Question

part b: determine whether each system of equations has no solution, one solution, or infinitely many solutions. place an x in the correct box for each system of equations. (6 points)

system of equations	no solution	one solution	infinitely many solutions

| \\(\

$$\begin{cases} 2x + y = 5 \\\\ y = 5 - 2x \\end{cases}$$

\\) | \\(\square\\) | \\(\square\\) | \\(\square\\) |
| \\(\

$$\begin{cases} y = -5x + 4 \\\\ 15x + 3y = 12 \\end{cases}$$

\\) | \\(\square\\) | \\(\square\\) | \\(\square\\) |
| \\(\

$$\begin{cases} y + 8 = 5x \\\\ 10x + 16 = 2y \\end{cases}$$

\\) | \\(\square\\) | \\(\square\\) | \\(\square\\) |

Explanation:

Step1: Analyze first system

Substitute $y=5-2x$ into $2x+y=5$:
$2x + (5-2x) = 5$
$2x + 5 - 2x = 5$
$5=5$
This is always true, so the equations are identical.

Step2: Analyze second system

Substitute $y=-5x+4$ into $15x+3y=12$:
$15x + 3(-5x+4) = 12$
$15x -15x +12 = 12$
$12=12$
This is always true, so the equations are identical.

Step3: Analyze third system

Rewrite $y+8=5x$ as $y=5x-8$. Substitute into $10x+16=2y$:
$10x+16=2(5x-8)$
$10x+16=10x-16$
$16=-16$
This is never true, so lines are parallel.

Answer:

1. For

$$\begin{cases}2x + y = 5 \\ y = 5 - 2x\end{cases}$$

: Mark $\boldsymbol{x}$ in the "Infinitely Many Solutions" box

1. For

$$\begin{cases} y = -5x + 4 \\ 15x + 3y = 12 \end{cases}$$

: Mark $\boldsymbol{x}$ in the "Infinitely Many Solutions" box

1. For

$$\begin{cases} y + 8 = 5x \\ 10x + 16 = 2y \end{cases}$$

: Mark $\boldsymbol{x}$ in the "No Solution" box