Algebra 1 · Unit 4
Show the solution to a linear inequality with two variables on a graph.
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≥/≤ solid · >/< dashed
Test a point such as (0, 0)
Solid or dashed
≥ or ≤ is a solid boundary (points on the line count). > or < is dashed (points on the line do not count).
Above or below
If the inequality is solved for y, shade above for > or ≥ and below for < or ≤.
Test a point
When the line does not go through the origin, test (0, 0). If the inequality is true, shade that side. If it is false, shade the other side.
Vertical and horizontal
x = a is vertical. x ≥ a shades to the right. y = b is horizontal. y ≥ b shades above.
Integer slopes already solved for y. Graph only in the answer.
Graph the boundary y = x − 2.
Draw the line solid.
Shade above the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Graph the boundary y = −x + 3.
Draw the line dashed.
Shade below the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Graph the boundary y = 2x + 1.
Draw the line solid.
Shade below the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Graph the boundary y = 2.
Draw the line solid.
Shade above the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Graph the boundary y = −2x + 4.
Draw the line dashed.
Shade above the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Graph the boundary y = x.
Draw the line solid.
Shade below the line.
Test (0, 0) if it is not on the boundary.
Answer: the shaded half-plane
Solve for y first, or graph a vertical line.
Solve for y.
2x + y < 6
y < −2x + 6
Draw y = −2x + 6 dashed.
Shade below.
Test (0, 0): 0 < 6, true, so shade toward the origin.
Answer: the shaded half-plane
y ≥ −x + 4
Draw y = −x + 4 solid.
Shade above.
Test (0, 0): 0 ≥ 4 is false, so do not shade toward the origin.
Answer: the shaded half-plane
x = 2 is vertical.
Solid, because of ≥.
Shade to the right.
Test (0, 0): 0 ≥ 2 is false, so the origin is not shaded.
Answer: the shaded half-plane
−x + y > 3
y > x + 3
Draw y = x + 3 dashed.
Shade above.
Answer: the shaded half-plane
x = −1 is vertical and dashed.
Shade to the left.
Test (0, 0): 0 < −1 is false, so the origin is not shaded.
Answer: the shaded half-plane
y ≤ −3x + 3
Solid, shade below.
Test (0, 0): 0 ≤ 3, true.
Answer: the shaded half-plane
Flip the inequality when you divide by a negative. Fractional slopes.
2x − y > 6
− y > −2x + 6
Divide by −1 and flip the inequality.
y < 2x − 6
Draw y = 2x − 6 dashed.
Shade below.
Test (0, 0): 0 > 6 is false, so the origin is not in the solution.
Answer: the shaded half-plane
Slope 12, y-intercept (0, −1).
Solid, shade above.
Answer: the shaded half-plane
−3x + 2y ≤ 6
2y ≤ 3x + 6
y ≤ 32x + 3
Solid, shade below.
Answer: the shaded half-plane
x − 2y ≥ −4
−2y ≥ −1x − 4
Divide by −2 and flip.
y ≤ 12x + 2
Solid, shade below.
Answer: the shaded half-plane
Vertical dashed line x = −3.
Shade to the right.
Test (0, 0): 0 > −3, true.
Answer: the shaded half-plane
Slope −23, y-intercept (0, 2).
Solid, shade below.
Answer: the shaded half-plane
> is dashed, not solid.
y > x + 1 shades above, not below.
Answer: dashed, shade above
3(2) − (−5) = 6 + 5 = 11
11 ≥ 10 is true, so yes.
The boundary is 3x − y = 10. The point is on the true side, not on the line, because 11 ≠ 10.
Answer: yes; the true side, not on the line
Vertical solid line x = 0, shade to the right.
In a tickets story, x is a count, so negative tickets are not allowed. The inequality x ≥ 0 is the first-quadrant restriction on x.
Answer: right half-plane; counts cannot be negative
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