Question

part a. 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. (3 points)

| system of equations | \\(\

$$\begin{cases} y = 5x + 12 \\\\ y = 5x + 10 \\end{cases}$$

\\) | \\(\

$$\begin{cases} y = 5x - 10 \\\\ y = 10 + x \\end{cases}$$

\\) | \\(\

$$\begin{cases} y = -5x + 6 \\\\ y = 6 - 5x \\end{cases}$$

\\) |

no solution	\\(\square\\)	\\(\square\\)	\\(\square\\)
one solution	\\(\square\\)	\\(\square\\)	\\(\square\\)
infinitely many solutions	\\(\square\\)	\\(\square\\)	\\(\square\\)

Explanation:

Step1: Analyze first system

First system:

$$\begin{cases} y=5x+12 \\ y=5x+10 \end{cases}$$

Set equal: $5x+12=5x+10$
Simplify: $12=10$ (false, parallel lines)

Step2: Analyze second system

Second system:

$$\begin{cases} y=5x-10 \\ y=10+x \end{cases}$$

Set equal: $5x-10=x+10$
Solve for $x$: $4x=20 \implies x=5$
Substitute $x=5$: $y=5(5)-10=15$ (unique solution)

Step3: Analyze third system

Third system:

$$\begin{cases} y=-5x+6 \\ y=6-5x \end{cases}$$

Rewrite second equation: $y=-5x+6$ (same as first, coinciding lines)

Answer:

System of Equations	No Solution	One Solution	Infinitely Many Solutions

|

$$\begin{cases} y=5x+12 \\ y=5x+10 \end{cases}$$

$\boldsymbol{\times}$

|

$$\begin{cases} y=5x-10 \\ y=10+x \end{cases}$$

$\boldsymbol{\times}$

|

$$\begin{cases} y=-5x+6 \\ y=6-5x \end{cases}$$

$\boldsymbol{\times}$