Year 11 Extended · Unit 6 Logarithms and non-linear inequalities · Lesson 8 of 8

How do you solve an inequality with the unknown on the bottom?

A fraction with x in the denominator changes sign at its numerator's zero and at its denominator's zero, and multiplying by the denominator can reverse the inequality. This lesson solves rational and cubic inequalities with sign diagrams and checks them on a graph.

Number and AlgebraGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From lesson 7: solve x2 − 9 < 0.
    Answer−3 < x < 3.
  2. From Year 10: write 1 ÷ x − 2 ÷ (x + 1) as a single fraction.
    Answer(1 − x) ÷ (x(x + 1)). The top is (x + 1) − 2x.
  3. From Year 10: for which x is 5 ÷ (x − 4) not defined?
    Answerx = 4.
Words used in this lesson
rational expression
a fraction with algebra on the top and bottom. For example: (x + 1) ÷ (x − 4).
denominator
the bottom of a fraction. For example: the x − 4 in a fraction.
excluded value
a value of x that makes an expression undefined. For example: x = 4 above.
asymptote
a line that a graph approaches but does not reach. For example: the dashed line in the figure.
sign diagram
a number line showing where an expression is positive or negative. For example: plus, minus, plus.

How many items before the cost per item is low enough?

A workshop in Kumasi makes school bags. It pays 500 cedis to set up the machine for a batch and then 20 cedis for the materials for each bag. For a batch of x bags the cost per bag is C = (500 + 20x) ÷ x cedis.

Predict first: how many bags must be made in a batch for the cost per bag to fall below 30 cedis? Then use the figure to see how a fraction changes sign, and why multiplying by x is safe here but not always.

−1
2
right
x < −1 or x > 1y is above zero when
x ≤ −1 or x > 1y is zero or above when

y = (x + 1) ÷ (x − 1): above zero when x < −1 or x > 1, and zero or above when x ≤ −1 or x > 1.

Show me the steps
  1. y = x ÷ (x + 4): zero at 0 and undefined at −4. Above zero when x < −4 or x > 0, and zero or above when x < −4 or x ≥ 0.
  2. y = (x − 2) ÷ (x − 3): above zero when x < 2 or x > 3, and zero or above when x ≤ 2 or x > 3, since the zero at 2 is included and the break at 3 never is.
  3. y = (x + 3) ÷ x: above zero when x < −3 or x > 0, and zero or above when x ≤ −3 or x > 0.
  4. Numerator zero at a and denominator zero at c: the sign changes at both, the value a belongs to a zero-or-above answer, and the value c never belongs.

Never multiply by an expression whose sign you do not know. If the multiplier is negative the inequality reverses, and if it can be zero the step is not allowed. To solve a rational inequality:

  1. Move everything to one side, so the other side is 0, and combine into a single fraction.
  2. Factorize the top and bottom. The critical values are the zeros of the top and the zeros of the bottom.
  3. Make a sign diagram. A value that makes the bottom zero is never included. A value that makes only the top zero is included when the inequality is ≤ or ≥.

The same method works for a product such as x(x − 2)(x + 2), and a graph of the two sides shows the same regions: the solution is where one graph is above the other.

Two common slips. Multiplying both sides by the denominator: 4 ÷ (x − 2) < 1 becomes 4 < x − 2 and x > 6, but x = 0 also works, because 4 ÷ (−2) = −2 is less than 1, and the multiplier x − 2 was negative there. And including the bottom's zero: in (x − 2) ÷ (x + 1) ≤ 0 the value x = −1 is never allowed, so the answer starts with −1 < x.

Worked examples

1. Solve (2x − 1) ÷ (x + 3) ≤ 0.

  1. The critical values are 0.5 (top zero) and −3 (bottom zero). The fraction is positive for x < −3, negative for −3 < x < 0.5, and positive for x > 0.5.
  2. Include 0.5 (the fraction is 0) but never −3: −3 < x ≤ 0.5.

2. Solve 6 ÷ (x − 1) < 3.

  1. Subtract 3 and combine: (6 − 3(x − 1)) ÷ (x − 1) = (9 − 3x) ÷ (x − 1) < 0.
  2. The critical values are 3 and 1. The numerator is positive for x < 3. The fraction is negative for x < 1 and for x > 3: x < 1 or x > 3.

3. Solve x3 − 4x > 0.

  1. Factorize: x(x − 2)(x + 2) > 0, with critical values −2, 0 and 2.
  2. For x > 2 all three factors are positive, and the sign alternates at each root: −2 < x < 0 or x > 2.

Your go. The first step is given. Solve x ÷ (x + 2) ≥ 1. Subtract 1 and combine: x ÷ (x + 2) − 1 = −2 ÷ (x + 2).

  1. The inequality is now −2 ÷ (x + 2) ≥ 0.
  2. The top is always negative, so the bottom must be negative. Remember the bottom's zero is never included.
Check the answer

x + 2 < 0, so x < −2. The value −2 is not included because the fraction does not exist there.

With a GDC (optional)

To check an inequality such as 6 ÷ x > x − 1, enter y = 6 ÷ x as the first graph and y = x − 1 as the second, and choose a window of about −8 to 8 on both axes. Use the intersect option to read the x-coordinates of the crossing points: these are the critical values. The inequality is true where the first graph lies above the second.

The graph gives the critical values and the regions, but not whether an end value is included, so state that yourself from the inequality sign. Remember that x = 0 is not allowed here: the calculator may draw a line through the break, which is not part of the graph.

Check before you go on

1. Which describes the solution of (x + 2) ÷ (x − 5) < 0?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Find the largest integer x with (x − 6) ÷ (x + 2) < 0.

3. For x > 0, find the largest integer x with 12 ÷ x > 5.

Level 2 · Building

4. Find the sum of all integers x that satisfy 10 ÷ (x + 1) ≥ 2.

5. How many integers x from −4 to 5 inclusive satisfy 9x − x3 ≥ 0?

Level 3 · Stretch

6. Find the smallest integer greater than 3 for which x ÷ (x − 3) ≤ 2.

7. The graphs of y = 8 ÷ x and y = x + 2 cross at x = −4 and x = 2. How many integers from −6 to 6 inclusive satisfy 8 ÷ x > x + 2?

Something to think about

The fraction 1 ÷ x is positive for x > 0 and negative for x < 0. Is 1 ÷ x ever zero, and is there a value of x where it is undefined? What does that say about the sign?

One way to think about it

1 ÷ x is never zero, because the numerator is 1, and it is undefined at x = 0. The sign can only change at a zero or at a break, so the break at 0 is the only place where the sign changes, which is why it is negative on one side and positive on the other.

Exit ticket

8. Which describes the solution of 15 ÷ (x + 3) > 3?