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Year 10 Maths Advanced Algebra Review

Graphing Inequalities

20 practice questions 1 video lesson Theory + worked examples

Graphing an inequality shows all of its solutions on a number line. A closed circle means the end value is included, an open circle means it is not, and the arrow or segment shows the rest of the range. In NSW Year 10 Advanced maths you solve the inequality first, then graph the result, as for \(-3\lt 2x+1\le 9\) (NESA outcome MA5‑EQU‑P‑01).

You will read inequalities from number lines, match a graph to the inequality that produces it, and solve and graph inequalities that involve brackets, fractions and dividing by a negative. You will graph double inequalities and pairs of conditions on one line, count the integers in a solution, and work backwards from a graph to find unknown constants. One question finds how many boxes a lift can safely carry. It is part of Year 10 Advanced Algebra Review.

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Theory

To graph an inequality on a number line, put a dot at the boundary number, open if it is not included and closed if it is, then shade the numbers that work. This Year 10 Advanced maths page covers reading a graph, solving then graphing, two conditions at once, whole-number solutions and finding unknown constants from a graph.

The solution of an inequality is a set of numbers, and a number line shows the whole set at once. The boundary is the number where the inequality changes from true to false.

An open dot (hollow circle) means the boundary is not included: use it for \(\lt \) and \(\gt \). A closed dot (filled circle) means it is included: use it for \(\le\) and \(\ge\). An arrow at the end of the shading means the solution goes on forever in that direction.

A solution between two numbers, such as \(-2 \lt x \le 3\), is a shaded segment with a dot at each end. It means \(x \gt -2\) and \(x \le 3\) at the same time, so it is where the two separate graphs overlap.

If the pronumeral can only be a whole number, such as a number of boxes, the graph is a set of separate dots rather than a shaded line. Linear inequalities prepare you for Advanced maths in Years 11 and 12.

Three inequalities on number linesTop: x greater than 1, open dot at 1, shaded to the right. Middle: x less than or equal to negative 1, closed dot, shaded to the left. Bottom: negative 2 less than x less than or equal to 3, open dot at negative 2, closed dot at 3, shaded between. βˆ’4βˆ’3βˆ’2βˆ’101234 x > 1 βˆ’4βˆ’3βˆ’2βˆ’101234 x ≀ βˆ’1 βˆ’4βˆ’3βˆ’2βˆ’101234 βˆ’2 < x ≀ 3
Open dot: the end point is not included. Closed dot: it is. An arrow means the solution keeps going.
A whole-number solutionA number line from negative 2 to 7 with separate filled dots at 0, 1, 2, 3 and 4 only, showing whole numbers n less than or equal to 4. βˆ’2βˆ’101234567 whole numbers n with n ≀ 4 Separate dots, not a shaded line: n = 2.5 boxes or people makes no sense.
When the pronumeral counts things, graph the solution as separate dots.
InequalityDot at the boundaryShade
\(x\gt a\)openright of \(a\)
\(x\ge a\)closedright of \(a\)
\(x\lt a\)openleft of \(a\)
\(x\le a\)closedleft of \(a\)
\(a\lt x\le b\)open at \(a\), closed at \(b\)between
a<x≤b
Pronumeral on the right. \(5 \gt x\) means \(x \lt 5\). Rewrite it with \(x\) first before you decide which way to shade.
Which inequality matches the graph? Solve each option and compare its solution with the graph. The boundary, the dot and the direction must all match.

How to graph the solution of an inequality

  1. Solve the inequality, reversing the sign if you multiply or divide by a negative.
  2. Rewrite it with the pronumeral on the left, such as \(x \ge -3\).
  3. Draw a number line with the boundary near the middle and a few numbers either side.
  4. Dot the boundary: open for \(\lt \) or \(\gt \), closed for \(\le\) or \(\ge\).
  5. Shade towards the numbers that work and finish with an arrow, or stop at a second dot for a segment.
  6. Test one number in the shaded part in the original inequality.

Other question types

  1. Reading a graph: note the boundary, the type of dot and the direction, then write the inequality (Example 1).
  2. Two conditions: graph each one, then keep only the overlap. If the graphs do not overlap, there is no solution.
  3. Counting integers: list the whole numbers in the shaded part. Include an end point only if its dot is closed.
  4. Whole-number contexts: solve, then round in the direction that keeps the inequality true, and graph separate dots.
  5. Unknown constants: read the solution from the graph, then either build the given expression from \(x\) (Example 4) or solve the inequality in terms of the constant and match the boundary.
Example 1 β€” Reading a number line
Write the inequality shown by each graph. For (ii), also list the integers in the solution.
(i)Graph for part (i)Open dot at 3, shaded to the left. βˆ’4βˆ’3βˆ’2βˆ’101234 (ii)Graph for part (ii)Closed dot at negative 4, open dot at 1, shaded between. βˆ’5βˆ’4βˆ’3βˆ’2βˆ’10123
Solution

(i) The open dot at \(3\) means \(3\) is not included, and the arrow points left (smaller numbers).

\(x\)\(\lt \)\(\boldsymbol{3}\)

(ii) Closed dot at \(-4\) (included), open dot at \(1\) (not included), shaded between.

\(-4\)\(\le\)\(\boldsymbol{x\lt 1}\)
−4≤x<1

Integers: \(\mathbf{-4,\ -3,\ -2,\ -1,\ 0}\). The \(1\) is left out because its dot is open.

Example 2 β€” Solve, then graph
Solve \(\dfrac{x+1}{-2}\le 1\) and graph the solution.
Solution

Multiply both sides by \(-2\). It is negative, so reverse the sign.

\(x+1\)\(\ge\)\(-2\)
\(x\)\(\ge\)\(\boldsymbol{-3}\)
x+1−2≤1⇒x≥−3

Closed dot at \(-3\) because of \(\ge\); shade to the right.

x greater than or equal to negative 3Closed dot at negative 3, shaded to the right. βˆ’5βˆ’4βˆ’3βˆ’2βˆ’10123

Check with \(x = 1\): \(\dfrac{2}{-2} = -1\le 1\). True, and \(1\) is in the shaded part.

Example 3 β€” Two conditions together
Graph the values of \(x\) that satisfy both \(2x+1\gt -5\) and \(4-x\ge 1\). How many integers satisfy both?
Solution

Solve each inequality on its own.

\(2x\)\(\gt \)\(-6\)
\(x\)\(\gt \)\(-3\)
\(-x\)\(\ge\)\(-3\)
\(x\)\(\le\)\(3\)

Both are true where the two graphs overlap:

\(-3\)\(\lt \)\(\boldsymbol{x\le 3}\)
−3<x≤3
negative 3 to 3Open dot at negative 3, closed dot at 3, shaded between. βˆ’5βˆ’4βˆ’3βˆ’2βˆ’1012345

Integers: \(-2,\ -1,\ 0,\ 1,\ 2,\ 3\), so \(\mathbf{6}\) integers.

Example 4 β€” Working backwards from a graph
The number line shows the solution of \(p\le 3x-2\lt q\). Find \(p\) and \(q\).
negative 1 to 4Closed dot at negative 1, open dot at 4, shaded between. βˆ’3βˆ’2βˆ’1012345
Solution

Read the graph first: \(-1\le x\lt 4\). Now build \(3x-2\) from \(x\), doing each step to all three parts.

\(-1\)\(\le\)\(x\lt 4\)
\(-3\)\(\le\)\(3x\lt 12\)
\(-5\)\(\le\)\(3x-2\lt 10\)
−5≤3x−2<10

So \(\boldsymbol{p = -5}\) and \(\boldsymbol{q = 10}\). Multiplying by \(3\) is positive, so the signs did not change.

Common pitfalls

Match the dot to the sign. Open for \(\lt \) and \(\gt \), closed for \(\le\) and \(\ge\). A closed dot on \(x \gt 2\) wrongly includes \(2\).
Shade the right way. \(-4 \ge x\) means \(x \le -4\), so shade to the left. Rewrite with \(x\) first to avoid shading the wrong side.
Reverse the sign before graphing. If you divided by a negative but forgot to reverse the sign, the graph will point the wrong way. Test a shaded number to catch it.
Check the end points when counting integers. For \(-1 \lt x \le 4\) the integers are \(0, 1, 2, 3, 4\): \(-1\) is not included, \(4\) is.
Do not shade for whole numbers. If \(n\) counts people or objects, graph separate dots. A shaded line would include values like \(2.5\).

Frequently asked questions

What does an open circle mean on a number line?

An open circle means the number at the circle is not part of the solution. It is used for strict inequalities, \(\lt \) and \(\gt \).

How do you graph an inequality on a number line?

Solve it first. Put a dot at the boundary number, open if it is not included and closed if it is. Then shade the side that makes the inequality true and add an arrow if it goes on forever.

How do you write the inequality from a number line?

Find the boundary and decide from the dot whether it is included. Then look at the direction of the shading: to the right means greater than, to the left means less than. A segment between two dots gives a double inequality.

How do you graph two inequalities on one number line?

Solve each inequality, then shade only the numbers that make both true, which is where their graphs overlap. The result is usually a segment between two dots.

When do you use dots instead of a shaded line?

When the pronumeral can only be a whole number, such as a number of boxes or tickets. Then each allowed value gets its own dot.