QUESTION IMAGE
Question
svlc algebra 1a - standard (15260)
solving systems of linear inequalities
determining two - variable linear inequalities with no
solution
which linear inequality will not have a shared solution set with the graphed linear inequality?
$y > \frac{5}{3}x + 2$
$y > -\frac{5}{3}x + 2$
$y < \frac{5}{3}x - 2$
$y < -\frac{5}{3}x + 1$
Step1: Analyze the graphed inequality
The graphed inequality is \( y < \frac{5}{3}x + 1 \). It has a slope of \( \frac{5}{3} \) and a y - intercept of 1, with the region below the dashed line shaded.
Step2: Analyze each option
- Option 1: \( y > \frac{5}{3}x + 2 \)
The line \( y=\frac{5}{3}x + 2 \) has the same slope as the graphed line (\( \frac{5}{3} \)) but a higher y - intercept (2 > 1). The region \( y > \frac{5}{3}x + 2 \) is above this line, and the region \( y < \frac{5}{3}x + 1 \) is below its line. Since the two lines are parallel (same slope) and the regions are on opposite sides of their respective parallel lines, there is no overlap between the solution sets.
- Option 2: \( y > -\frac{5}{3}x + 2 \)
The slope of this line is \( -\frac{5}{3} \), which is different from the slope of the graphed line (\( \frac{5}{3} \)). The two lines will intersect, so their solution sets will have a shared region.
- Option 3: \( y < \frac{5}{3}x - 2 \)
The line \( y = \frac{5}{3}x-2 \) has the same slope as the graphed line (\( \frac{5}{3} \)) but a lower y - intercept (- 2<1). The region \( y < \frac{5}{3}x - 2 \) is below this line, and the region \( y < \frac{5}{3}x + 1 \) is below its line. Since the lines are parallel and both regions are below their respective lines (with the second line being "lower" in terms of y - intercept), there will be a shared region (the region below both lines).
- Option 4: \( y < -\frac{5}{3}x + 1 \)
The slope of this line is \( -\frac{5}{3} \), different from the graphed line's slope. The two lines will intersect, so their solution sets will have a shared region.
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\( y > \frac{5}{3}x + 2 \) (the first option: \( y > \frac{5}{3}x + 2 \))