Inequalities

The Case Method

Consider the following example: $\displaystyle \dfrac{x-3}{x-1}<10$

You might be tempted to cross multiply, but be careful! The quantity $x-1$ is not always positive. If we multiply by $x-1$ , the inequality needs to flip for some values of $x$ How might we deal with this?

\displaystyle \frac{x-3}{x-1}<10

The default threshold is the worked example below. Change it, then test a point on each side of x = 1.

Input x0Right-hand threshold10

Solution set:

\displaystyle (-\infty,\,\frac{7}{9})\cup(1,\,\infty)

Case 2: the denominator is negative. Reverse < to > when multiplying.

\displaystyle -3>-10

Original left-hand value: 3.000 · Satisfies the inequality.

A case solution must also satisfy the sign assumption used to obtain it.

1. Separate into 2 cases.

Case 1

\begin{aligned} x-1&>0\\[.5em]x&>1 \end{aligned}

Case 2

\begin{aligned} x-1&<0\\[.5em]x&<1 \end{aligned}

2. Solve the original problem under each assumption.

Case 1

\begin{aligned} \dfrac{x-3}{x-1}&<10\\[.5em]x-3&<10(x-1)\\[.5em]-9x&<-7\\[.5em]x&>\dfrac79 \end{aligned}

Case 2

\begin{aligned} \dfrac{x-3}{x-1}&<10\\[.5em]x-3&>10(x-1)\\[.5em]-9x&>-7\\[.5em]x&<\dfrac79 \end{aligned}

3. Find all common points between assumption and solution.

Case 1

\begin{aligned} x&>1\quad\text{and}\quad x>\dfrac79\\[.5em]x&\in(1,\infty) \end{aligned}

Case 2

\begin{aligned} x&<1\quad\text{and}\quad x<\dfrac79\\[.5em]x&\in\left(-\infty,\dfrac79\right) \end{aligned}

4. Consolidate the 2 cases by taking the union.

x\in(1,\infty)\cup\left(-\infty,\dfrac79\right)

QUESTION

Why must

\displaystyle x=1

be excluded from

\displaystyle \dfrac{x-3}{x-1}<10

Find all values of $\displaystyle x$ such that $\displaystyle \displaystyle \dfrac{7x-2}{1-2x}\ge4$

◀ Previous SectionIntervals and Solving Inequalities Next Section ▶The Number Line Method

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Inequalities

The Case Method

Consider the following example: $\displaystyle \dfrac{x-3}{x-1}<10$

\displaystyle \frac{x-3}{x-1}<10

The default threshold is the worked example below. Change it, then test a point on each side of x = 1.

Input x0Right-hand threshold10

Solution set:

\displaystyle (-\infty,\,\frac{7}{9})\cup(1,\,\infty)

Case 2: the denominator is negative. Reverse < to > when multiplying.

\displaystyle -3>-10

Original left-hand value: 3.000 · Satisfies the inequality.

A case solution must also satisfy the sign assumption used to obtain it.

1. Separate into 2 cases.

Case 1

\begin{aligned} x-1&>0\\[.5em]x&>1 \end{aligned}

Case 2

\begin{aligned} x-1&<0\\[.5em]x&<1 \end{aligned}

2. Solve the original problem under each assumption.

Case 1

\begin{aligned} \dfrac{x-3}{x-1}&<10\\[.5em]x-3&<10(x-1)\\[.5em]-9x&<-7\\[.5em]x&>\dfrac79 \end{aligned}

Case 2

\begin{aligned} \dfrac{x-3}{x-1}&<10\\[.5em]x-3&>10(x-1)\\[.5em]-9x&>-7\\[.5em]x&<\dfrac79 \end{aligned}

3. Find all common points between assumption and solution.

Case 1

\begin{aligned} x&>1\quad\text{and}\quad x>\dfrac79\\[.5em]x&\in(1,\infty) \end{aligned}

Case 2

\begin{aligned} x&<1\quad\text{and}\quad x<\dfrac79\\[.5em]x&\in\left(-\infty,\dfrac79\right) \end{aligned}

4. Consolidate the 2 cases by taking the union.

x\in(1,\infty)\cup\left(-\infty,\dfrac79\right)

QUESTION

Why must

\displaystyle x=1

be excluded from

\displaystyle \dfrac{x-3}{x-1}<10

Find all values of $\displaystyle x$ such that $\displaystyle \displaystyle \dfrac{7x-2}{1-2x}\ge4$

◀ Previous SectionIntervals and Solving Inequalities Next Section ▶The Number Line Method

The Case Method

The Case Method

Watch the denominator change sign

KNOWLEDGE QUIZ

Solve a rational inequality by cases

The Case Method

The Case Method

Watch the denominator change sign

KNOWLEDGE QUIZ

Solve a rational inequality by cases