Question

fluency and skills practice | name:
lesson 12
understanding the number of solutions to a
system of linear equations continued
3 one equation in a system of equations is y - 8x = 1. write a second equation so
that the system of equations has the number of solutions stated.
no solution one solution infinitely many solutions

4 two runners are racing against each other. jeri graphs a linear equation for
each runner that shows the runner’s distance from the starting line over time.
the two equations form a system that has infinitely many solutions. describe
the intersection point(s) of the lines and explain what the solution means
in this situation.

Explanation:

Question 3 (no solution, one solution, infinitely many solutions)

No Solution

Step1: Recall parallel lines condition

For a system \(

$$\begin{cases} y - 8x = 1 \\ y - 8x = c \end{cases}$$

\) (where \( c
eq 1 \)), lines are parallel (same slope, different y - intercepts), so no solution. Rewrite \( y - 8x = 1 \) as \( y = 8x + 1 \). A parallel line has the same slope (\( m = 8 \)) and different y - intercept. Let's choose \( c = 2 \), so the equation is \( y - 8x = 2 \) (or \( y = 8x + 2 \)).

Step2: Verify

The first equation \( y - 8x = 1 \) has slope \( 8 \) and y - intercept \( 1 \). The second equation \( y - 8x = 2 \) has slope \( 8 \) and y - intercept \( 2 \). Parallel lines never intersect, so no solution.

One Solution

Step1: Recall intersecting lines condition

For a system to have one solution, the lines must have different slopes. The first equation \( y - 8x = 1 \) can be written as \( y = 8x + 1 \) (slope \( m_1 = 8 \)). Let's choose a line with a different slope, say slope \( m_2 = 1 \). Using point - slope or slope - intercept form, let's take \( y = x + 1 \) (or \( y - x = 1 \)).

Step2: Verify

The first equation \( y = 8x + 1 \) has slope \( 8 \), the second equation \( y = x + 1 \) has slope \( 1 \). Since \( 8
eq1 \), the lines intersect at one point, so one solution.

Infinitely Many Solutions

Step1: Recall coinciding lines condition

For a system to have infinitely many solutions, the two equations must be equivalent (same line). The first equation is \( y - 8x = 1 \). Multiply both sides by a non - zero constant, say \( 2 \). We get \( 2(y - 8x)=2\times1 \), which simplifies to \( 2y - 16x = 2 \) (or we can also write it as \( y - 8x = 1 \) itself, but to show a different form, multiplying by a constant is better).

Step2: Verify

If we solve \( 2y - 16x = 2 \) for \( y \), we divide both sides by \( 2 \): \( y - 8x = 1 \), which is the same as the original equation. So the two equations represent the same line, and there are infinitely many solutions.

Question 4

Answer:

• The two lines representing the runners' distance - time graphs are coincident (they overlap completely). So, every point on the line is an intersection point (infinitely many intersection points).
• In the race situation, this means that at every moment in time, the two runners are at the same distance from the starting line. They are running together, with the same speed and same (or equivalent) starting conditions, so their distance from the start is identical at all times during the race.

Final Answers for Question 3

• No solution: \( \boldsymbol{y - 8x = 2} \) (or other equations with same slope and different y - intercept as \( y - 8x = 1 \))
• One solution: \( \boldsymbol{y - x = 1} \) (or other equations with different slope as \( y - 8x = 1 \))
• Infinitely many solutions: \( \boldsymbol{2y - 16x = 2} \) (or other equivalent equations to \( y - 8x = 1 \))

Final Answer for Question 4

The lines are coincident (all points on the line are intersection points). The solution means the two runners are at the same distance from the starting line at every time during the race (they run together with the same speed and same relative position over time).