Solving Inequalities Flashcards

All 11 cards in this deck

What are the steps to solve a linear inequality with the unknown on both sides?

e.g. $14n > 11n + 6$

1. Collect the unknown on one side: $3n > 6$
2. Divide both sides by the coefficient: $n > 2$

What must you do to the inequality sign when you multiply or divide both sides by a negative number?

Reverse it.
e.g. $-2x < 6$ gives $x > -3$

What are the steps to solve a double-sided inequality?

e.g. $-2 < x + 3 \le 4$

1. Do the same operation to all three parts: $-2 - 3 < x \le 4 - 3$
2. Simplify: $-5 < x \le 1$

When showing an inequality on a number line, when do you use an open circle and when a closed (filled) circle?

Open circle for $<$ or $>$ (value not included); closed circle for $\le$ or $\ge$ (value included).

How do you write a solution in set notation?

e.g. the solution $-5 < x \le 1$

$\{x : -5 < x \le 1\}$

True or false? If $n$ is an integer and $-2 \le n < 5$, the greatest possible value of $n$ is $5$.

False. It is $4$, because $5$ is not included by the strict $<$ sign.

What are the steps to solve a quadratic inequality?

e.g. $x^2 - x - 6 < 0$

1. Write it as $ax^2+bx+c$ compared with $0$: already done
2. Factorise to get the critical values: $(x-3)(x+2)$, so $x = 3$ and $x = -2$
3. Choose the region using a sketch: $-2 < x < 3$

For $a < b$, what is the solution set of $(x-a)(x-b) < 0$?

e.g. $(x+5)(x-2) < 0$

$a < x < b$ (the values between the critical values).
e.g. $-5 < x < 2$

For $a < b$, what is the solution set of $(x-a)(x-b) > 0$?

e.g. $(x+2)(x-\tfrac{1}{2}) > 0$

$x < a$ or $x > b$ (the values outside the critical values).
e.g. $x < -2$ or $x > \frac{1}{2}$

True or false? The solution of $x^2 > 9$ is $x > 3$.

False. The solution is $x < -3$ or $x > 3$.

What are the steps to find the set of possible values of $x$ when one area must be greater than another?

e.g. rectangle area $(x-1)(3x-2)$ greater than triangle area $\tfrac{1}{2}(2x \times x)$

1. Write the inequality: $(x-1)(3x-2) > x^2$
2. Expand and rearrange to $ax^2+bx+c>0$: $2x^2-5x+2>0$
3. Solve and reject values making a length negative: critical values $\frac{1}{2}$ and $2$, so $x > 2$