Solved Examples on Cramer's Rule for 2 * 2 Linear System

Samuel Dominic Chukwuemeka (SamDom For Peace)

Pre-requisite: Solved Examples on the Determinant of a Matrix
For the first twenty questions, begin with the word problems translated here: $2 * 2$ Linear Systems

You may verify your answers as applicable with: Matrices Calculators

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 any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve each question using the Method of Determinants
Show all work
Interpret your solutions

(1.) $ x + y = -4 ...eqn.(1) \\[3ex] x - y = 28 ...eqn.(2) \\[3ex] $

$ \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -4 \\[3ex] 28 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -4 & 1 \\[3ex] 28 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{4 - 28}{-1 - 1} = \dfrac{-24}{-2} \\[5ex] x = 12 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -4 \\[3ex] 1 & 28 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{28 + 4}{-1 - 1} = \dfrac{32}{-2} \\[5ex] y = -16 \\[3ex] $ The sum of the numbers: $12 + -16 = 12 - 16$ is $-4$
The difference between the numbers: $12 - (-16) = 12 + 16$ is $28$

(2.) $ x + y = 54 ...eqn.(1) \\[3ex] y = 2 + 3x ...eqn.(2) \\[3ex] $

Rearrange $eqn.(2)$ to "line up"

$ -2 = 3x - y \\[3ex] 3x - y = -2 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 3 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 54 \\[3ex] -2 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 54 & 1 \\[3ex] -2 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-54 + 2}{-1 - 3} = \dfrac{-52}{-4} \\[5ex] x = 13 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 54 \\[3ex] 3 & -2 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-2 - 162}{-1 - 3} = \dfrac{-164}{-4} \\[5ex] y = 41 \\[3ex] $ There are $13$ commercial launches
There are $41$ non-commercial launches

(3.) $ x = 7y - 6 ...eqn.(1) \\[3ex] x - y = 6 ...eqn.(2) \\[3ex] $

Rearrange $eqn.(1)$ to "line up"

$ x - 7y = -6 ...new\:\: eqn.(1) \\[3ex] \begin{bmatrix} 1 & -7 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -6 \\[3ex] 6 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -6 & -7 \\[3ex] 6 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 42}{-1 + 7} = \dfrac{48}{6} \\[5ex] x = 8 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -6 \\[3ex] 1 & 6 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 6}{-1 + 7} = \dfrac{12}{6} \\[5ex] y = 2 \\[3ex] $ The two-digit number is $82$

(4.) $ x + y = 4802 ...eqn.(1) \\[3ex] y = x - 1116 ...eqn.(2) \\[3ex] $

Rearrange $eqn.(2)$ to "line up"

$ 1116 = x - y \\[3ex] x - y = 1116 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 4802 \\[3ex] 1116 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 4802 & 1 \\[3ex] 1116 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{-4802 - 1116}{-1 - 1} = \dfrac{-5918}{-2} \\[5ex] x = 2959 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 4802 \\[3ex] 1 & 1116 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{1116 - 4802}{-1 - 1} = \dfrac{-3686}{-2} \\[5ex] y = 1843 \\[3ex] $ The average apartment rent in the City of Santa Monica is $\$2959.00$
The average apartment rent in the City of San Francisco is $\$1843.00$

(5.) $ x - y = 141 ...eqn.(1) \\[3ex] x + y = 183 ...eqn.(2) \\[3ex] $

$ \begin{bmatrix} 1 & -1 \\[3ex] 1 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 141 \\[3ex] 183 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 141 & -1 \\[3ex] 183 & 1 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{141 + 183}{1 + 1} = \dfrac{324}{2} \\[5ex] x = 162 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 141 \\[3ex] 1 & 183 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{183 - 141}{1 + 1} = \dfrac{42}{2} \\[5ex] y = 21 \\[3ex] $ The speed of the plane is $162\: mph$
The speed of the wind is $21\: mph$

(6.) $ 10 + x = y ...eqn.(1) \\[3ex] 4y = 6x + 25 ...eqn.(2) \\[3ex] $

Rearrange $eqns. (1.) \:\:and\:\: (2)$ to "line up"

$ x - y = -10 ...new\:\: eqn.(1) \\[3ex] -25 = 6x - 4y \\[3ex] 6x - 4y = -25 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & -1 \\[3ex] 6 & -4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -10 \\[3ex] -25 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -10 & -1 \\[3ex] -25 & -4 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{40 - 25}{-4 + 6} = \dfrac{15}{2} \\[5ex] x = 7.5 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -10 \\[3ex] 6 & -25 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{-25 + 60}{-4 + 6} = \dfrac{35}{2} \\[5ex] y = 17.5 \\[3ex] $ $7.5$ pounds of cashews needs to be mixed with $10$ pounds of peanuts to give $17.5$ pounds of cashews and peanuts so that the mixture will produce the same revenue as would selling the nuts separately.

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