex] Yes\;\;on\;\;the\;\;LHS \\[3ex] No\;\;on\;\;the\;\;RHS \\[3ex] The\;\;system\;\;is\;\;inconsistent \\[3ex] It\;\;has\;\;no\;\;solution \\[5ex] (iii) \\[3ex] x = 4 - y \\[3ex] 5 = -y + 5x \\[5ex] Arrange\;\;in\;\;standard\;\;form \implies \\[3ex] x + y = 4 ...eqn.(1) \\[3ex] 5x - y = 5 ...eqn.(2) \\[3ex] \underline{Check\;\;for\;\;relationships} \\[3ex] 1 * 5 = 5 \\[3ex] 1 * 5 \ne -1 \\[3ex] 4 * 5 \ne 5 \\[3ex] No\;\;relationship \\[3ex] The\;\;system\;\;is\;\;consistent-independent \\[3ex] It\;\;has\;\;one\;\;solution \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(1) \\[3ex] x = 4 - y ...eqn.(3) \\[3ex] Substitute\;\;(4 - y)\;\;for\;\;x\;\;in\;\;eqn.(2) \\[3ex] 5(4 - y) - y = 5 \\[3ex] 20 - 5y - y = 5 \\[3ex] 20 - 5 = 5y + y \\[3ex] 15 = 6y \\[3ex] 6y = 15 \\[3ex] y = \dfrac{15}{6} \\[5ex] y = \dfrac{5}{2} \\[5ex] Substitute\;\;\dfrac{5}{2}\;\;for\;\;y\;\;in\;\;eqn.(3) \\[5ex] x = 4 - \dfrac{5}{2} \\[5ex] x = \dfrac{8}{2} - \dfrac{5}{2} \\[5ex] x = \dfrac{8 - 5}{2} \\[5ex] x = \dfrac{3}{2} \\[5ex] $ Check
$ x = \dfrac{3}{2},\;\;\; y = \dfrac{5}{2} $
LHS RHS
$ x \\[3ex] \dfrac{3}{2} $
$5$
$ 4 - y \\[3ex] 4 - \dfrac{5}{2} \\[5ex] \dfrac{8}{2} - \dfrac{5}{2} \\[5ex] \dfrac{8 - 5}{2} \\[5ex] \dfrac{3}{2} $
$ -y + 5x \\[3ex] -\dfrac{5}{2} + 5\left(\dfrac{3}{2}\right) \\[5ex] -\dfrac{5}{2} + \dfrac{15}{2} \\[5ex] \dfrac{-5 + 15}{2} \\[5ex] \dfrac{10}{2} \\[5ex] 5 $
(4.) SAT-NC $$ x = 4 \\[3ex] \hspace{2.2em}y = \dfrac{x}{4} + 2 $$ What is the solution (x, y) to the given system of equations?

$ A)\;\; (4, 6) \\[3ex] B)\;\; (4, 3) \\[3ex] C)\;\; (4, 2) \\[3ex] D)\;\; (4, 1) \\[3ex] $

$ x = 4 \\[3ex] y = \dfrac{x}{4} + 2 \\[5ex] y = \dfrac{4}{4} + 2 \\[5ex] y = 1 + 2 \\[3ex] y = 3 \\[3ex] (x, y) = (4, 3) $
(5.) For each ordered pair (p, q), determine whether it is a solution to the system of equations.

$ -6q + 3 = -9p \\[3ex] -3p = 1 - 2q \\[5ex] (a.)\;\; (-1, -1) \\[3ex] (b.)\;\; (-2, 3) \\[3ex] (c.)\;\; (1, 2) \\[3ex] (d.)\;\; (2, 1) \\[3ex] (e.)\;\; \left(-\dfrac{5}{3}, -2\right) \\[5ex] (f.)\;\; \left(2, \dfrac{7}{2}\right) \\[5ex] $

LHS RHS
$Eqn.(1.)\;\;-6q + 3$
$Eqn.(2.)\;\;-3q$
$-9p$
$1 - 2q$
$ (a.)\;\; (-1, -1) \\[3ex] -6(-1) + 3 \\[3ex] 6 + 3 \\[3ex] 9 $
$ -3(-1) \\[3ex] 3 $
$ p = -1,\;\;\;q = -1 \\[3ex] -9(-1) \\[3ex] 9 $
$ 1 - 2(-1) \\[3ex] 1 + 2 \\[3ex] 3 $
$(-1, -1)$ is a solution.
$ (b.)\;\; (-2, 3) \\[3ex] -6(3) + 3 \\[3ex] -18 + 3 \\[3ex] -15 $
$ -3(-2) \\[3ex] 6 $
$ p = -2,\;\;\;q = -3 \\[3ex] -9(-2) \\[3ex] 18 $
$ 1 - 2(-3) \\[3ex] 1 + 6 \\[3ex] 7 $
$(-2, 3)$ is NOT a solution.
$ (c.)\;\; (1, 2) \\[3ex] -6(2) + 3 \\[3ex] -12 + 3 \\[3ex] -9 $
$ -3(1) \\[3ex] -3 $
$ p = 1,\;\;\;q = 2 \\[3ex] -9(1) \\[3ex] -9 $
$ 1 - 2(2) \\[3ex] 1 - 4 \\[3ex] -3 $
$(1, 2)$ is a solution.
$ (d.)\;\; (2, 1) \\[3ex] -6(1) + 3 \\[3ex] -6 + 3 \\[3ex] -3 $
$ -3(2) \\[3ex] -6 $
$ p = 2,\;\;\;q = 1 \\[3ex] -9(2) \\[3ex] -18 $
$ 1 - 2(1) \\[3ex] 1 - 2 \\[3ex] -1 $
$(2, 1)$ is NOT a solution.
$ (e.)\;\; \left(-\dfrac{5}{3}, -2\right) \\[5ex] -6(-2) + 3 \\[3ex] 12 + 3 \\[3ex] 15 $
$ -3\left(-\dfrac{5}{3}\right) \\[5ex] 5 $
$ p = -\dfrac{5}{3},\;\;\;q = -2 \\[5ex] -9\left(-\dfrac{5}{3}\right) \\[5ex] 3(5) \\[3ex] 15 $
$ 1 - 2(-2) \\[3ex] 1 + 4 \\[3ex] 5 $
$\left(-\dfrac{5}{3}, -2\right)$ is a solution.
$ (f.)\;\; \left(2, \dfrac{7}{2}\right) \\[5ex] -6\left(\dfrac{7}{2}\right) + 3 \\[5ex] -3(7) + 3 \\[3ex] -21 + 3 \\[3ex] -18 $
$ -3(2) \\[3ex] -6 $
$ p = 2,\;\;\;q = \dfrac{7}{2} \\[5ex] -9(2) \\[3ex] -18 $
$ 1 - 2\left(\dfrac{7}{2}\right) \\[5ex] 1 - 7 \\[3ex] -6 $
$\left(2, \dfrac{7}{2}\right)$ is a solution.
(6.) ACT For what value of a would the following system of equations have an infinite number of solutions? $$ x - 2y = 8 \\[3ex] 3x - 6y = 4a $$ $ F.\;\; 2 \\[3ex] G.\;\; 6 \\[3ex] H.\;\; 8 \\[3ex] J.\;\; 24 \\[3ex] K.\;\; 32 \\[3ex] $

$ \underline{Check\;\;for\;\;relationships} \\[3ex] x:\;\; 1 * \color{darkblue}{3} = 3 \\[3ex] y:\;\; -2 * \color{darkblue}{3} = -6 \\[3ex] $ There is a relationship on the LHS
Multiply the coeffient on eqn.(1) by 3 to get the coeffient on eqn.(2)
Infinite number of solutions implies that the system is consistent-dependent
This implies that the same relationship should exist on the RHS

$ 8 * 3 = 4a \\[3ex] 24 = 4a \\[3ex] 4a = 24 \\[3ex] a = \dfrac{24}{4} \\[5ex] a = 6 $
(7.) SAT-NC $$ y = 3x + 5 \\[3ex] y = px + 8 $$ In the given system of equations, p is a constant.
The system has no solution.
What is the value of p?

$ A)\;\; -3 \\[3ex] B)\;\; -\dfrac{1}{3} \\[5ex] C)\;\; \dfrac{1}{3} \\[5ex] D)\;\; 3 \\[3ex] $

$ Arrange\;\;in\;\;standard\;\;form \\[3ex] -3x + y = 5 ...eqn.(1) \\[3ex] -px + y = 8 ...eqn.(2) \\[3ex] No\;\;solution \implies Relationship\;\;on\;\;the\;\;LHS;\;\;No\;\;relationship\;\;on\;\;the\;\;RHS \\[3ex] \underline{LHS} \\[3ex] y:\;\; 1 * \color{darkblue}{1} = 1 \\[3ex] x:\;\; -3 * \color{darkblue}{1} = -p \\[3ex] -3 = -p \\[3ex] -p = -3 \\[3ex] p = \dfrac{-3}{-1} \\[5ex] p = 3 $
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