match each system of linear equations with the graph of its solution, represented by the shaded regions of the graphs.

(left column, top to bottom)
1. ( y geq 5x - 1 )
( y geq -2x + 3 )
2. ( y leq 5x - 1 )
( y geq -2x + 3 )
3. ( y geq 5x - 1 )
( y leq -2x + 3 )
4. ( y leq 5x - 1 )
( y leq -2x + 3 )

(right column, top to bottom: graphs with shaded regions, coordinate axes x and y, grid lines, and lines ( y = 5x - 1 ) (dashed or solid) and ( y = -2x + 3 ) (dashed or solid))

Question

match each system of linear equations with the graph of its solution, represented by the shaded regions of the graphs.

(left column, top to bottom)
1. ( y geq 5x - 1 )
   ( y geq -2x + 3 )
2. ( y leq 5x - 1 )
   ( y geq -2x + 3 )
3. ( y geq 5x - 1 )
   ( y leq -2x + 3 )
4. ( y leq 5x - 1 )
   ( y leq -2x + 3 )

(right column, top to bottom: graphs with shaded regions, coordinate axes x and y, grid lines, and lines ( y = 5x - 1 ) (dashed or solid) and ( y = -2x + 3 ) (dashed or solid))

Accepted Answer

# Explanation:
## Step1: Analyze the first system $ y \geq 5x - 1 $ and $ y \geq -2x + 3 $
For $ y \geq 5x - 1 $, the region is above the line $ y = 5x - 1 $ (since the inequality is "greater than or equal to", and the line has a positive slope). For $ y \geq -2x + 3 $, the region is above the line $ y=-2x + 3 $ (this line has a negative slope). The intersection of these two "above" regions should be a region that is above both lines. Looking at the graphs, the third graph (bottom - middle right) has the shaded region above both lines (the line with positive slope $ y = 5x-1 $ and the line with negative slope $ y=-2x + 3 $)? Wait, no, let's re - check. Wait, the first system: $ y\geq5x - 1 $ (slope 5, y - intercept - 1) and $ y\geq - 2x+3 $ (slope - 2, y - intercept 3). The intersection of two "greater than or equal to" inequalities will be the region that satisfies both, so above both lines. The third graph (the one with the shaded region at the top, between the two lines) - wait, maybe I mixed up. Let's check the second system: $ y\leq5x - 1 $ and $ y\leq - 2x + 3 $. For $ y\leq5x - 1 $, the region is below the line $ y = 5x-1 $, and for $ y\leq - 2x + 3 $, the region is below the line $ y=-2x + 3 $. The intersection of two "less than or equal to" regions is below both lines. The second graph (middle - right) has the shaded region on the left, below both lines? Wait, no, the second system $ y\leq5x - 1 $ and $ y\leq - 2x + 3 $: the line $ y = 5x-1 $ has a positive slope, so below it is to the left of the line (for positive slope, as x increases, y increases, so below is where y is smaller for a given x). The line $ y=-2x + 3 $ has a negative slope, so below it is to the right of the line (as x increases, y decreases, so below is where y is smaller for a given x). The intersection of these two "below" regions is the region that is below both lines. The second graph (middle - right) has the shaded region on the left, which is below $ y = 5x-1 $ (since for $ x = 0 $, $ y\leq - 1 $ for $ y\leq5x - 1 $, but wait, no, $ y = 5x-1 $ at $ x = 0 $ is $ y=-1 $, so $ y\leq - 1 $ when $ x = 0 $, but the second graph at $ x = 0 $ has $ y\leq3 $ for $ y\leq - 2x + 3 $ (at $ x = 0 $, $ y\leq3 $) and $ y\leq - 1 $ for $ y\leq5x - 1 $ (at $ x = 0 $, $ y\leq - 1 $). Wait, maybe a better approach is to find the intersection point of the two lines $ y = 5x-1 $ and $ y=-2x + 3 $. Set $ 5x-1=-2x + 3 $, then $ 5x+2x=3 + 1 $, $ 7x = 4 $, $ x=\frac{4}{7}\approx0.57 $, and $ y=5\times\frac{4}{7}-1=\frac{20}{7}-1=\frac{13}{7}\approx1.86 $.

## Step2: Analyze the third system $ y\geq5x - 1 $ and $ y\leq - 2x + 3 $
For $ y\geq5x - 1 $ (above the line with positive slope) and $ y\leq - 2x + 3 $ (below the line with negative slope). The intersection of these two regions is between the two lines, above $ y = 5x-1 $ and below $ y=-2x + 3 $. The first graph (top - right) has the shaded region between the two lines, above the line with positive slope ( $ y = 5x-1 $) and below the line with negative slope ( $ y=-2x + 3 $)? Wait, the first graph (top - right) has the shaded region between the two lines. Let's check the intersection point. The two lines $ y = 5x-1 $ and $ y=-2x + 3 $ intersect at $ x=\frac{4}{7},y=\frac{13}{7}\approx1.86 $. The first graph (top - right) has the shaded region between the two lines, which is the region that satisfies $ y\geq5x - 1 $ (since it's above the line with positive slope) and $ y\leq - 2x + 3 $ (since it's below the line with negative slope). So the system $ y\geq5x - 1 $ and $ y\leq - 2x + 3 $ matches the first graph (top - right).

## Step3: Analyze the fourth system $ y\leq5x - 1 $ and $ y\geq - 2x + 3 $
For $ y\leq5x - 1 $ (below the line with positive slope) and $ y\geq - 2x + 3 $ (above the line with negative slope). The intersection of these two regions is between the two lines, below $ y = 5x-1 $ and above $ y=-2x + 3 $. The fourth graph (bottom - right) has the shaded region to the right of the line with positive slope and above the line with negative slope, which is the region that satisfies $ y\leq5x - 1 $ (below the line with positive slope, since for large x, $ y = 5x-1 $ is large, so below it is to the right? Wait, no, for a line with positive slope $ y = mx + b $ ($ m>0 $), the region $ y\leq mx + b $ is to the left of the line (as x increases, y increases, so for a given y, x is smaller when $ y\leq mx + b $). The region $ y\geq - 2x+3 $ (for $ m=-2<0 $) is above the line, which is to the left of the line (since as x increases, y decreases, so for a given y, x is smaller when $ y\geq - 2x + 3 $). Wait, no, this is getting confusing. Let's use the process of elimination.

The first system $ y\geq5x - 1 $, $ y\geq - 2x + 3 $: shaded region above both lines. The third graph (bottom - middle right) has the shaded region above both lines? Wait, the third graph (bottom - middle right) has the shaded region at the top, between the two lines? No, the third graph (bottom - middle right) has the shaded region above the line with positive slope and above the line with negative slope? Wait, no, the line $ y=-2x + 3 $ at $ x = 0 $ is $ y = 3 $, and $ y = 5x-1 $ at $ x = 0 $ is $ y=-1 $. So the region $ y\geq5x - 1 $ and $ y\geq - 2x + 3 $ will be the region that is above the higher of the two lines at each x. Since $ -2x + 3$ is above $ 5x - 1 $ when $ -2x+3\geq5x - 1$, $ 3 + 1\geq7x $, $ x\leq\frac{4}{7}\approx0.57 $, and $ 5x - 1$ is above $ -2x + 3 $ when $ x>\frac{4}{7}$. So the region is above $ -2x + 3 $ when $ x\leq\frac{4}{7}$ and above $ 5x - 1 $ when $ x>\frac{4}{7}$, which is the region above both lines, and the third graph (bottom - middle right) has the shaded region at the top, which is above both lines.

The fourth system $ y\leq5x - 1 $ and $ y\geq - 2x + 3 $: the region is below $ 5x - 1 $ and above $ -2x + 3 $, which is the region between the two lines, below the line with positive slope and above the line with negative slope, and the fourth graph (bottom - right) has the shaded region to the right of the line with positive slope and above the line with negative slope, which matches this.

Let's summarize:

- System 1 ($ y\geq5x - 1,y\geq - 2x + 3 $): Shaded region above both lines (third graph, bottom - middle right)
- System 2 ($ y\leq5x - 1,y\leq - 2x + 3 $): Shaded region below both lines (second graph, middle - right)
- System 3 ($ y\geq5x - 1,y\leq - 2x + 3 $): Shaded region between the lines, above $ y = 5x-1 $ and below $ y=-2x + 3 $ (first graph, top - right)
- System 4 ($ y\leq5x - 1,y\geq - 2x + 3 $): Shaded region between the lines, below $ y = 5x-1 $ and above $ y=-2x + 3 $ (fourth graph, bottom - right)

# Answer:
- $ y\geq5x - 1,y\geq - 2x + 3 $: The third graph (bottom - middle right)
- $ y\leq5x - 1,y\leq - 2x + 3 $: The second graph (middle - right)
- $ y\geq5x - 1,y\leq - 2x + 3 $: The first graph (top - right)
- $ y\leq5x - 1,y\geq - 2x + 3 $: The fourth graph (bottom - right)

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the first system \( y \geq 5x - 1 \) and \( y \geq -2x + 3 \)

Step2: Analyze the third system \( y\geq5x - 1 \) and \( y\leq - 2x + 3 \)

Answer: