Question

1. draw a second line on the graph in order to demonstrate a system of equations with no solution. record your equation.
2. draw a second line on the graph in order to demonstrate a system of equations with one solution. record your equation.

Explanation:

Step1: Analyze Problem 10 (No Solution System)

A system with no solution has parallel lines (same slope, different y - intercepts). First, find the slope of the given line. Let's take two points on the given line, e.g., \((-2, 3)\) and \((2, -1)\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-1 - 3}{2-(-2)}=\frac{-4}{4}=-1\). So we need a line with slope \(-1\) and different y - intercept. Let's choose y - intercept \(b = 2\), so the equation is \(y=-x + 2\).

Step2: Analyze Problem 11 (One Solution System)

A system with one solution has lines with different slopes. The given line has slope \(m = 1\) (e.g., points \((-4,-4)\) and \((4,4)\), \(m=\frac{4-(-4)}{4 - (-4)}=\frac{8}{8}=1\)). Let's choose a different slope, say \(m = 2\), and y - intercept \(b = 0\), so the equation is \(y = 2x\). (Note: For drawing, the first line in problem 10 has slope - 1, so a parallel line like \(y=-x + 3\) would also work. For problem 11, any line with slope not equal to 1, e.g., \(y=-x\) would intersect the given line.)

Answer:

Problem 10 Equation: \(y=-x + 2\) (or other parallel line with slope - 1, different y - intercept)
Problem 11 Equation: \(y = 2x\) (or other line with slope ≠1)