Solving linear inequalities

A lin­ear inequal­ity is like a lin­ear equa­tion with the equals sign replaced by $<$, $>$, $\le$ or $\ge$. It is solved by the same steps as an equa­tion, except that the sign must be reversed when­ever both sides are mul­ti­plied or divided by a neg­a­tive num­ber. The answer is usu­ally an inter­val rather than a sin­gle num­ber. This les­son solves sin­gle inequal­i­ties, dou­ble inequal­i­ties and sys­tems, and fin­ishes with prob­lems from every­day life.

Solv­ing one inequal­ity#

Use the rules from the last les­son:

• add or sub­tract the same num­ber on both sides;
• mul­ti­ply or divide both sides by a pos­i­tive num­ber;
• if you mul­ti­ply or divide by a neg­a­tive num­ber, reverse the sign.

Exam­ple 1#

Solve $3x - 7 > 5$.

$$3x - 7 > 5 \implies 3x > 12 \implies x > 4$$

The solu­tion set is $(4, \infty)$. On the num­ber line: an open cir­cle at 4, shaded to the right.

Exam­ple 2#

Solve $5 - 2x \le 11$.

Sub­tract 5: $-2x \le 6$. Divide by $-2$, revers­ing the sign: $x \ge -3$. The solu­tion set is $[-3, \infty)$.

Check with a value: $x = 0$ gives $5 \le 11$, true; $x = -4$ gives $13 \le 11$, false. This agrees with $x \ge -3$.

Exam­ple 3#

Solve $\displaystyle \frac{x - 2}{3} - \frac{x + 1}{4} \ge 1$.

Mul­ti­ply both sides by 12 (pos­i­tive):

$$4(x - 2) - 3(x + 1) \ge 12 \implies 4x - 8 - 3x - 3 \ge 12 \implies x - 11 \ge 12 \implies x \ge 23$$

The solu­tion set is $[23, \infty)$.

Exam­ple 4#

Solve $2(x + 3) < 2x + 1$.

Expand­ing, $2x + 6 < 2x + 1$, so $6 < 1$, which is false what­ever $x$ is. There is no solu­tion: the solu­tion set is $\phi$.

Sim­i­larly, $2(x + 3) > 2x + 1$ reduces to $6 > 1$, which is always true, so its solu­tion set is $R$.

Restrict­ing to inte­gers or nat­ural num­bers#

Exam­ple 5#

Solve $4x + 3 < 19$ when (a) $x$ is a nat­ural num­ber (b) $x$ is an inte­ger (c) $x$ is real.

$4x < 16$, so $x < 4$.

(a) Nat­ural num­bers: $\{1, 2, 3\}$. (b) Inte­gers: $\{\ldots, -1, 0, 1, 2, 3\}$. (c) Real num­bers: $(-\infty, 4)$.

Dou­ble inequal­i­ties#

A dou­ble inequal­ity such as $-3 < 2x + 1 \le 7$ says that $2x + 1$ lies between two num­bers. Do the same oper­a­tion to all three parts at once.

Exam­ple 6#

Solve $-3 < 2x + 1 \le 7$.

Sub­tract 1 from all three parts: $-4 < 2x \le 6$. Divide by 2: $-2 < x \le 3$. The solu­tion set is $(-2, 3]$.

Exam­ple 7#

Solve $-5 \le 1 - 3x < 10$.

Sub­tract 1: $-6 \le -3x < 9$. Divide by $-3$, revers­ing both signs: $2 \ge x > -3$, that is $-3 < x \le 2$. The solu­tion set is $(-3, 2]$.

Sys­tems of inequal­i­ties#

When sev­eral inequal­i­ties must all hold, solve each one and take the inter­sec­tion of the solu­tion sets.

Exam­ple 8#

Solve the sys­tem $2x - 1 > 3$ and $x + 4 \le 10$.

The first gives $x > 2$; the sec­ond gives $x \le 6$. Both must hold, so $x \in (2, 6]$.

Exam­ple 9#

Solve the sys­tem $3x + 2 \ge 2x - 1$ and $5x - 4 < 3x - 10$.

The first gives $x \ge -3$; the sec­ond gives $2x < -6$, so $x < -3$. No num­ber is both at least $-3$ and less than $-3$, so the sys­tem has no solu­tion: $\phi$.

Prob­lems from every­day life#

Exam­ple 10#

Ravi scored 62 and 71 marks in his first two tests. How much must he score in the third test so that his aver­age is at least 70?

Let the third score be $x$. Then

$$\displaystyle \frac{62 + 71 + x}{3} \ge 70 \implies 133 + x \ge 210 \implies x \ge 77$$

So he must score at least 77 marks (and, since tests are out of 100, at most 100).

Exam­ple 11#

A taxi in Vijayawada charges ₹50 plus ₹15 per km. Sita wants to spend at least ₹200 but not more than ₹350. Find the range of dis­tances she can travel.

The fare for $x$ km is $50 + 15x$ rupees. We need $200 \le 50 + 15x \le 350$. Sub­tract 50: $150 \le 15x \le 300$. Divide by 15: $10 \le x \le 20$. She can travel from 10 km to 20 km.

Exam­ple 12#

The length of a rec­tan­gle is 4 cm more than its width, and its perime­ter is at most 60 cm. Find the pos­si­ble widths.

If the width is $w$ cm, the perime­ter is $2(w + w + 4) = 4w + 8$. So $4w + 8 \le 60$, giv­ing $w \le 13$. A width must also be pos­i­tive, so $0 < w \le 13$, that is $w \in (0, 13]$.

Prac­tice#

1. Solve $5x + 2 \ge 17$.
2. Solve $7 - 4x > 19$.
3. Solve $\displaystyle \frac{2x + 1}{3} < \frac{x - 2}{2} + 3$.
4. Solve $3(x - 1) \ge 3x + 2$.
5. Find the nat­ural num­bers sat­is­fy­ing $2x - 5 \le 7$.
6. Solve $-7 \le 3x - 1 < 8$.
7. Solve $\displaystyle 2 < 4 - \frac{x}{2} \le 6$.
8. Solve the sys­tem $4x - 3 > 1$ and $2x + 5 \le 15$.
9. Solve the sys­tem $x - 2 < 0$ and $3x + 1 > 7$.
10. Sita's marks in three tests are 55, 68 and 74. What must she score in the fourth test to have an aver­age of at least 65?

Answers#