Problem 8 - Elementary Algebra

Question

Determine Whether an Ordered Pair is a Solution of a System of Equations. In the following exercises, determine if the following points are solutions to the given system of equations. $$ \left\\{\begin{array}{l} x+3 y=9 \\ y=\frac{2}{3} x-2 \end{array}\right. $$ (a) (-6,5) (b) \(\left(5, \frac{4}{3}\right)\)

Step-by-Step Solution

Answer

Only \((5, \frac{4}{3})\) is a solution.

Step 1: Understand the Given System

The system of equations given is: \[\left\{\begin{array}{l}x + 3y = 9 \y = \frac{2}{3}x - 2\end{array}\right.\]We need to check if the points (-6, 5) and \(\left(5, \frac{4}{3}\right)\) satisfy both equations.

Step 2: Check Point (-6, 5)

Substitute \(x = -6\) and \(y = 5\) into the first equation: \[-6 + 3(5) = -6 + 15 = 9\]The point (-6, 5) satisfies the first equation. Now substitute into the second equation:\[5 = \frac{2}{3}(-6) - 2 = -4 - 2 = -6\]The point (-6, 5) does not satisfy the second equation, so it is not a solution.

Step 3: Check Point (5, 4/3)

Substitute \(x = 5\) and \(y = \frac{4}{3}\) into the first equation: \[5 + 3\left(\frac{4}{3}\right) = 5 + 4 = 9\]The point \(\left(5, \frac{4}{3}\right)\) satisfies the first equation. Now substitute into the second equation:\[\frac{4}{3} = \frac{2}{3}(5) - 2 = \frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3}\]The point \(\left(5, \frac{4}{3}\right)\) satisfies the second equation, so it is a solution.

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. In our case, we have two linear equations involving variables x and y. The goal is to find the value of these variables that satisfy all equations in the system simultaneously.
There can be three types of solutions for a system of equations:

• A single solution, where the lines intersect at one point.
• No solution, where the lines are parallel and never meet.
• Infinitely many solutions, where the lines are coincident and lie on top of each other.

For this exercise, we were given the system of equations:
\ \ \ \[ \begin{aligned} x + 3y &= 9 \ y &= \frac{2}{3}x - 2 \end{aligned} \]
And we needed to check if specific points are solutions to this system.

Substitution Method

The substitution method is a technique to solve a system of equations by substituting one equation into another. This method involves three main steps:

• Solve one equation for one variable.
• Substitute this expression into the other equation.
• Solve the resulting equation.

In our example, the second equation is already solved for y:
\ \[ y = \frac{2}{3}x - 2 \]
We substitute this expression into the first equation to find out if the point (x, y) is a solution. We began by plugging in the given points (\(-6, 5\)) and (\(5, \frac{4}{3}\)) into both equations to see if both equations are satisfied.

Linear Equations

Linear equations are algebraic expressions that represent straight lines when graphed on a coordinate plane. They have the general form: \ \[ ax + by = c \]
where a, b, and c are constants.
In our given system, both equations are linear:
\ \ \ \[ \begin{aligned} 1. x + 3y &= 9 \ 2. y &= \frac{2}{3}x - 2 \end{aligned} \]
These equations form straight lines when plotted.
To determine if a point like (\(-6, 5\)) or (\(5, \frac{4}{3}\)) is a solution, we substitute x and y into each equation. If the point satisfies both equations, it is considered a solution to the system. Through a step-by-step process in the solution, we found that \((5, \frac{4}{3})\) is a valid solution as it satisfies both equations, whereas \(-6, 5\) does not.