Question

b. consider the system below and answer the following questions: \\(\

$$\begin{cases} y = -\\frac{4}{3}x + 5 \\\\ 4x + 3y = 15 \\end{cases}$$

\\) the lines will: be parallel or be infinite (circle one) the system will have: infinite solutions or no solution (circle one) the solution is (\underline{\hspace{1cm}}, \underline{\hspace{1cm}}).

Explanation:

Step1: Rewrite the second equation

Rewrite \(4x + 3y = 15\) in slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept).
Subtract \(4x\) from both sides: \(3y=-4x + 15\).
Divide both sides by 3: \(y=-\frac{4}{3}x + 5\).

Step2: Compare the two equations

The first equation is \(y = -\frac{4}{3}x+5\) and the second equation (after rewriting) is also \(y=-\frac{4}{3}x + 5\).
Two lines with the same slope (\(m = -\frac{4}{3}\)) and the same y - intercept are the same line (they are coincident, which means they are not parallel in the sense of distinct parallel lines, but we can say they are "infinite" in the sense of being the same line).
For a system of linear equations \(a_1x + b_1y=c_1\) and \(a_2x + b_2y = c_2\), if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (in our case, \(\frac{4}{4}=\frac{3}{3}=\frac{15}{15}\) when we consider the two equations \(y = -\frac{4}{3}x + 5\) (or \(4x+3y = 15\)) and \(4x + 3y=15\)), the system has infinite solutions.

Answer:

The lines will: \(\boldsymbol{\text{Be Infinite}}\) (Circle one)
The system will have: \(\boldsymbol{\text{Infinite Solutions}}\) (Circle one)