Question

svlc algebra 1a - standard (15260)
modeling with two - variable linear inequalities
using a graph to solve a problem
the graph represents the ways janelle can win the beanbag toss game. which describes a way janelle can win the game?
land on round 5 times;
land on square 3 times
land on round 7 times;
land on square 1 time
land on round 4 times;
land on square 4 times
land on round 6 times;
land on square 2 times

Explanation:

To solve this, we first find the equation of the dashed line. The line passes through (0, 6) and (10, 0). The slope \( m=\frac{0 - 6}{10 - 0}=-\frac{3}{5} \). Using slope - intercept form \( y = mx + b \) (where \( b = 6 \)), the equation is \( y=-\frac{3}{5}x + 6 \), or \( 3x+5y = 30 \). The solution region is above the dashed line (since the shading is above). We check each option:

Option 1: Land on round 5 times, land on square 3 times (\(x = 5,y = 3\))

Substitute into \( 3x + 5y \): \( 3(5)+5(3)=15 + 15=30 \). Since the line is dashed, the points on the line are not included. And we need the region above the line, so this point is on the line and not a winning point.

Option 2: Land on round 7 times, land on square 1 time (\(x = 7,y = 1\))

Substitute into \( 3x + 5y \): \( 3(7)+5(1)=21 + 5 = 26\). Since \( 26<30 \), this point is below the line and not in the solution region.

Option 3: Land on round 4 times, land on square 4 times (\(x = 4,y = 4\))

Substitute into \( 3x + 5y \): \( 3(4)+5(4)=12 + 20 = 32\). Since \( 32>30 \), this point is above the line and in the solution region.

Option 4: Land on round 6 times, land on square 2 times (\(x = 6,y = 2\))

Substitute into \( 3x + 5y \): \( 3(6)+5(2)=18+10 = 28\). Since \( 28<30 \), this point is below the line and not in the solution region.

Answer:

land on round 4 times; land on square 4 times