Worked examples
The questions the video works, written out: the setup, each step, the answer and the trap.
0:43Example 1
Graph the system and and shade the solution region. Is (0, 0) a solution?
- Take the first inequality, . Its boundary is the line y = x - 2, slope 1 and y-intercept -2. The symbol is greater-than-or-equal, so the line is solid: points on it count. Test (0, 0): 0 is at least -2, true, so shade the side containing (0, 0), which is above the line.
- Take the second, . Boundary y = -x + 4, slope -1 and y-intercept 4. Strict less-than, so the line is dashed: points on it do not count. Test (0, 0): 0 is less than 4, true, so shade the side containing (0, 0), below the line.
- The solution set is only the overlap: the wedge above the solid line and below the dashed line. The two lines cross where x - 2 = -x + 4, so 2x = 6, x = 3 and y = 1; the wedge opens leftward from (3, 1).
- Check (0, 0) in both: 0 is at least -2 and 0 is less than 4, both true, so (0, 0) sits in the overlap and is a solution.
Answer: solid line y = x - 2 shaded above, dashed line y = -x + 4 shaded below; the overlap is the solution set; (0, 0) is a solution
A point solves a system only when it satisfies every inequality at once, which on the graph is where the two shadings overlap. The traps are drawing the strict boundary solid, and shading above by habit rather than testing a point; the second inequality shades below even though y is on the left.
1:12Example 2
For the inequality , is the boundary line solid or dashed, and which side gets shaded?
- Look at the symbol first. Strict less-than means points on the line make the two sides equal, not less, so they are not solutions. Dashed line.
- Pick a test point off the line. (0, 0) is easiest: substituting gives 0 < 4, true, so the side containing the origin is shaded. The origin sits below the line, since at x = 0 the line is at height 4.
- Check with a point on the other side, (5, 5): 5 < -5 + 4 = -1 is false, so that side stays blank.
Answer: dashed line, shade below
Two decisions turn any inequality into a region: the symbol decides solid or dashed, and one test point decides the side. The trap is assuming less-than always means below; it happens to be true here, but the test point is what proves it, and it still works when the inequality is written in a messier form.
1:41Example 3
Is the point (4, 3) a solution to the system and ?
- Substitute x = 4 and y = 3 into the first: , that is, 3 is at least 2. True.
- Substitute into the second: , that is, 3 is less than 0. False.
- A solution must pass every inequality. (4, 3) passes one and fails the other, so it is not a solution.
- Check on the graph: (4, 3) is above the solid line, which is at height 2 when x = 4, but it is also above the dashed line, which is at height 0 there, so it is outside the overlap.
Answer: No, (4, 3) is not a solution
Passing one inequality is not enough; the solution region is the overlap, where both are true at the same time. The trap is checking only the first inequality, seeing that 3 is at least 2, and answering yes. Test every inequality every time.
Lesson transcript
The narration of the video, word for word, under its chapter headings.
Welcome to SAT Climb. A system of inequalities gives you two rules at once. Each one shades half the plane — and the answer is only where the two shadings overlap. Let's make that overlap obvious. A system stacks two inequalities: y ≥ x − 2, and y < −x + 4. Each inequality on its own shades half the plane. To be a solution, a point has to satisfy both at the same time. Graph them one at a time. y ≥ x − 2 — draw a solid line and shade above. y < −x + 4 — draw a dashed line and shade below. The solution is where the two shaded regions overlap. Test (0,0): it's in both, so it's a solution. Two decisions turn each inequality into a region. First the line: a strict is dashed; ≤ or ≥ is solid. Second the side: pick a test point — (0,0) is easiest — plug it in, and if it's true, shade that side. Your turn. Is the point (4, 3) a solution? Check it against every inequality. First, 3 ≥ 2 — true. Second, is 3 < 0? No. It passes one and fails the other, so (4, 3) is not a solution. Three traps. One: a strict inequality uses a dashed line, not solid. Two: don't assume you shade above — test a point. Three: the answer is only the overlap, where every inequality is true. Systems of inequalities: solved. Graph each one, then take the overlap — that's your solution set. Start your free trial at satclimb.com. Keep climbing.
Read the written version: the Linear inequalities strategy guide, then try 25 hard Linear inequalities questions with full explanations. Both are free, no account needed.