Question

y < x - 1\
-x + 2y \leq 1\
in which region of the graph is the solution to this system found?\
section __\
\bigcirc i\
\bigcirc ii\
\bigcirc iii\
\bigcirc iv

Explanation:

Step1: Analyze \( y < x - 1 \)

The inequality \( y < x - 1 \) is a dashed line (since it's \( < \)) with slope 1 and y - intercept - 1. The solution to this inequality is the region below the dashed line.

Step2: Analyze \( -x + 2y \leq 1 \)

First, rewrite it in slope - intercept form (\( y=mx + b\)):
Add \( x \) to both sides: \( 2y\leq x + 1 \)
Divide by 2: \( y\leq\frac{1}{2}x+\frac{1}{2} \)
This is a solid line with slope \( \frac{1}{2} \) and y - intercept \( \frac{1}{2} \). The solution to this inequality is the region below or on the solid line.

Step3: Find the intersection of regions

We need to find the region that satisfies both inequalities. Let's test a point in each region. Let's take the origin \((0,0)\) for testing:

• For \( y < x - 1 \): \( 0<0 - 1\) i.e., \( 0 < - 1\) which is false. So the origin is not in the region of \( y < x - 1 \).
• Let's test a point in region II, say \((3,0)\):
• For \( y < x - 1 \): \( 0<3 - 1=2\), which is true.
• For \( y\leq\frac{1}{2}x+\frac{1}{2} \): \( 0\leq\frac{1}{2}(3)+\frac{1}{2}=\frac{3 + 1}{2}=2\), which is true.
• Test a point in region I, say \((3,2)\):
• For \( y < x - 1 \): \( 2<3 - 1 = 2\), which is false (since it's \( < \), not \( \leq \)).
• Test a point in region III, say \((-1,-2)\):
• For \( y < x - 1 \): \( - 2<-1 - 1=-2\), which is false.
• Test a point in region IV, say \((-1,1)\):
• For \( y < x - 1 \): \( 1<-1 - 1=-2\), which is false.

Answer:

II