Linear Inequalities in One Variable, Worked Solutions

Ques­tion 1: Solv­ing three lin­ear inequal­i­ties over the real num­bers#

The prob­lem#

Solve for $x \in \mathbf{R}$:

(a) $4x + 7 > 2x - 5$ (b) $3 - 5x \le 18$ (c) $\displaystyle \frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5}$

Under­stand­ing the prob­lem#

You are given three sep­a­rate inequal­i­ties. For each one you must find every real num­ber $x$ that makes it true, and write the answer as an inequal­ity and as an inter­val. "$x \in \mathbf{R}$" means $x$ may be any real num­ber, so the answer is a whole stretch of the num­ber line, not a list.

The idea#

Solve exactly as you would solve an equa­tion: col­lect the $x$ terms on one side and the num­bers on the other. Adding or sub­tract­ing keeps the sign. Divid­ing by a pos­i­tive num­ber keeps it; divid­ing by a neg­a­tive num­ber reverses it. In part (c), clear the frac­tions first by mul­ti­ply­ing by the LCM of the denom­i­na­tors.

Step-by-step solu­tion#

Part (a)

Step 1. Bring the $x$ terms to the left by sub­tract­ing $2x$ from both sides. Sub­tract­ing keeps the direc­tion.

$$4x - 2x + 7 > -5 \quad\Longrightarrow\quad 2x + 7 > -5$$

Step 2. Remove the $7$ by sub­tract­ing $7$ from both sides.

$$2x > -5 - 7 = -12$$

Step 3. Divide both sides by $2$. Since $2$ is pos­i­tive, the sign stays as $>$.

$$x > -6$$

Part (b)

Step 1. Sub­tract $3$ from both sides.

$$-5x \le 18 - 3 = 15$$

Step 2. Divide both sides by $-5$. This is a neg­a­tive num­ber, so $\le$ turns into $\ge$.

$$\displaystyle x \ge \frac{15}{-5} = -3$$

Part (c)

Step 1. The denom­i­na­tors are $4$, $3$ and $5$; their LCM is $60$. Mul­ti­ply every term by $60$. Since $60$ is pos­i­tive, the sign stays.

$$\displaystyle 60 \cdot \frac{x}{4} < 60 \cdot \frac{5x - 2}{3} - 60 \cdot \frac{7x - 3}{5}$$

Step 2. Sim­plify each term. Keep the brack­ets, because the minus sign in front of the last frac­tion applies to the whole numer­a­tor.

$$15x < 20(5x - 2) - 12(7x - 3)$$

Step 3. Expand the brack­ets care­fully. Note $-12 \times (-3) = +36$.

$$15x < 100x - 40 - 84x + 36$$

Step 4. Com­bine like terms on the right: $100x - 84x = 16x$ and $-40 + 36 = -4$.

$$15x < 16x - 4$$

Step 5. Sub­tract $16x$ from both sides.

$$-x < -4$$

Step 6. Mul­ti­ply both sides by $-1$ (neg­a­tive), so $<$ turns into $>$.

$$x > 4$$

Check­ing the answer#

• (a) Try $x = 0$ (which is $> -6$): $7 > -5$ is true. Try $x = -7$: $-21 > -19$ is false. Good.
• (b) Try $x = -3$: $3 + 15 = 18 \le 18$ is true, so $-3$ is included. Try $x = -4$: $23 \le 18$ is false. Good.
• (c) Try $x = 5$: left side $1.25$; right side $\displaystyle \tfrac{23}{3} - \tfrac{32}{5} = 7.667 - 6.4 = 1.267$, and $1.25 < 1.267$ is true. At $x = 4$ both sides equal $1$, so $4$ itself is excluded. Good.

Answer#

(a) $x > -6$, i.e. $(-6, \infty)$; (b) $x \ge -3$, i.e. $[-3, \infty)$; (c) $x > 4$, i.e. $(4, \infty)$.

Com­mon mis­take to avoid#

In part (c), writ­ing $-12 \cdot 7x - 3$ instead of $-12(7x - 3)$ loses the $+36$. Always put each numer­a­tor in brack­ets before you mul­ti­ply out.

Ques­tion 2: A dou­ble inequal­ity over the inte­gers#

The prob­lem#

Solve for $x \in \mathbf{Z}$: $-3 < 2x + 1 \le 9$.

Under­stand­ing the prob­lem#

This is two inequal­i­ties writ­ten together: $2x + 1 > -3$ and $2x + 1 \le 9$. The con­di­tion $x \in \mathbf{Z}$ means $x$ must be an inte­ger, so the final answer is a list of inte­gers, not an inter­val. Watch the two signs: the left one is strict ($<$), the right one allows equal­ity ($\le$).

The idea#

Work on all three parts at once: what­ever you do to the mid­dle, do to both ends. Once $x$ stands alone in the mid­dle, list the inte­gers in that range.

Step-by-step solu­tion#

Step 1. Sub­tract $1$ from all three parts to remove the $+1$ in the mid­dle.

$$-3 - 1 < 2x \le 9 - 1 \quad\Longrightarrow\quad -4 < 2x \le 8$$

Step 2. Divide all three parts by $2$ (pos­i­tive, so both signs stay).

$$-2 < x \le 4$$

Step 3. List the inte­gers. $-2$ is excluded (strict sign) and $4$ is included.

$$x \in \{-1, 0, 1, 2, 3, 4\}$$

Check­ing the answer#

At $x = -1$: $2x + 1 = -1$, and $-3 < -1 \le 9$ is true. At $x = 4$: $2x + 1 = 9$, and $9 \le 9$ is true. At $x = -2$: $2x + 1 = -3$, and $-3 < -3$ is false, so $-2$ is rightly left out. At $x = 5$: $11 \le 9$ is false.

Answer#

$x \in \{-1, 0, 1, 2, 3, 4\}$.

Ques­tion 3: A dou­ble inequal­ity with a neg­a­tive coef­fi­cient, shown on the num­ber line#

The prob­lem#

Solve and draw on the num­ber line: $\displaystyle -5 \le \frac{2 - 3x}{4} \le 9$.

Under­stand­ing the prob­lem#

The mid­dle expres­sion $\displaystyle \frac{2 - 3x}{4}$ must lie between $-5$ and $9$, both ends included. You must find all real $x$ for which this holds and then show the set on a num­ber line. The coef­fi­cient of $x$ is $-3$, which is neg­a­tive, so at some point you will divide by a neg­a­tive num­ber.

The idea#

Treat all three parts together. Clear the frac­tion by mul­ti­ply­ing by $4$, move the con­stant, then divide by $-3$ and reverse both signs.

Step-by-step solu­tion#

Step 1. Mul­ti­ply all three parts by $4$ (pos­i­tive, signs stay).

$$-20 \le 2 - 3x \le 36$$

Step 2. Sub­tract $2$ from all three parts.

$$-22 \le -3x \le 34$$

Step 3. Divide all three parts by $-3$. This is neg­a­tive, so both $\le$ signs turn into $\ge$.

$$\displaystyle \frac{-22}{-3} \ge x \ge \frac{34}{-3} \quad\Longrightarrow\quad \frac{22}{3} \ge x \ge -\frac{34}{3}$$

Step 4. Rewrite with the smaller num­ber on the left, which is the usual way to read an inter­val.

$$\displaystyle -\frac{34}{3} \le x \le \frac{22}{3}$$

In dec­i­mals, $\displaystyle -\tfrac{34}{3} \approx -11.33$ and $\displaystyle \tfrac{22}{3} \approx 7.33$.

Step 5 (the draw­ing). Draw a num­ber line marked at least from $-12$ to $8$. Put a filled (solid) cir­cle at $\displaystyle -\tfrac{34}{3}$, just to the right of $-12$, and a filled cir­cle at $\displaystyle \tfrac{22}{3}$, just to the right of $7$. Both are filled because both ends use $\le$. Shade the seg­ment between the two cir­cles, and noth­ing out­side it.

Check­ing the answer#

At $\displaystyle x = -\tfrac{34}{3}$: $2 - 3x = 2 + 34 = 36$ and $\displaystyle \tfrac{36}{4} = 9$, the right end. At $\displaystyle x = \tfrac{22}{3}$: $2 - 3x = 2 - 22 = -20$ and $\displaystyle \tfrac{-20}{4} = -5$, the left end. At $x = 0$ (inside): $\displaystyle \tfrac{2}{4} = 0.5$, which is between $-5$ and $9$. Good.

Answer#

$\displaystyle -\frac{34}{3} \le x \le \frac{22}{3}$, i.e. $\displaystyle \left[-\frac{34}{3}, \frac{22}{3}\right]$, drawn with filled cir­cles at both ends and the seg­ment between them shaded.

Com­mon mis­take to avoid#

Revers­ing only one of the two signs when divid­ing by $-3$. Both signs flip.

Ques­tion 4: A sys­tem of two inequal­i­ties#

The prob­lem#

Solve the sys­tem: $3x - 7 < 5 + x$ and $11 - 5x \le 1$.

Under­stand­ing the prob­lem#

A value of $x$ is a solu­tion of the sys­tem only if it sat­is­fies both inequal­i­ties at the same time. So you solve each one sep­a­rately and then keep only the val­ues com­mon to both, which is the inter­sec­tion of the two solu­tion sets.

The idea#

Solve each inequal­ity with the usual rules, draw both solu­tion sets on one num­ber line (in your head or on paper), and read off the over­lap.

Step-by-step solu­tion#

Step 1. Solve the first inequal­ity. Sub­tract $x$ from both sides and add $7$ to both sides.

$$3x - x < 5 + 7 \quad\Longrightarrow\quad 2x < 12 \quad\Longrightarrow\quad x < 6$$

Step 2. Solve the sec­ond inequal­ity. Sub­tract $11$ from both sides.

$$-5x \le 1 - 11 = -10$$

Step 3. Divide by $-5$ (neg­a­tive), so $\le$ becomes $\ge$.

$$\displaystyle x \ge \frac{-10}{-5} = 2$$

Step 4. Inter­sect. The first gives every­thing to the left of $6$ (not includ­ing $6$); the sec­ond gives every­thing from $2$ onwards (includ­ing $2$). The over­lap is from $2$ up to, but not includ­ing, $6$.

$$2 \le x < 6$$

Check­ing the answer#

$x = 2$: $3(2) - 7 = -1 < 7$ true, and $11 - 10 = 1 \le 1$ true. $x = 6$: $11 < 11$ is false, so $6$ is excluded. $x = 1$: $11 - 5 = 6 \le 1$ is false, so val­ues below $2$ fail. Good.

Answer#

$2 \le x < 6$, i.e. $[2, 6)$.

Ques­tion 5: An inequal­ity with three frac­tions#

The prob­lem#

Solve: $\displaystyle \frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$.

Under­stand­ing the prob­lem#

You must find all real $x$ for which the left frac­tion is at least as large as the right side. There are three denom­i­na­tors ($3$, $4$, $5$), and a minus sign sits in front of the last frac­tion, which affects its whole numer­a­tor.

The idea#

Mul­ti­ply every term by the LCM $60$ to clear frac­tions (this keeps the sign because $60 > 0$), expand with brack­ets, col­lect terms, and divide at the end, watch­ing for a neg­a­tive coef­fi­cient.

Step-by-step solu­tion#

Step 1. Mul­ti­ply each term by $60$: $60 \div 3 = 20$, $60 \div 4 = 15$, $60 \div 5 = 12$.

$$20(2x - 1) \ge 15(3x - 2) - 12(2 - x)$$

Step 2. Expand. Note $-12 \times 2 = -24$ and $-12 \times (-x) = +12x$.

$$40x - 20 \ge 45x - 30 - 24 + 12x$$

Step 3. Com­bine like terms on the right: $45x + 12x = 57x$ and $-30 - 24 = -54$.

$$40x - 20 \ge 57x - 54$$

Step 4. Move $x$ terms to the left (sub­tract $57x$) and num­bers to the right (add $20$).

$$40x - 57x \ge -54 + 20 \quad\Longrightarrow\quad -17x \ge -34$$

Step 5. Divide by $-17$ (neg­a­tive), so $\ge$ becomes $\le$.

$$\displaystyle x \le \frac{-34}{-17} = 2$$

Check­ing the answer#

At $x = 2$: left $\displaystyle = \tfrac{3}{3} = 1$; right $\displaystyle = \tfrac{4}{4} - \tfrac{0}{5} = 1$. Equal, so $2$ is included. At $x = 0$: left $\displaystyle = -\tfrac{1}{3} \approx -0.33$; right $\displaystyle = -\tfrac{2}{4} - \tfrac{2}{5} = -0.9$; $-0.33 \ge -0.9$ is true. Good.

Answer#

$x \le 2$, i.e. $(-\infty, 2]$.

Com­mon mis­take to avoid#

Writ­ing $-12(2 - x)$ as $-24 - 12x$. The prod­uct of two neg­a­tives is pos­i­tive, so it is $-24 + 12x$.

Ques­tion 6: The score needed for a required aver­age#

The prob­lem#

A stu­dent scored $62$, $71$ and $79$ in three tests. What must the fourth score be for an aver­age of at least $70$?

Under­stand­ing the prob­lem#

Let the unknown fourth score be $x$. The aver­age of four scores is their sum divided by $4$. "At least $70$" means the aver­age must be $70$ or more, so the sign is $\ge$. You must find the small­est value $x$ can take.

The idea#

Write the con­di­tion as an inequal­ity in $x$ and solve it.

Step-by-step solu­tion#

Step 1. Write the con­di­tion on the aver­age.

$$\displaystyle \frac{62 + 71 + 79 + x}{4} \ge 70$$

Step 2. Mul­ti­ply both sides by $4$ (pos­i­tive, sign stays).

$$62 + 71 + 79 + x \ge 280$$

Step 3. Add the known scores: $62 + 71 + 79 = 212$.

$$212 + x \ge 280$$

Step 4. Sub­tract $212$ from both sides.

$$x \ge 68$$

Check­ing the answer#

If the fourth score is exactly $68$, the total is $280$ and the aver­age is $\displaystyle \tfrac{280}{4} = 70$, which meets the con­di­tion. A score of $67$ gives $\displaystyle \tfrac{279}{4} = 69.75$, which falls short. Good.

Answer#

The fourth score must be at least $68$.

Ques­tion 7: Con­sec­u­tive mul­ti­ples of 4#

The prob­lem#

Find all pairs of con­sec­u­tive mul­ti­ples of $4$, both greater than $10$, whose sum is less than $60$.

Under­stand­ing the prob­lem#

"Con­sec­u­tive mul­ti­ples of $4$" are num­bers like $12$ and $16$: the sec­ond is $4$ more than the first. If the smaller num­ber is $x$, the larger is $x + 4$. Two con­di­tions apply: both num­bers exceed $10$, and their sum is below $60$. Also, $x$ itself must be a mul­ti­ple of $4$.

The idea#

Turn each con­di­tion into an inequal­ity in $x$, solve both, inter­sect them, and then pick out the mul­ti­ples of $4$ in the range.

Step-by-step solu­tion#

Step 1. "Both greater than $10$": since $x + 4 > x$, it is enough to ask for the smaller one.

$$x > 10$$

Step 2. "Sum less than $60$":

$$x + (x + 4) < 60 \quad\Longrightarrow\quad 2x + 4 < 60 \quad\Longrightarrow\quad 2x < 56 \quad\Longrightarrow\quad x < 28$$

Step 3. Com­bine both con­di­tions.

$$10 < x < 28$$

Step 4. List the mul­ti­ples of $4$ strictly between $10$ and $28$: $12, 16, 20, 24$. ($28$ is excluded because the sign is strict.)

Step 5. Write each pair $(x, x + 4)$.

$$(12, 16),\ (16, 20),\ (20, 24),\ (24, 28)$$

Check­ing the answer#

The last pair $(24, 28)$ has sum $52 < 60$. The next pair $(28, 32)$ has sum $60$, which is not less than $60$, so it is rightly excluded. The pair before the first, $(8, 12)$, fails because $8$ is not greater than $10$.

Answer#

$(12, 16),\ (16, 20),\ (20, 24),\ (24, 28)$.

Ques­tion 8: The longest pos­si­ble short­est side of a tri­an­gle#

The prob­lem#

In a tri­an­gle, one side is twice the short­est side and the third side is $4$ cm longer than the short­est. The perime­ter is at most $52$ cm. Find the great­est pos­si­ble length of the short­est side.

Under­stand­ing the prob­lem#

Let the short­est side be $x$ cm. Then the other two sides are $2x$ cm and $(x + 4)$ cm. The perime­ter is the sum of the three sides, and "at most $52$" means $\le 52$. You want the largest $x$ allowed.

The idea#

Write the perime­ter con­di­tion as an inequal­ity, solve it, and then check that the extreme case really gives a tri­an­gle.

Step-by-step solu­tion#

Step 1. Write the perime­ter con­di­tion.

$$x + 2x + (x + 4) \le 52$$

Step 2. Col­lect like terms: $x + 2x + x = 4x$.

$$4x + 4 \le 52$$

Step 3. Sub­tract $4$ and divide by $4$ (pos­i­tive, sign stays).

$$4x \le 48 \quad\Longrightarrow\quad x \le 12$$

Step 4. Check that $x = 12$ gives a gen­uine tri­an­gle. The sides are $12$, $24$ and $16$ cm. The two smaller sides must add to more than the largest: $12 + 16 = 28 > 24$. So it is a valid tri­an­gle.

Check­ing the answer#

Perime­ter at $x = 12$: $12 + 24 + 16 = 52$ cm, exactly the limit. Any larger $x$ would push the perime­ter above $52$.

Answer#

The great­est pos­si­ble length of the short­est side is $12$ cm.

Ques­tion 9: Dilut­ing a salt solu­tion#

The prob­lem#

How many litres of water must be added to $600$ L of a $40\%$ salt solu­tion so that the result­ing mix­ture is between $20\%$ and $30\%$ salt?

Under­stand­ing the prob­lem#

Adding water does not change the amount of salt; it only increases the total vol­ume, which low­ers the per­cent­age. Let $w$ litres of water be added. You need the salt frac­tion of the new mix­ture to lie strictly between $0.20$ and $0.30$.

The idea#

Find the amount of salt, write the new con­cen­tra­tion as $\displaystyle \frac{\text{salt}}{\text{new volume}}$, set up a dou­ble inequal­ity, and solve its two halves sep­a­rately. Because the vol­ume $600 + w$ is pos­i­tive, you may mul­ti­ply through by it with­out revers­ing any sign.

Step-by-step solu­tion#

Step 1. Amount of salt in the orig­i­nal solu­tion.

$$40\% \text{ of } 600 = 0.4 \times 600 = 240 \text{ L}$$

Step 2. After adding $w$ L of water, the vol­ume is $600 + w$ L and the salt is still $240$ L. The con­di­tion is

$$\displaystyle 0.2 < \frac{240}{600 + w} < 0.3$$

Step 3. Right half: $\displaystyle \frac{240}{600 + w} < 0.3$. Mul­ti­ply by $600 + w$ (pos­i­tive).

$$240 < 0.3(600 + w) = 180 + 0.3w \quad\Longrightarrow\quad 60 < 0.3w \quad\Longrightarrow\quad w > 200$$

Step 4. Left half: $\displaystyle \frac{240}{600 + w} > 0.2$. Mul­ti­ply by $600 + w$ (pos­i­tive).

$$240 > 0.2(600 + w) = 120 + 0.2w \quad\Longrightarrow\quad 120 > 0.2w \quad\Longrightarrow\quad w < 600$$

Step 5. Com­bine the two halves.

$$200 < w < 600$$

Check­ing the answer#

With $w = 200$: $\displaystyle \tfrac{240}{800} = 0.30$, exactly $30\%$ (the bound­ary). With $w = 600$: $\displaystyle \tfrac{240}{1200} = 0.20$, exactly $20\%$. With $w = 400$: $\displaystyle \tfrac{240}{1000} = 0.24 = 24\%$, which is between. Good.

Answer#

More than $200$ L but less than $600$ L of water must be added.

Com­mon mis­take to avoid#

Think­ing the salt amount changes when water is added. Only the total vol­ume changes.

Ques­tion 10: Absolute value inequal­i­ties#

The prob­lem#

Explain why $\lvert x \rvert < 3$ means $-3 < x < 3$, and solve $\lvert 2x - 1 \rvert \le 5$.

Under­stand­ing the prob­lem#

There are two tasks: first, an expla­na­tion (a short argu­ment in words), and sec­ond, a cal­cu­la­tion. The key term is $\lvert x \rvert$, the absolute value of $x$, which is the dis­tance of $x$ from $0$ on the num­ber line, always $\ge 0$.

The idea#

Use the dis­tance mean­ing. A num­ber whose dis­tance from $0$ is less than $3$ must lie between $-3$ and $3$. The same rea­son­ing turns $\lvert 2x - 1 \rvert \le 5$ into a dou­ble inequal­ity you can solve as in Ques­tion 2.

Step-by-step solu­tion#

Part (a): the expla­na­tion

Step 1. $\lvert x \rvert$ is the dis­tance from $x$ to $0$. For exam­ple, $\lvert -2 \rvert = 2$ and $\lvert 2 \rvert = 2$.

Step 2. On the num­ber line, the points exactly $3$ units from $0$ are $-3$ and $3$. Points closer to $0$ than that lie strictly between them; points out­side this gap are $3$ or more units away.

Step 3. So "dis­tance less than $3$" is the same as "strictly between $-3$ and $3$":

$$\lvert x \rvert < 3 \iff -3 < x < 3$$

Part (b): solv­ing $\lvert 2x - 1 \rvert \le 5$

Step 1. By the same rea­son­ing (with $\le$, so the ends are included), the quan­tity $2x - 1$ must lie between $-5$ and $5$.

$$-5 \le 2x - 1 \le 5$$

Step 2. Add $1$ to all three parts.

$$-4 \le 2x \le 6$$

Step 3. Divide all three parts by $2$ (pos­i­tive, signs stay).

$$-2 \le x \le 3$$

Check­ing the answer#

At $x = -2$: $\lvert -5 \rvert = 5 \le 5$, true. At $x = 3$: $\lvert 5 \rvert = 5 \le 5$, true. At $x = 4$: $\lvert 7 \rvert = 7$, which is not $\le 5$. At $x = 0$: $\lvert -1 \rvert = 1 \le 5$, true. Good.

Answer#

$\lvert x \rvert < 3$ means $x$ is less than $3$ units from $0$, i.e. $-3 < x < 3$; and $\lvert 2x - 1 \rvert \le 5$ gives $-2 \le x \le 3$, i.e. $[-2, 3]$.