Inequalities on graphs: worked examples and method

Compare heights

Work out the line’s y at that x. A larger point-y is above the line.

On the line

The point (1, 3) and the line y=2x+1 lands on the line, because 2×1+1 = 3.

Strict inequality

A point must satisfy every inequality to be inside a region. A point satisfies y < x when its y is less than its x. A point on the line y = x does not satisfy y < x, so a boundary for a strict inequality is drawn dashed. For y ≤ x the boundary is solid and points on the line do count.

Boundary intercept

The y-intercept of the boundary y=2x+1 is 1.

Boundary gradient

The gradient of the boundary line y=2x is 2.

Yes or no

Substitute. If the inequality is true, say yes.

Draw the boundary, then decide the side

Treat y>2x+1 as the line y=2x+1 first. It crosses the y-axis at 1 and has gradient 2, so it also goes through (1, 3). The sign is strict, so the line is dashed. The origin gives 0 on the left and 2(0)+1=1 on the right. 0>1 is false, so the origin is not in the region. Shade the other side of the line, or label R on the side that does not contain (0, 0), depending on what the question asks.

A solid line is a different statement. y≤ 2x+1 uses the same line and the boundary is included. The origin now passes, because 0≤ 1 is true, so the region is the other side of the line, the side with the origin, and points on the line count. Mixing these up loses the mark for the line style even when the shading is on the correct side. Write "dashed" or "solid" next to the line while you draw it. Do not wait until the end.

Method: 1. Draw the boundary from its equation, ignoring the inequality sign for that line. 2. Use a solid line for ≤ or ≥, and a dashed line for < or >. 3. Pick a test point that is not on the line. 4. Shade the side where the test point makes the inequality true.

A test point beats a guess

If the line does not pass through the origin, (0, 0) is the quickest test. If it does, pick (1, 0) or (0, 1) instead. Substitute carefully: for x+2y≥ 4, the origin gives 0≥ 4, which is false, so the origin is on the unwanted side. For x+2y≤ 4 the same arithmetic says the origin is on the wanted side. The line is identical. Only the shading changes. That is why guessing from the direction of the sign is unreliable once there are two variables.

When several inequalities appear together, test once for each line, or test one point against all of them at the end. A point is in the final region only if every inequality is true. (2, 3) in x+y<6, y>1, x≥ 0: the sum is 5, which is less than 6; y is greater than 1; x is not negative. All three hold. (4, 2) has sum 6, so it fails the strict inequality even though the other two hold. One failure is enough to reject the point.

Method: 1. Choose a point with easy coordinates, often (0,0) if it is not on the line. 2. Substitute it into the inequality, not into the equation alone. 3. If it fails, shade the other side. 4. Write the test on the page so the side is justified.

Listing integer points

For x≥ 0, y≥ 0 and x+y≤ 3, list systematically. When x=0, y can be 0, 1, 2, 3. When x=1, y can be 0, 1, 2. When x=2, y can be 0, 1. When x=3, y can be 0. That is 4+3+2+1=10 points. The formula for x+y≤ n in the first quadrant, axes included, is ((n+1)(n+2))/2. For n=3 that is 10. Use the list in the answer and the formula as a check, not the other way round, so a missed point is visible.

A strict edge removes points. x+y<3 with x≥ 0, y≥ 0 drops (0, 3), (1, 2), (2, 1) and (3, 0). Six points remain. If the question draws a dashed line, those four points are on the boundary and you must leave them out. Writing them "because they look close" is the usual error. Deal with one inequality at a time: find the candidates, then cross out anything on a dashed edge.

Method: 1. Mark the boundary lines and the region they enclose. 2. List whole-number coordinates inside the region, including a solid edge. 3. Leave out a dashed edge. 4. Count the list and check one point just outside.

The overlap, and the instruction about shading

Three lines make a polygon only if you keep the correct side of each. Take y≤ x+1, y>2 and x<4. The line y=x+1 is solid, y=2 is dashed, and x=4 is dashed. Integers with y≥ 3, x≤ 3 and y≤ x+1 are (2, 3), (3, 3) and (3, 4). Check (2, 3): it sits on y=x+1, and ≤ allows that. Check (3, 4): 4=3+1, so it is on the solid line, x=3<4, and y>2. Check (4, 3): x=4 is not strictly less than 4, so it is out. Check (1, 3): 3≤ 1+1 is false.

Some questions say "shade the region that satisfies all the inequalities". Others say "shade the unwanted region and label R". Those are opposite shadings of the same maths. Underline the instruction. If you have shaded the wanted region by habit, label it and say so, rather than leaving a reader to guess. On a list of integer points the shading instruction does not change the list. The solid and dashed rules still do.

Method: 1. Sketch each inequality on the same diagram. 2. The solution is the overlap, where every inequality holds. 3. A point has to pass every test, not just one. 4. If the question says "unshaded", read which region was actually required.

Going backwards: writing inequalities for a region

Some questions give you the shaded region and ask for the inequalities that define it. Look at each boundary line in turn. A vertical line x=5 with the region on its left gives x≤ 5 if the line is solid and x<5 if it is dashed. A horizontal line works the same way with y. Always say which side by picking a point clearly inside the region and checking it, rather than trusting your first impression of the picture.

For a sloping boundary, first find its equation from two points it passes through. A boundary through (1, 4) and (5, 6) has gradient (6-4)/(5-1)=1/2, so y=1/2x+7/2. Then test a point inside the region, such as (3, 3). Since 3<1/2× 3+7/2=5, the region is below the line and the inequality is yle1/2x+7/2. A shaded triangle needs three inequalities, a quadrilateral needs four.

Method: 1. List the boundary lines one at a time. 2. Find the equation of any sloping boundary from two points on it. 3. Test one point inside the region against each boundary to choose the sign. 4. Use ≤ or ≥ for a solid line and < or > for a dashed line.

Tip: Count the boundaries on the diagram, then check you have written exactly that many inequalities.

Sloping boundaries and a minus sign on y

To draw x+2y≤ 8 quickly, find where the line cuts the axes. Put x=0 to get 2y=8, so y=4. Put y=0 to get x=8. Join (0, 4) and (8, 0). If an inequality such as 3x-2y<6 has a negative y term, you can still test a point such as the origin: 0<6 is true, so the origin is in the region. This is safer than rearranging, because rearranging can reverse the inequality.

If you do rearrange, remember that multiplying or dividing both sides by a negative number reverses the sign. From 3x-2y<6, subtract 3x to get -2y<6-3x. Dividing by -2 gives y>(3x-6)/2, which is y>3/2x-3. The "less than" has become "greater than". Check with (0, 0) again: 0>-3 is true, which agrees with the first test. A dashed boundary stays dashed after rearranging.

Method: 1. Find the axis crossings by putting x=0 and then y=0. 2. Test the origin, or another easy point if the line passes through the origin. 3. If you rearrange, reverse the inequality sign when you divide by a negative. 4. Confirm the rearranged form with the same test point.

Integer pairs and the greatest or least value

A common calculator-free question gives bounds such as -3≤ x≤ 2 and 7≤ x+y≤ 10, with x and y integers, and asks for the greatest possible value of y-x. Do not try every pair. Rewrite the target using the quantities you know: y-x=(x+y)-2x. This is largest when x+y is as big as possible and x is as small as possible, so take x+y=10 and x=-3.

That gives y=13 and y-x=10+6=16. Always check that your chosen pair obeys every condition: x=-3 is allowed, and x+y=10 is allowed. For the least value, do the opposite, with x+y as small as possible and x as large as possible. Writing the target in terms of the given sums and differences is a standard way to turn a long search into two lines of working.

Method: 1. Write the target as a combination of the given expressions. 2. Decide which bound to use for each part to make the target as large or as small as possible. 3. Work out the matching integer pair. 4. Check that the pair satisfies every inequality.

Is the claim correct? Points on boundaries

Exam questions sometimes show a region and a statement such as "Geoffrey says the point (2, 1) is not in the region." You must say whether he is right and give a reason. The deciding factor is usually the boundary. A point on a solid line is in the region, since ≤ and ≥ include the line. A point on a dashed line is not, since < and > exclude it. Check the point in every inequality and say which one fails, if any.

A reason has to refer to the actual numbers. "It is on the line" is weak. "y=1 is not greater than 1, so (2, 1) fails y>1" is a full reason. Where a context is given, such as numbers of adults and children in a hall, the point must also make sense: a negative number of people, or a fraction of a child, is not allowed even if it satisfies the inequalities. Always read the question for words like "whole number".

Method: 1. Substitute the point into every inequality in turn. 2. Write each result as a true or false statement with the numbers shown. 3. Mention whether a boundary is solid or dashed when the point lies on it. 4. In a context, check that the answer is a whole number and not negative.