Question

learning objectives:

1. solve a system of equations by graphing and check the solutions by substitution.
2. identify a system with infinite or no solutions by their graphs

for #1 - 6: solve each system of equations using the graphing method.

1. \\(\

$$\begin{cases} y = -x - 5 \\\\ y = x + 1 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} -2x + y = 6 \\\\ y = -x \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} 4x - 2y = 6 \\\\ y = 2x + 3 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} 2x + y = 6 \\\\ y = -2x + 6 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} x = -3 \\\\ y = 2 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} y = 3x - 4 \\\\ y = -5x + 1 \\end{cases}$$

\\)

Explanation:

Problem 1:

Step1: Set equations equal

Since \( y = -x - 5 \) and \( y = x + 1 \), set \( -x - 5 = x + 1 \).

Step2: Solve for \( x \)

\( -x - x = 1 + 5 \) → \( -2x = 6 \) → \( x = -3 \).

Step3: Find \( y \)

Substitute \( x = -3 \) into \( y = x + 1 \): \( y = -3 + 1 = -2 \).

Step1: Rewrite first equation

From \( -2x + y = 6 \), get \( y = 2x + 6 \).

Step2: Analyze slopes

\( y = 2x + 6 \) (slope 2) and \( y = -x \) (slope -1). Different slopes, so one solution.

Step3: Set equal and solve

\( 2x + 6 = -x \) → \( 3x = -6 \) → \( x = -2 \). Then \( y = -(-2) = 2 \).

Step1: Rewrite first equation

From \( 4x - 2y = 8 \), divide by 2: \( 2x - y = 4 \) → \( y = 2x - 4 \).

Step2: Analyze slopes

\( y = 2x - 4 \) (slope 2) and \( y = 2x + 3 \) (slope 2). Same slope, different y - intercepts: no solution.

Answer:

\((-3, -2)\)

Problem 2: