Inequalities in Two Variables and Mixed Practice, Worked Solutions

Ques­tion 1: A lin­ear inequal­ity with frac­tions#

The prob­lem#

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

Under­stand­ing the prob­lem#

You must find every real num­ber $x$ that makes the left side greater than or equal to the right side. The answer will be a set of num­bers, not a sin­gle value.

The idea#

Clear the frac­tions by mul­ti­ply­ing both sides by the LCM of the denom­i­na­tors, $4$. Mul­ti­ply­ing by a pos­i­tive num­ber keeps the direc­tion of the inequal­ity. Then col­lect $x$ terms on one side.

Step-by-step solu­tion#

Step 1. Mul­ti­ply both sides by $4$ (pos­i­tive, so $\ge$ stays $\ge$).

$$\displaystyle 4 \cdot \frac{3x - 4}{2} \ge 4 \cdot \frac{x+1}{4} - 4 \quad\Longrightarrow\quad 2(3x - 4) \ge x + 1 - 4$$

Step 2. Open the bracket and sim­plify the right side.

$$6x - 8 \ge x - 3$$

Step 3. Sub­tract $x$ from both sides and add $8$ to both sides.

$$6x - x \ge -3 + 8 \quad\Longrightarrow\quad 5x \ge 5$$

Step 4. Divide by $5$ (pos­i­tive).

$$x \ge 1$$

Check­ing the answer#

Try $x = 1$: left $\displaystyle = \frac{-1}{2} = -0.5$, right $\displaystyle = \frac{2}{4} - 1 = -0.5$; equal, so $x = 1$ is included. Try $x = 0$ (not in the answer): left $= -2$, right $= -0.75$, and $-2 \ge -0.75$ is false. Good.

Answer#

$x \ge 1$, that is, $x \in [1, \infty)$.

Com­mon mis­take to avoid#

When you mul­ti­ply by $4$, every term must be mul­ti­plied, includ­ing the $-1$. For­get­ting it gives $6x - 8 \ge x + 1 - 1$, which is wrong.

Ques­tion 2: A dou­ble inequal­ity#

The prob­lem#

Solve: $\displaystyle 7 \le \frac{3x + 11}{2} \le 11$.

Under­stand­ing the prob­lem#

This is two inequal­i­ties at once: $\displaystyle \frac{3x+11}{2} \ge 7$ and $\displaystyle \frac{3x+11}{2} \le 11$. You want every $x$ that sat­is­fies both.

The idea#

Do the same oper­a­tion to all three parts together, so that $x$ is left alone in the mid­dle.

Step-by-step solu­tion#

Step 1. Mul­ti­ply all three parts by $2$.

$$14 \le 3x + 11 \le 22$$

Step 2. Sub­tract $11$ from all three parts.

$$3 \le 3x \le 11$$

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

$$\displaystyle 1 \le x \le \frac{11}{3}$$

Check­ing the answer#

At $x = 1$: $\displaystyle \frac{14}{2} = 7$, the lower end. At $\displaystyle x = \frac{11}{3}$: $\displaystyle \frac{11 + 11}{2} = 11$, the upper end. A mid­dle value, $x = 2$: $\displaystyle \frac{17}{2} = 8.5$, which lies between $7$ and $11$.

Answer#

$\displaystyle 1 \le x \le \frac{11}{3}$, that is, $\displaystyle x \in \left[1, \frac{11}{3}\right]$.

Ques­tion 3: A sys­tem of two inequal­i­ties in one vari­able#

The prob­lem#

Solve the sys­tem $2(x - 1) < x + 5$, $3(x + 2) > 2 - x$.

Under­stand­ing the prob­lem#

A "sys­tem" means both inequal­i­ties must hold at the same time. Solve each one sep­a­rately and then take the num­bers com­mon to both.

The idea#

Solve each lin­ear inequal­ity, then find the over­lap (inter­sec­tion) of the two solu­tion sets on the num­ber line.

Step-by-step solu­tion#

Step 1. Solve the first inequal­ity.

$$\begin{aligned} 2x - 2 &< x + 5 \\ 2x - x &< 5 + 2 \\ x &< 7 \end{aligned}$$

Step 2. Solve the sec­ond inequal­ity.

$$\begin{aligned} 3x + 6 &> 2 - x \\ 3x + x &> 2 - 6 \\ 4x &> -4 \\ x &> -1 \end{aligned}$$

Step 3. Take the com­mon part: $x$ must be greater than $-1$ and less than $7$. Both ends are strict, so nei­ther end­point is included.

$$-1 < x < 7$$

Check­ing the answer#

$x = 0$: $-2 < 5$ true and $6 > 2$ true. $x = 7$: $12 < 12$ is false, so $7$ is cor­rectly excluded. $x = -1$: $3 > 3$ is false, so $-1$ is cor­rectly excluded.

Answer#

$-1 < x < 7$, that is, $x \in (-1, 7)$.

Ques­tion 4: Graph­ing three sin­gle inequal­i­ties#

The prob­lem#

Graph: (a) $3x - 2y \ge 6$; (b) $x + y < 4$; (c) $y \le 3$.

Under­stand­ing the prob­lem#

Each inequal­ity describes a half-plane. For each you must draw the bound­ary line (solid or dashed) and shade the cor­rect side.

The idea#

Use the lesson's method: draw the line $ax + by = c$, solid for $\ge$ or $\le$, dashed for $>$ or $<$; then test the point $(0, 0)$ and shade the side it lies on if it sat­is­fies the inequal­ity, oth­er­wise the other side.

Step-by-step solu­tion#

Part (a)

Step 1. Bound­ary line $3x - 2y = 6$. Find its inter­cepts: put $y = 0$ to get $x = 2$; put $x = 0$ to get $-2y = 6$, so $y = -3$. The line passes through $(2, 0)$ and $(0, -3)$.

Step 2. The sign is $\ge$, so the line is included: draw it solid.

Step 3. Test $(0, 0)$: $3(0) - 2(0) = 0$, and $0 \ge 6$ is false. So the ori­gin is not in the region.

Step 4. Shade the side of the line away from the ori­gin (below and to the right of the line).

Part (b)

Step 1. Bound­ary line $x + y = 4$, through $(4, 0)$ and $(0, 4)$.

Step 2. The sign is $<$, so the line is not included: draw it dashed.

Step 3. Test $(0, 0)$: $0 < 4$ is true.

Step 4. Shade the side con­tain­ing the ori­gin (below and to the left of the line).

Part (c)

Step 1. Bound­ary $y = 3$ is a hor­i­zon­tal line through $(0, 3)$.

Step 2. The sign is $\le$: draw it solid.

Step 3. Test $(0, 0)$: $0 \le 3$ is true.

Step 4. Shade below the line (the side con­tain­ing the ori­gin).

Check­ing the answer#

Pick a point in each shaded region and test it. (a) $(4, 0)$: $12 \ge 6$ true. (b) $(1, 1)$: $2 < 4$ true. (c) $(0, -5)$: $-5 \le 3$ true.

Answer#

(a) Solid line through $(2, 0)$ and $(0, -3)$, shade the side away from the ori­gin; (b) dashed line through $(4, 0)$ and $(0, 4)$, shade the side con­tain­ing the ori­gin; (c) solid line $y = 3$, shade below it.

Ques­tion 5: A sys­tem and its cor­ner points#

The prob­lem#

Graph the sys­tem $2x + y \ge 4$, $x + y \le 5$, $x \ge 0$, $y \ge 0$ and list its cor­ner points.

Under­stand­ing the prob­lem#

You need the region where all four inequal­i­ties hold. $x \ge 0$ and $y \ge 0$ keep you in the first quad­rant. The cor­ner points are where two bound­ary lines of the region meet.

The idea#

Graph each bound­ary line, shade each cor­rect side using a test point, and find the region com­mon to all. Then find each cor­ner by solv­ing the pair of line equa­tions that meet there.

Step-by-step solu­tion#

Step 1. Line $2x + y = 4$: inter­cepts $(2, 0)$ and $(0, 4)$; solid. Test $(0,0)$: $0 \ge 4$ false, so shade away from the ori­gin.

Step 2. Line $x + y = 5$: inter­cepts $(5, 0)$ and $(0, 5)$; solid. Test $(0,0)$: $0 \le 5$ true, so shade towards the ori­gin.

Step 3. $x \ge 0$ and $y \ge 0$: stay in the first quad­rant.

Step 4. The com­mon region is the band in the first quad­rant that lies above the line $2x + y = 4$ and below the line $x + y = 5$. It is a quadri­lat­eral.

Step 5. Find the cor­ners.

$$\begin{aligned} 2x + y = 4 \text{ and } y = 0 &: \ x = 2, \ \text{corner } (2, 0) \\ x + y = 5 \text{ and } y = 0 &: \ x = 5, \ \text{corner } (5, 0) \\ x + y = 5 \text{ and } x = 0 &: \ y = 5, \ \text{corner } (0, 5) \\ 2x + y = 4 \text{ and } x = 0 &: \ y = 4, \ \text{corner } (0, 4) \end{aligned}$$

Step 6. Do the two slanted lines meet inside the first quad­rant? Sub­tract­ing $x + y = 5$ from $2x + y = 4$ gives $x = -1$, which is out­side, so they do not give a cor­ner.

Check­ing the answer#

Test each cor­ner in all four inequal­i­ties. For exam­ple $(0, 4)$: $4 \ge 4$ ✓, $4 \le 5$ ✓, $x, y \ge 0$ ✓. An inside point such as $(2, 2)$: $6 \ge 4$ ✓ and $4 \le 5$ ✓.

Answer#

The region is a quadri­lat­eral with cor­ner points $(2, 0)$, $(5, 0)$, $(0, 5)$ and $(0, 4)$.

Ques­tion 6: An unbounded region#

The prob­lem#

Graph the sys­tem $x - 2y \le 3$, $3x + 4y \ge 12$, $x \ge 0$, $y \ge 1$.

Under­stand­ing the prob­lem#

You need the region where all four inequal­i­ties hold. Note that $y \ge 1$ (not $y \ge 0$) is the bot­tom bound­ary. Watch whether the region closes up or goes on for ever.

The idea#

Draw the four bound­aries, shade using test points, then find the cor­ner points by solv­ing pairs of equa­tions, keep­ing only the cor­ners that sat­isfy all four inequal­i­ties.

Step-by-step solu­tion#

Step 1. Line $x - 2y = 3$: inter­cepts $(3, 0)$ and $(0, -1.5)$; solid. Test $(0, 0)$: $0 \le 3$ true, so shade the side con­tain­ing the ori­gin (to the upper left of the line).

Step 2. Line $3x + 4y = 12$: inter­cepts $(4, 0)$ and $(0, 3)$; solid. Test $(0, 0)$: $0 \ge 12$ false, so shade away from the ori­gin (above and to the right).

Step 3. $x \ge 0$: right of the $y$-axis. $y \ge 1$: on or above the hor­i­zon­tal line $y = 1$.

Step 4. The two slanted lines meet where

$$\begin{aligned} x &= 3 + 2y \\ 3(3 + 2y) + 4y &= 12 \\ 10y &= 3 \quad\Rightarrow\quad y = 0.3,\ x = 3.6 \end{aligned}$$

But $y = 0.3$ is below $y = 1$, so this point is cut off by $y \ge 1$ and is not a cor­ner.

Step 5. Find the cor­ners on $y = 1$.

$$\displaystyle \begin{aligned} 3x + 4(1) = 12 &\Rightarrow x = \tfrac{8}{3}, \ \text{corner } \left(\tfrac{8}{3}, 1\right) \\ x - 2(1) = 3 &\Rightarrow x = 5, \ \text{corner } (5, 1) \end{aligned}$$

Step 6. Find the cor­ner on $x = 0$: $3(0) + 4y = 12$ gives $y = 3$, cor­ner $(0, 3)$.

Step 7. Describe the region. From $(0, 3)$ the bound­ary runs down along $3x + 4y = 12$ to $\displaystyle \left(\tfrac{8}{3}, 1\right)$, along $y = 1$ to $(5, 1)$, then up along $x - 2y = 3$ for ever. On the left it runs up the $y$-axis from $(0, 3)$ for ever. So the region is unbounded upwards.

Check­ing the answer#

The point $(1, 5)$ lies inside: $1 - 10 = -9 \le 3$ ✓, $3 + 20 = 23 \ge 12$ ✓, $x \ge 0$ ✓, $y \ge 1$ ✓. Larger $y$ keeps all four true, con­firm­ing the region goes on upward.

Answer#

An unbounded region with cor­ner points $\displaystyle \left(\tfrac{8}{3}, 1\right)$, $(5, 1)$ and $(0, 3)$, extend­ing upward between the $y$-axis and the line $x - 2y = 3$.

Com­mon mis­take to avoid#

Do not list $(3.6, 0.3)$ as a cor­ner. The slanted lines do meet there, but that point breaks $y \ge 1$.

Ques­tion 7: A sys­tem with no solu­tion#

The prob­lem#

Show that $x + y \ge 7$ and $2x + 2y \le 10$ have no com­mon solu­tion.

Under­stand­ing the prob­lem#

You must prove that no point $(x, y)$ sat­is­fies both inequal­i­ties at once.

The idea#

Sim­plify the sec­ond inequal­ity. Then com­pare: both are about the same quan­tity $x + y$, and the two con­di­tions con­tra­dict each other. Graph­i­cally, the bound­ary lines are par­al­lel and the shaded sides face away from each other, as in Exam­ple 6.

Step-by-step solu­tion#

Step 1. Divide $2x + 2y \le 10$ by $2$ (pos­i­tive).

$$x + y \le 5$$

Step 2. Sup­pose a point sat­is­fied both. Then $x + y \ge 7$ and $x + y \le 5$, so

$$7 \le x + y \le 5,$$

which would mean $7 \le 5$. That is false.

Step 3. So no point sat­is­fies both. On a graph, $x + y = 7$ and $x + y = 5$ are par­al­lel; the first region lies above the upper line and the sec­ond below the lower line, so they never over­lap.

Check­ing the answer#

Try any point, say $(4, 4)$: $8 \ge 7$ true, but $16 \le 10$ false. Try $(2, 2)$: sec­ond true, first false. No test point passes both.

Answer#

The sec­ond inequal­ity says $x + y \le 5$, which con­tra­dicts $x + y \ge 7$; the sys­tem has no solu­tion.

Ques­tion 8: Labour hours for chairs and tables#

The prob­lem#

A com­pany makes chairs ($4$ h of labour each) and tables ($6$ h each), with at most $240$ labour hours. Write the inequal­ity and graph it for $x, y \ge 0$.

Under­stand­ing the prob­lem#

Let $x$ be the num­ber of chairs and $y$ the num­ber of tables. The total labour used must not exceed $240$ hours. "At most" means $\le$. Num­bers of items can­not be neg­a­tive.

The idea#

Total labour $=$ (hours per chair)$\times x$ $+$ (hours per table)$\times y$. Then graph as in Exam­ple 4 of the les­son.

Step-by-step solu­tion#

Step 1. Write the inequal­ity.

$$4x + 6y \le 240, \qquad x \ge 0, \quad y \ge 0$$

Step 2. Bound­ary $4x + 6y = 240$. Inter­cepts: $y = 0$ gives $x = 60$; $x = 0$ gives $y = 40$. Draw a solid line through $(60, 0)$ and $(0, 40)$.

Step 3. Test $(0, 0)$: $0 \le 240$ true, so shade the side con­tain­ing the ori­gin.

Step 4. With $x \ge 0$ and $y \ge 0$, the region is the tri­an­gle with ver­tices $(0, 0)$, $(60, 0)$ and $(0, 40)$.

Check­ing the answer#

$(30, 20)$: $120 + 120 = 240$, on the line, so allowed. $(30, 30)$: $120 + 180 = 300 > 240$, out­side the tri­an­gle, as expected.

Answer#

$4x + 6y \le 240$ with $x, y \ge 0$: the tri­an­gle with ver­tices $(0, 0)$, $(60, 0)$ and $(0, 40)$.

Ques­tion 9: Range of men­tal age#

The prob­lem#

IQ $\displaystyle = \frac{MA}{CA} \times 100$. For $12$-year-olds the IQ ranges over $80 \le IQ \le 140$. Find the range of men­tal age $MA$.

Under­stand­ing the prob­lem#

$MA$ is men­tal age and $CA$ is chrono­log­i­cal (actual) age. Here $CA = 12$. You are given bounds on IQ and must turn them into bounds on $MA$.

The idea#

Sub­sti­tute the for­mula into the dou­ble inequal­ity and solve for $MA$ by doing the same oper­a­tion to all three parts.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $CA = 12$.

$$\displaystyle 80 \le \frac{MA}{12} \times 100 \le 140$$

Step 2. Mul­ti­ply all parts by $12$.

$$960 \le 100 \cdot MA \le 1680$$

Step 3. Divide all parts by $100$.

$$9.6 \le MA \le 16.8$$

Check­ing the answer#

$MA = 9.6$: IQ $\displaystyle = \frac{9.6}{12} \times 100 = 80$. $MA = 16.8$: IQ $\displaystyle = \frac{16.8}{12} \times 100 = 140$. Both ends match.

Answer#

$9.6 \le MA \le 16.8$ (men­tal age from $9.6$ to $16.8$ years).

Ques­tion 10: An absolute-value con­di­tion with another con­di­tion#

The prob­lem#

Find all real $x$ sat­is­fy­ing both $\lvert x - 3 \rvert < 4$ and $x \ge 2$.

Under­stand­ing the prob­lem#

$\lvert x - 3 \rvert < 4$ means the dis­tance of $x$ from $3$ is less than $4$. You need num­bers that sat­isfy this and are also at least $2$.

The idea#

Use $\lvert u \rvert < a \iff -a < u < a$ (for $a > 0$), then inter­sect with $x \ge 2$.

Step-by-step solu­tion#

Step 1. Remove the absolute value.

$$-4 < x - 3 < 4$$

Step 2. Add $3$ to all three parts.

$$-1 < x < 7$$

Step 3. Com­bine with $x \ge 2$. The lower limit becomes $2$ (included) because $2 > -1$; the upper limit stays $7$ (excluded).

$$2 \le x < 7$$

Check­ing the answer#

$x = 2$: $\lvert -1 \rvert = 1 < 4$ ✓ and $2 \ge 2$ ✓. $x = 7$: $\lvert 4 \rvert = 4 < 4$ false, so $7$ is cor­rectly left out.

Answer#

$2 \le x < 7$, that is, $x \in [2, 7)$.