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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

Solution of Linear Systems by Graph

Calculators: Calculators

Prerequisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals, and Percents
(3.) Expressions and Equations
(4.) Introduction to Relations and Functions

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.
The Graphical Method of solving linear systems must be used.
Check all solutions where applicable.
Show all work.

Please NOTE:
(1.) Only the variables: x and y are used.
(2.) The graphical solution is expressed as (x, y)

(1.) ACT If x + y = 26 and x − y = 14, then y = ?

$ F.\;\; 6 \\[3ex] G.\;\; 7 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 20 \\[3ex] K.\;\; 40 \\[3ex] $

$y = 6$

(2.) ACT When y = x and x + y = 10 are graphed in the standard (x, y) coordinate plane, at which point do they intersect?

$ A.\;\; (-5, 5) \\[3ex] B.\;\; (0, 10) \\[3ex] C.\;\; (5, -5) \\[3ex] D.\;\; (5, 5) \\[3ex] E.\;\; (10, 0) \\[3ex] $

$(x, y) = (5, 5)$

(3.) Graph the system below on a learning management system software (ALEKS, Pearson MML, WebAssign, etc.) rather than on a graph paper.
Write its solution.

$ x + 3y = 1 \\[3ex] y = 2x - 2 \\[3ex] $ Hint: Because you are using a learning management system, decimals and fractions are not allowed.
This implies that you need only two points for each equation and both points should be integers only.

$ x + 3y = 1...eqn.(1) \\[3ex] y = 2x - 2...eqn.(2) \\[3ex] $ First, we need to isolate y
In other words, let us make y the subject of the formula
Let us begin with equation (1).
Let us isolate y for equation (1.)

$ x + 3y = 1 \\[3ex] 3y = 1 - x \\[3ex] 3y = -x + 1 \\[3ex] y = \dfrac{-x + 1}{3} \\[5ex] $ Second, we need to get our two points for eqn.(1)
We need to find the values of x such that when we multiply it by -1, add 1, and divide by 3, it gives a result of an integer as our value of y
We also need to make sure that both values of x and y are within the boundary values of the graph.

$ \underline{Equation\;(1)} \\[3ex] y = \dfrac{-x + 1}{3}...modified\;\;eqn.(1) \\[5ex] When\;\;x = -2 \\[3ex] y = \dfrac{-(-2) + 1}{3} \\[5ex] y = \dfrac{2 + 1}{3} \\[5ex] y = \dfrac{3}{3} \\[5ex] y = 1 \\[3ex] 1st\;\;Point = (x_1,y_1) = (-2, 1) \\[3ex] When\;\;x = 4 \\[3ex] y = \dfrac{-4 + 1}{3} \\[5ex] y = \dfrac{-3}{3} \\[5ex] y = -1 \\[3ex] 2nd\;\;Point = (x_2,y_2) = (4, -1) \\[5ex] \underline{Equation\;(2)} \\[3ex] y = 2x - 2...eqn.(2) \\[3ex] When\;\;x = 0 \\[3ex] y = 2(0) - 2 \\[3ex] y = 0 - 2 \\[3ex] y = -2 \\[3ex] 1st\;\;Point = (x_1,y_1) = (0, -2) \\[3ex] When\;\;x = 1 \\[3ex] y = 2(1) - 2 \\[3ex] y = 2 - 2 \\[3ex] y = 0 \\[3ex] 2nd\;\;Point = (x_2,y_2) = (1, 0) \\[3ex] $ We can graph the linear system using these four points:
two points for equation (1)
two points for equation (2)

Number 3

$(x, y) = (1, 0)$

Check

$x = 1,\;\;\; y = 0$
LHS	RHS
$ x + 3y \\[3ex] 1 + 3(0) \\[3ex] 1 + 0 \\[3ex] 1 $ $ y \\[3ex] 0 $	$1$ $ 2x - 2 \\[3ex] 2(1) - 2 \\[3ex] 2 - 2 \\[3ex] 0 $

(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] $

Let:
a = x
b = y

$ 3x - 2y = 21 \\[3ex] x + 3y = -4 \\[5ex] $

$y = b = -3$

(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 Graph

Number 5

$(x, y) = (-2, 3)$

(6.) GCSE Solve the following simultaneous equations. $$ 2x + y = 8 \\[3ex] x - y = 1 $$

$(x, y) = (3, 2)$

Check

$(x, y) = (3, 2)$
LHS	RHS
$ 2x + y \\[3ex] 2(3) + 2 \\[3ex] 6 + 2 \\[3ex] 8 $ $ x - y \\[3ex] 3 - 2 \\[3ex] 1 $	$8$ $1$

(7.) Solve the following system of equations. $$ 5x - 4y = 22 \\[3ex] -7x - 3y = -5 $$

$(x, y) = (2, -3)$

Check

$x = 2,\;\;\; y = -3$
LHS	RHS
$ 5x - 4y \\[3ex] 5(2) - 4(-3) \\[3ex] 10 + 12 \\[3ex] 22 $ $ -7x - 3y \\[3ex] -7(2) - 3(-3) \\[3ex] -14 + 9 \\[3ex] -5 $	$22$ $-5$

(8.) curriculum.gov.mt Solve the simultaneous equations:

$ 6a - 5b = 19 \\[3ex] 2a + b = 9 \\[3ex] $

Let:
a = x
b = y

$ 6x - 5y = 19 \\[3ex] 2x + y = 9 \\[5ex] $

$(x, y) = (a, b) = (4, 1)$

Check

$(a, b) = (4, 1)$
LHS	RHS
$ 6a - 5b \\[3ex] 6(4) - 5(1) \\[3ex] 24 - 5 \\[3ex] 19 $ $ 2a + b \\[3ex] 2(4) + 1 \\[3ex] 8 + 1 \\[3ex] 9 $	$19$ $9$

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

$ A.\:\: (-4, 3) \\[3ex] B.\:\: (-1, 1.5) \\[3ex] C.\:\: (1, 0.5) \\[3ex] D.\:\: (0, 1) \\[3ex] E.\:\: (2, 0) \\[3ex] $

We can solve this question using at least two methods.

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

LHS	RHS
$ (-4, 3) \implies x = -4, \;\;\; y = 3 $ $ x + 2y \\[3ex] -4 + 2(3) \\[3ex] -4 + 6 \\[3ex] 2 \checkmark $ $ -x + y \\[3ex] -(-4) + 3 \\[3ex] 4 + 3 \\[3ex] 7 \checkmark $	2 7 Option A is YES
$ (-1, 1.5) \implies x = -1, \;\;\; y = 1.5 $ $ x + 2y \\[3ex] -1 + 2(1.5) \\[3ex] -1 + 3 \\[3ex] 2 \checkmark $ $ -x + y \\[3ex] -(-1) + 1.5 \\[3ex] 1 + 1.5 \\[3ex] 2.5 \large\unicode{x2718} $	2 7 Option B is NO
$ (1, 0.5) \implies x = 1, \;\;\; y = 0.5 $ $ x + 2y \\[3ex] 1 + 2(0.5) \\[3ex] 1 + 1 \\[3ex] 2 \checkmark $ $ -x + y \\[3ex] -1 + 0.5 \\[3ex] -0.5 \large\unicode{x2718} $	2 7 Option C is NO
$ (0, 1) \implies x = 0, \;\;\; y = 1 $ $ x + 2y \\[3ex] 0 + 2(1) \\[3ex] 0 + 2 \\[3ex] 2 \checkmark $ $ -x + y \\[3ex] -0 + 1 \\[3ex] 1 \large\unicode{x2718} $	2 7 Option D is NO
$ (2, 0) \implies x = 2, \;\;\; y = 0 $ $ x + 2y \\[3ex] 2 + 2(0) \\[3ex] 2 + 0 \\[3ex] 2 \checkmark $ $ -x + y \\[3ex] -2 + 0 \\[3ex] -2 \large\unicode{x2718} $	2 7 Option E is NO

Second Method: Solution by Graph

$(x, y) = (-4, 3)$

(17.) GCSE Solve the simultaneous equations $$ 2x + 4y = -9 \\[3ex] 2y = 4x - 7 $$

$(x, y) = (0.5, -2.5)$

Check

$(x, y) = (0.5, -2.5)$
LHS	RHS
$ 2x + 4y \\[3ex] 2(0.5) + 4(-2.5) \\[3ex] 1 - 10 \\[3ex] -9 $ $ 2y \\[3ex] 2(-2.5) \\[3ex] -5 $	$-9$ $ 4x - 7 \\[3ex] 4(0.5) - 7 \\[5ex] 2 - 7 \\[3ex] -5 $

(19.) JAMB Solve the pair of equations for x and y respectively

$ 2x^{-1} - 3y^{-1} = 4 \\[3ex] 4x^{-1} + y^{-1} = 1 \\[5ex] A.\;\; -1, 2 \\[3ex] B.\;\; 1, 2 \\[3ex] C.\;\; 2, 1 \\[3ex] D.\;\; 2, -1 \\[3ex] $

$(x, y) = (2, -1)$

Check

$x = 2,\;\;\;y = -1$
LHS	RHS
$ 2x^{-1} - 3y^{-1} \\[3ex] \dfrac{2}{x} - \dfrac{3}{y} \\[5ex] \dfrac{2}{2} - \dfrac{3}{-1} \\[5ex] 1 - -3 \\[3ex] 1 + 3 \\[3ex] 4 $ $ 4x^{-1} + y^{-1} \\[3ex] \dfrac{4}{x} + \dfrac{1}{y} \\[5ex] \dfrac{4}{2} + \dfrac{1}{-1} \\[5ex] 2 + - 1 \\[3ex] 2 - 1 \\[3ex] 1 $	$4$ $1$

(20.) SAT-C $$ y = 3x + 6 \\[3ex] y = -3x + 9 $$ The solution to the given system of equations is (x, y)
What is the value of y?

$ A)\;\; 15 \\[3ex] B)\;\; 7.5 \\[3ex] C)\;\; 1.5 \\[3ex] D)\;\; 0.5 \\[3ex] $

$ y = 3x + 6 \\[3ex] y = -3x + 9 \\[5ex] $

$y = 7.5$

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(23.) GCSE Here is line L and the graph of y = x − 1
The scales of the axes are not shown.

Work out the equation of line L.

First: Let us find the scale for both axes

$ \underline{Line:\;\;y = x - 1} \\[3ex] \boldsymbol{y-intercept} \\[3ex] x = 0,\;\;y = ? \\[3ex] y = 0 - 1 \\[3ex] y = -1 \\[3ex] y-intercept = (0, -1) \\[3ex] \boldsymbol{x-intercept} \\[3ex] y = 0,\;\; x = ? \\[3ex] 0 = x - 1 \\[3ex] x - 1 = 0 \\[3ex] x = 1 \\[3ex] x-intercept = (1, 0) \\[3ex] $ This implies that:
5 lines = 1 unit on the y-axis
10 lines = 1 unit on he x-axis
Next: let us find the intercepts for Line L, then find the slope, and then write the equation for line L

$ \underline{Line:\;\;L} \\[3ex] y-intercept = (0, 3) \\[3ex] x-intercept = (2, 0) \\[3ex] \implies \\[3ex] Point\;1 = (0, 3) \\[3ex] x_1 = 0 \\[3ex] y_1 = 3 \\[3ex] Point\;2 = (2, 0) \\[3ex] x_2 = 2 \\[3ex] y_2 = 0 \\[3ex] slope = m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{0 - 3}{2 - 0} \\[5ex] m = -\dfrac{3}{2} \\[5ex] b = y-intercept = 3 \\[3ex] y = mx + b ...equation\;\;of\;\;a\;\;straight\;\;line\;\;in\;\;slope-intercept\;\;form \\[3ex] y = -\dfrac{3}{2}x + 3 $

(27.) Some people say that taxes are a necessary good.
Some say that taxes are a necessary evil.
Be it as it may, there seems to be a common agreement that taxes are necessary.
So, why would someone not pay taxes? ☺
Taxes can be flat, progressive, or regressive.
Person A prefers a progressive tax rate (Plan A).
Person B prefers a flat tax rate (Plan B).
Person C prefers a regressive tax rate (Plan C).
A graph showing an example of the tax plans from Persons A and B is shown below.
The exact taxes rather than the tax rates and the weekly income are used for this example.
The unit for both variables (axis) is the dollar.

(a.) Identify the independent variable
(Hint: Does the tax depend on the weekly income or does the weekly income depend on the tax?)

(b.) Identify the dependent variable
(Hint: Does the tax depend on the weekly income or does the weekly income depend on the tax?)

(c.) Which tax plan is indicated with the blue color?

(d.) Which tax plan is indicated with the red color?

(e.) Samuel makes a weekly income of $250.
Which tax plan is favorable for him?

(f.) Chukwuemeka makes a weekly income of $300
Which tax plan is favorable for him?

(g.) For what weekly income are the two tax plans the same?

The tax depends on the weekly income.
Hence:
(b.) Tax is the dependent variable.
It is the y-axis

(a.) The weekly income is the independent variable.
It is the x-axis

(c.) Plan A is the blue color because it is the progressive tax rate.

(d.) Plan B is the red color because it is a flat tax.

For a weekly income below $275, Plan A is better because the person will pay less than $22 in taxes.
(e.) Plan A is favorable for Samuel.

(g.) For a weekly income of $275, both Plans (Plan A and Plan B) are the same because the person will pay $22 in taxes.

For a weekly income above $275, Plan B is better because the person will pay more than $22 in taxes.
(f.) Plan B is favorable for Chukwuemeka.

(29.) GCSE

Use the diagram to solve the simultaneous equations $$ y - x = 3, \\[3ex] x + 2y = 1 $$ Give your answers correct to 1 decimal place.

Based on the diagram, the scales are:
10 lines to 1 unit on the y-axis
10 lines to 1 unit on the x-axis

This implies that:

$ 1\;line = \dfrac{1}{10} = 0.1 \;\;for\;\;both\;\;axis \\[3ex] x = 16.8\;lines = 16.8(0.1) = -1.68 \approx -1.7...negative\;\;x-axis \\[3ex] y = 13.4\;lines = 13.4(0.1) = 1.34 \approx 1.3 ...positive\;\;y-axis \\[3ex] $ Check

$x \approx -1.68,\;\;\; y \approx 1.34$
LHS	RHS
$ y - x \\[3ex] 1.34 - -1.68 \\[3ex] 1.34 + 1.68 \\[3ex] 3.02 \approx 3 $ $ x + 2y \\[3ex] -1.68 + 2(1.34) \\[3ex] -1.68 + 2.68 \\[3ex] 1 $	$3$ $1$

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