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Solution of Linear Systems by Elimination

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.
The Elimination Method of solving linear systems must be used.
You may use Elimination-by-Addition or Elimination-by-Subtraction
However, I shall use Elimination-by-Addition to prepare you for the Gauss-Jordan Method (Gaussian Elimination Method)
Check all solutions where applicable.
Show all work.

(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] $

$ x + y = 26 ...eqn.(1) \\[3ex] x - y = 14 ...eqn.(2) \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] -1 * eqn.(1) + eqn.(2) \implies \\[3ex] -1(x + y) + (x - y) = -1(26) + 14 \\[3ex] -x - y + x - y = -26 + 14 \\[3ex] -2y = -12 \\[3ex] y = \dfrac{-12}{-2} \\[5ex] 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] $

$ y = x \\[3ex] y - x = 0 \\[3ex] -x + y = 0 ...eqn.(1) \\[3ex] x + y = 10 ...eqn.(2) \\[3ex] To\;\;find\;\;x,\;\;eliminate\;\;y \\[3ex] -1 * eqn.(1) + eqn.(2) \implies \\[3ex] -1(-x + y) + (x + y) = -1(0) + 10 \\[3ex] x - y + x + y = 0 + 10 \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[5ex] x = 5 \\[3ex] To\;\;find\;\;y,\;\;eliminate\;\;x \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (-x + y) + (x + y) = 0 + 10 \\[3ex] -x + y + x + y = 10 \\[3ex] 2y = 10 \\[3ex] y = \dfrac{10}{2} \\[5ex] y = 5 \\[3ex] \therefore (x, y) = (5, 5) \\[3ex] $ The two functions intersect at (5, 5)

(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] To\;\;find\;\;b;\;\;eliminate\;\;a \\[3ex] -3 * eqn.(2) + eqn.(1) \implies \\[3ex] -3(a + 3b) + (3a - 2b) = -3(-4) + 21 \\[3ex] -3a - 9b + 3a - 2b = 12 + 21 \\[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 Elimination

$ x + 2y = 4 ...eqn.(1) \\[3ex] -2x + y = 7 ...eqn.(2) \\[3ex] To\;\;find\;\;x;\;\;eliminate\;\;y \\[3ex] -2 * eqn.(2) + eqn.(1) \implies \\[3ex] -2(-2x + y) + (x + 2y) = -2(7) + 4 \\[3ex] 4x - 2y + x + 2y = -14 + 4 \\[3ex] 5x = -10 \\[3ex] x = -\dfrac{10}{5} \\[5ex] x = -2 \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] 2 * eqn.(1) + eqn.(2) \implies \\[3ex] 2(x + 2y) + (-2x + y) = 2(4) + 7 \\[3ex] 2x + 4y - 2x + y = 8 + 7 \\[3ex] 5y = 15 \\[3ex] y = \dfrac{15}{5} \\[5ex] y = 3 \\[3ex] (x, y) = (-2, 3) $

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

$ 2x + y = 8 ...eqn.(1) \\[3ex] x - y = 1 ...eqn.(2) \\[3ex] To\;\;find\;\;x,\;\;eliminate\;\;y \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (2x + y) + (x - y) = 8 + 1 \\[3ex] 2x + y + x - y = 9 \\[3ex] 3x = 9 \\[3ex] x = \dfrac{9}{3} \\[5ex] x = 3 \\[3ex] To\;\;find\;\;y,\;\;eliminate\;\;x \\[3ex] -2 * eqn.(2) + eqn.(1) \implies \\[3ex] -2(x - y) + (2x + y) = -2(1) + 8 \\[3ex] -2x + 2y + 2x + y = -2 + 8 \\[3ex] 3y = 6 \\[3ex] y = \dfrac{6}{3} \\[5ex] y = 2 \\[3ex] $ 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. $$ 5x - 4y = 22 \\[3ex] -7x - 3y = -5 $$

$ 5x - 4y = 22 ...eqn.(1) \\[3ex] -7x - 3y = -5 ...eqn.(2) \\[3ex] To\;\;find\;\;x,\;\;eliminate\;\;y \\[3ex] -3 * eqn.(1) + 4 * eqn.(2) \implies \\[3ex] -3(5x - 4y) + 4(-7x - 3y) = -3(22) + 4(-5) \\[3ex] -15x + 12y - 28x - 12y = -66 - 20 \\[3ex] -43x = -86 \\[3ex] x = \dfrac{-86}{-43} \\[5ex] x = 2 \\[3ex] To\;\;find\;\;y,\;\;eliminate\;\;x \\[3ex] 7 * eqn.(1) + 5 * eqn.(2) \implies \\[3ex] 7(5x - 4y) + 5(-7x - 3y) = 7(22) + 5(-5) \\[3ex] 35x - 28y - 35x - 15y = 154 - 25 \\[3ex] -43y = 129 \\[3ex] y = \dfrac{129}{-43} \\[5ex] y = -3 \\[3ex] $ 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] $

$ 6a - 5b = 19 ...eqn.(1) \\[3ex] 2a + b = 9 ...eqn.(2) \\[3ex] To\;\;find\;\;a,\;\;eliminate\;\;b \\[3ex] eqn.(1) + 5 * eqn.(2) \implies \\[3ex] (6a - 5b) + 5(2a + b) = 19 + 5(9) \\[3ex] 6a - 5b + 10a + 5b = 19 + 45 \\[3ex] 16a = 64 \\[3ex] a = \dfrac{64}{16} \\[5ex] a = 4 \\[3ex] To\;\;find\;\;b,\;\;eliminate\;\;a \\[3ex] -1 * eqn.(1) + 3 * eqn.(2) \implies \\[3ex] -1(6a - 5b) + 3(2a + b) = -1(19) + 3(9) \\[3ex] -6a + 5b + 6a + 3b = -19 + 27 \\[3ex] 8b = 8 \\[3ex] b = \dfrac{8}{8} \\[5ex] b = 1 \\[3ex] (a, b) = (4, 1) \\[3ex] $ 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 Elimination

$ x + 2y = 2 ...eqn.(1) \\[3ex] -x + y = 7 ...eqn.(2) \\[3ex] To\;\;find\;\;x;\;\;eliminate\;\;y \\[3ex] -2 * eqn.(2) + eqn.(1) \implies \\[3ex] -2(-x + y) + (x + 2y) = -2(7) + 2 \\[3ex] 2x - 2y + x + 2y = -14 + 2 \\[3ex] 3x = -12 \\[3ex] x = -\dfrac{12}{3} \\[5ex] x = -4 \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (x + 2y) + (-x + y) = 2 + 7 \\[3ex] x + 2y - x + y = 9 \\[3ex] 3y = 9 \\[3ex] y = \dfrac{9}{3} \\[5ex] y = 3 \\[3ex] (x, y) = (-4, 3) $

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

$ 2x + 4y = -9 ...eqn.(1) \\[3ex] 2y = 4x - 7 ...eqn.(2) \\[3ex] Arrange\;\;eqn.(2)\;\;to\;\;be\;\;in\;\;standard\;\;form \\[3ex] 4x - 7 = 2y \\[3ex] 4x - 2y = 7 ...modified\;\;eqn.(2) \\[3ex] To\;\;find\;\;x,\;\;eliminate\;\;y \\[3ex] eqn.(1) + 2 * eqn.(2) \implies \\[3ex] (2x + 4y) + 2(4x - 2y) = -9 + 2(7) \\[3ex] 2x + 4y + 8x - 4y = -9 + 14 \\[3ex] 10x = 5 \\[3ex] x = \dfrac{5}{10} \\[5ex] x = \dfrac{1}{2} \\[5ex] To\;\;find\;\;y,\;\;eliminate\;\;x \\[3ex] -2 * eqn.(1) + eqn.(2) \implies \\[3ex] -2(2x + 4y) + (4x - 2y) = -2(-9) + 7 \\[3ex] -4x - 8y + 4x - 2y = 18 + 7 \\[3ex] -10y = 25 \\[3ex] y = \dfrac{25}{-10} \\[5ex] y = -\dfrac{5}{2} \\[5ex] (x, y) = \left(\dfrac{1}{2}, -\dfrac{5}{2}\right) \\[5ex] $ Check

$(x, y) = \left(\dfrac{1}{2}, -\dfrac{5}{2}\right)$
LHS	RHS
$ 2x + 4y \\[3ex] 2\left(\dfrac{1}{2}\right) + 4\left(-\dfrac{5}{2}\right) \\[5ex] 1 + 2(-5) \\[3ex] 1 - 10 \\[3ex] -9 $ $ 2y \\[3ex] 2\left(-\dfrac{5}{2}\right) \\[5ex] -5 $	$-9$ $ 4x - 7 \\[3ex] 4\left(\dfrac{1}{2}\right) - 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] $

$ Let: \\[3ex] c = \dfrac{1}{x} \\[5ex] d = \dfrac{1}{y} \\[5ex] \implies \\[3ex] 2x^{-1} = \dfrac{2}{x} = 2\left(\dfrac{1}{x}\right) = 2c \\[5ex] 3y^{-1} = \dfrac{3}{y} = 3\left(\dfrac{1}{y}\right) = 2d \\[5ex] 4x^{-1} = \dfrac{4}{x} = 4\left(\dfrac{1}{x}\right) = 4c \\[5ex] y^{-1} = \dfrac{1}{y} = d \\[5ex] \implies \\[3ex] 2c - 3d = 4 ...eqn.(1) \\[3ex] 4c + d = 1 ...eqn.(2) \\[3ex] To\;\;find\;\;c,\;\;eliminate\;\;d \\[3ex] eqn.(1) + 3 * eqn.(2) \implies \\[3ex] 2c - 3d + 3(4c + d) = 4 + 3(1) \\[3ex] 2c - 3d + 12c + 3d = 4 + 3 \\[3ex] 14c = 7 \\[3ex] c = \dfrac{7}{14} \\[5ex] c = \dfrac{1}{2} \\[5ex] To\;\;find\;\;d,\;\;eliminate\;\;c \\[3ex] -2 * eqn.(1) + eqn.(2) \implies \\[3ex] -2(2c - 3d) + (4c + d) = -2(4) + 1 \\[3ex] -4c + 6d + 4c + d = -8 + 1 \\[3ex] 7d = -7 \\[3ex] d = -\dfrac{7}{7} \\[5ex] d = -1 \\[3ex] \underline{Recall} \\[3ex] c = \dfrac{1}{x} \implies x = \dfrac{1}{c} \\[5ex] x = 1 \div c \\[3ex] x = 1 \div \dfrac{1}{2} \\[5ex] x = 1 * \dfrac{2}{1} \\[5ex] x = 2 \\[3ex] d = \dfrac{1}{y} \implies y = \dfrac{1}{d} \\[5ex] y = \dfrac{1}{-1} \\[5ex] y = -1 \\[3ex] $ 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...eqn.(1) \\[3ex] y = -3x + 9...eqn.(2) \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] y + y = (3x + 6) + (-3x + 9) \\[3ex] 2y = 3x + 6 - 3x + 9 \\[3ex] 2y = 15 \\[3ex] y = \dfrac{15}{2} \\[5ex] y = 7.5 $

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(23.) GCSE Solve the simultaneous equations $$ 2x + 3y = 5p \\[3ex] \;\;\;\;\; y = 2x + p $$ where p is a constant.
Give your answers in terms of p in their simplest form.

$ 2x + 3y = 5p ...eqn.(1) \\[3ex] y = 2x + p ...eqn.(2) \\[3ex] Put\;\;eqn.(2) \;\;in\;\;standard\;\;form \\[3ex] -2x + y = p ...new\;\;eqn.(2) \\[3ex] To\;\;find\;\;y,\;\;eliminate\;\; x \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (2x + 3y) + (-2x + y) = 5p + p \\[3ex] 2x + 3y - 2x + y = 6p \\[3ex] 4y = 6p \\[3ex] y = \dfrac{6p}{4} \\[5ex] y = \dfrac{3p}{2} \\[5ex] To\;\;find\;\;x,\;\;eliminate\;\;y \\[3ex] -1 * eqn.(1) + 3 * eqn.(2) \implies \\[3ex] -1(2x + 3y) + 3(-2x + y) = -1(5p) + 3(p) \\[3ex] -2x - 3y - 6x + 3y = -5p + 3p \\[3ex] -8x = -2p \\[3ex] x = \dfrac{-2p}{-8} \\[5ex] x = \dfrac{p}{4} \\[5ex] $ Check

$ x = \dfrac{p}{4} \\[5ex] y = \dfrac{3p}{2} $
LHS	RHS
$ 2x + 3y \\[3ex] 2\left(\dfrac{p}{4}\right) + 3\left(\dfrac{3p}{2}\right) \\[5ex] \dfrac{p}{2} + \dfrac{9p}{2} \\[5ex] \dfrac{p + 9p}{2} \\[5ex] \dfrac{10p}{2} \\[5ex] 5p \\[3ex] $ $ y \\[3ex] \dfrac{3p}{2} $	$ 5p \\[3ex] $ $ 2x + p \\[3ex] 2\left(\dfrac{p}{4}\right) + p \\[5ex] \dfrac{p}{2} + \dfrac{2p}{2} \\[5ex] \dfrac{p + 2p}{2} \\[5ex] \dfrac{3p}{2} $

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