Math 7 · 7.EE.B.4 · 7.EE.B.4B

Two-step inequalities

A total is a fixed amount plus so much for each item, and it must stay on one side of a limit. Which numbers of items keep it there?

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One-step inequalities and the sign flipMath 7 Two-step equationsMath 7

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Compound inequalities: and, orAlgebra 1 Multi-step inequalitiesAlgebra 1

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One-step inequalities and the sign flip

Two-step inequalities

Multi-step inequalities

When you reach for it

In a two-step equation an unknown was multiplied, then shifted, and the result was known exactly: 3x + 5 = 26, so x = 7. Very often the result is not pinned to one number but held on one side of a limit. A school trip must cost at most its budget. A fundraiser must raise at least its target. The clay a potter uses must stay within what is in the bin. Each time the total is a fixed amount plus a rate times the unknown, and the sentence is a two-step inequality:

px + q > r

or the same with <, \ge or \le in place of >. Here p, q and r are numbers, p is not 0, and the unknown appears exactly once. The answer is not one number but a whole range of them, the solution set, which one-step inequalities introduced together with its graph on a number line.

This lesson takes two days. Part 1 solves px + q > r and its relatives, watching for the one step that turns the inequality sign round, and graphs the answer. Part 2 writes the inequality from a story, solves it, and reads the answer back in the story, including the case where only whole numbers make sense.

The procedure

Part 1: two steps, and the one place the sign can turn

Solving an inequality uses the same two undoing steps as solving the matching equation, in the same reverse order: the last thing done to x is the first thing undone. What is different is what each step must protect. A step is allowed only if it leaves the solution set exactly as it was, and one-step inequalities showed which steps do:

adding or subtracting the same number on both sides never changes the direction of the sign;
multiplying or dividing both sides by the same positive number keeps the direction, and by the same negative number reverses it, so < becomes > and \le becomes \ge.

Take 3x + 5 > 26. Reading the left side by the order of operations, x was multiplied by 3 and then 5 was added.

Undo the addition: subtract 5 from both sides. Subtracting never changes the direction:

3x > 21

Undo the multiplication: divide both sides by 3. The number 3 is positive, so the direction stays:

x > 7

Every number greater than 7 is a solution, and no other number is. On a number line the boundary 7 gets an open circle, because 7 itself fails: 3(7) + 5 = 26, and 26 > 26 is false. Everything to the right of it is shaded. (The conventions are those of inequalities on the number line: an open circle for < and >, a filled dot for \le and \ge.)

Figure 1The solution of 3x + 5 > 26. The open circle says that 7 itself is left out.

Check, in two parts. First the boundary: putting 7 into the left side gives exactly 26, the number on the right, so 7 is where the truth of the inequality changes. Then one test number on each side of it: x = 10 gives 35 > 26, true, and x = 0 gives 5 > 26, false. So the shading belongs on the right. Each part catches something the other cannot. A test number alone cannot tell x > 7 from x > 6, since 10 satisfies both; the boundary alone cannot tell which side to shade.

When the coefficient is negative. Take 20 - 4x \ge 8. Written as -4x + 20 \ge 8, it says x was multiplied by -4 and then 20 was added. Subtract 20 from both sides:

-4x \ge -12

The right side has become negative, and that changes nothing: subtracting never turns the sign. Now divide both sides by -4. This is the one step that turns it:

x \le 3

Check the boundary: 20 - 4(3) = 8, and 8 \ge 8 is true, so 3 is included and gets a filled dot. A test number on the left, x = 0, gives 20 \ge 8, true; one on the right, x = 5, gives 0 \ge 8, false. The shading goes left.

Figure 2The solution of 20 - 4x \ge 8. The filled dot says that 3 is included.

A second route never divides by a negative number, and it confirms the turn. Add 4x to both sides of the original inequality, which is allowed, since it adds the same amount to both:

20 \ge 8 + 4x

Subtract 8 to get 12 \ge 4x, and divide by the positive number 4 to get 3 \ge x. “3 is at least x” says exactly what “x is at most 3” says. Both routes reach the same set, as they must, since every step on both was allowed.

Reading the answer with the variable first. If x ends up on the right, as in \displaystyle \frac{5}{2} < y, read the statement from the variable’s side: \displaystyle y > \frac{5}{2}. Either way the pointed end of the symbol faces the smaller number.

One bracket. In p(x + q) > r the bracket was multiplied by p last, so you may divide both sides by p first (turning the sign if p is negative), or multiply out with the distributive property and then take the two steps. Example 4 does it both ways.

Part 2: writing the inequality from a story

The total in a story is built the same way as in an equation: a fixed amount plus a rate times the unknown. What is new is the word that compares the total with the limit, so name the unknown with its unit first (“let s be the number of students”) and then read the comparing word.

Words	Symbol
more than, greater than, exceeds, over	>
at least, no less than, a minimum of	\ge
less than, fewer than, under, below	<
at most, no more than, cannot exceed, a maximum of	\le

“At least 500” includes 500 itself; “more than 500” does not. The phrase “less than” does two different jobs. After “is” it compares: “a number is less than 13” is n < 13. After a number it subtracts: “eight less than three times a number” is 3n - 8, eight taken away. The same goes for “more than”.

Solve exactly as in Part 1. Then comes the step a bare inequality does not have: read the answer back in the story. The algebra returns every real number that satisfies the inequality, and the story usually allows only some of them. A count of students or tickets must be a whole number (0, 1, 2, 3 and so on). A number of weeks cannot be negative. An amount of money can be any number of cents.

When only whole numbers make sense and the boundary is not whole, the direction of the inequality decides which way to go, not the usual rounding rule. From s \le 38\tfrac{2}{3} the largest whole number allowed is 38, although 38\tfrac{2}{3} is nearer 39. From c \ge 84\tfrac{4}{9} the smallest allowed is 85, although 84\tfrac{4}{9} is nearer 84. In each case the nearest whole number lies outside the solution set. Finish by trying the answer and its neighbor in the words of the story: one should work and the other fail.

Worked examples

Examples 1 to 4 belong to Part 1 and Examples 5 to 7 to Part 2.

Example 1

A positive coefficient, and the unknown on the right

Solve each inequality and describe its graph on a number line.

(a) 4x - 9 \le 15 \qquad (b) 17 < 6y + 2

Solution

(a) Add 9 to both sides, which never changes the direction: 4x \le 24. Divide both sides by 4, a positive number, so the direction stays:

x \le 6

The graph is a filled dot at 6 with everything to its left shaded. The dot is filled because 6 is included: 4(6) - 9 = 15, and 15 \le 15 is true. A test number, x = 0, gives -9 \le 15, true, and 0 is on the left, as the graph says.

(b) The unknown sits on the right side, which changes nothing about the moves. Subtract 2 from both sides, then divide by the positive number 6:

\begin{aligned} 15 &< 6y \\ \frac{5}{2} &< y \end{aligned}

Read from the variable’s side, \displaystyle y > \frac{5}{2}. The graph is an open circle at \displaystyle \frac{5}{2}, halfway between 2 and 3, with everything to its right shaded. Check: \displaystyle y = \frac{5}{2} gives \displaystyle 6 \cdot \frac{5}{2} + 2 = 17, and 17 < 17 is false, so the circle is open; y = 3 gives 20, and 17 < 20 is true. The solutions need not be whole numbers: 2.6 works, and so does 100.

Example 2

A negative coefficient

Solve -3n + 4 > 19 and describe its graph.

Solution

Subtract 4 from both sides. Then divide both sides by -3, a negative number, so the sign turns from > to <:

\begin{aligned} -3n + 4 &> 19 \\ -3n &> 15 \\ n &< -5 \end{aligned}

The graph is an open circle at -5 with everything to its left shaded. Check the boundary: -3(-5) + 4 = 19, and 19 > 19 is false, so -5 is left out. Test n = -6: 18 + 4 = 22, and 22 > 19 is true. Test n = 0: 4 > 19 is false. The solutions lie to the left of -5, as solved.

Example 3

Fraction coefficients

Solve (a) \displaystyle 7 - \frac{m}{2} \ge 3 \qquad (b) \displaystyle \frac{2}{3}k + 5 < 1

Solution

(a) Read the left side as \displaystyle -\frac{1}{2}m + 7. Subtract 7 from both sides:

-\frac{m}{2} \ge -4

Dividing by -2 is undone by multiplying by -2. That number is negative, so the sign turns:

m \le 8

Check: m = 8 gives 7 - 4 = 3, and 3 \ge 3 is true, so 8 is included. m = 0 gives 7 \ge 3, true, and m = 10 gives 2 \ge 3, false. The graph is a filled dot at 8, shaded to the left.

(b) Subtract 5 from both sides:

\frac{2}{3}k < -4

Multiply both sides by \displaystyle \frac{3}{2}, the reciprocal of \displaystyle \frac{2}{3}. It is positive, so the sign stays:

k < -6

The right side was negative before this step and the sign did not turn, because only the number you multiply by matters. Check: k = -9 gives -6 + 5 = -1, and -1 < 1 is true; k = -6 gives -4 + 5 = 1, and 1 < 1 is false, so the circle at -6 is open; k = 0 gives 5 < 1, false.

Example 4

One bracket, two ways

Solve -2(x + 3) \le 8.

Solution

Dividing first. The bracket was multiplied by -2, so divide both sides by -2 and turn the sign. Then subtract 3:

\begin{aligned} x + 3 &\ge -4 \\ x &\ge -7 \end{aligned}

Distributing first. The distributive property gives -2x - 6 \le 8. Add 6, then divide by -2 and turn the sign:

\begin{aligned} -2x &\le 14 \\ x &\ge -7 \end{aligned}

Each route turns the sign exactly once, because each divides by -2 exactly once. Check: x = -7 gives -2(-4) = 8, and 8 \le 8 is true, so -7 gets a filled dot. x = 0 gives -6 \le 8, true, so the shading runs to the right. x = -10 gives -2(-7) = 14, and 14 \le 8 is false.

Example 5

A school trip budget

Gerda is organizing a school trip. The coach costs 420 dollars for the day, and theater tickets cost 15 dollars for each student. The whole trip may cost at most 1{,}000 dollars. How many students can go?

Solution

Let s be the number of students. The cost is the fixed 420 plus 15 for each student, and “at most” is \le:

420 + 15s \le 1000

Subtract 420 from both sides, then divide by the positive number 15:

\begin{aligned} 15s &\le 580 \\ s &\le \frac{580}{15} = 38\tfrac{2}{3} \end{aligned}

Now read the answer back. s counts students, so only whole numbers make sense. The inequality is \le, so the allowed numbers lie at or below the boundary: every whole number up to 38 is less than 38\tfrac{2}{3}, and 39 is not, so at most 38 students can go.

Figure 3Example 5. The inequality allows every number up to 38\tfrac{2}{3}. The story allows only whole numbers of students, which carry on down to 0 off the picture, so 38 is the most.

Check in the story: 38 students cost 420 + 570 = 990 dollars, inside the budget, while 39 cost 420 + 585 = 1005 dollars, five over. Ordinary rounding would have picked 39, and the budget rules it out.

Example 6

A fundraiser target

Hamish’s club needs to raise at least 500 dollars for new equipment. A local shop has donated 120 dollars, and the club makes 4.50 dollars on each candle it sells. How many candles must it sell?

Solution

Let c be the number of candles sold. The money raised is 120 + 4.5c, and “at least” is \ge:

120 + 4.5c \ge 500

Subtract 120, then divide by the positive number 4.5:

\begin{aligned} 4.5c &\ge 380 \\ c &\ge \frac{380}{4.5} = 84\tfrac{4}{9} \end{aligned}

Candles are sold whole. The smallest whole number that is at least 84\tfrac{4}{9} is 85, so the club must sell at least 85 candles. Check: 84 candles raise 120 + 378 = 498 dollars, two short, and 85 raise 120 + 382.50 = 502.50 dollars. This time the answer goes up from the boundary, because the inequality is \ge.

Example 7

Spending down savings

Konrad has 300 dollars in savings and pays for music lessons out of it, 35 dollars each week. He wants to keep at least 80 dollars in the account. For how many weeks can he pay?

Solution

Let w be the number of weeks. After w weeks he has 300 - 35w dollars, and that must be at least 80:

300 - 35w \ge 80

Subtract 300 from both sides, then divide by -35. That number is negative, so the sign turns:

\begin{aligned} -35w &\ge -220 \\ w &\le \frac{220}{35} = 6\tfrac{2}{7} \end{aligned}

Read back: w is a whole number of weeks, and the largest whole number not above 6\tfrac{2}{7} is 6. He can pay for at most 6 weeks. Check: after 6 weeks he has 300 - 210 = 90 dollars, enough, and after 7 weeks 300 - 245 = 55 dollars, too little.

The turned sign has a meaning here. A lower limit on the money (“at least 80”) became an upper limit on the weeks (“at most 6”) because every extra week leaves less money.

Boundaries

The unknown appears once, inside at most one bracket. If it appears on both sides, or the expression needs several brackets multiplied out and terms collected, that is multi-step inequalities.

The coefficient p is not 0. If it were, the unknown would vanish and there would be nothing to divide by. 0x + 5 > 3 is true for every x, and 0x + 5 > 7 for none.

The sign turns exactly when both sides are multiplied or divided by a negative number. Adding or subtracting any number, positive or negative, never turns it, and a negative number appearing on one side never turns it by itself. Multiplying both sides by 0 is never allowed: it turns the true 2 < 5 into the false 0 < 0, and every solution set into nothing or everything.

Strict or not decides the boundary. With > or < the boundary number is left out and drawn as an open circle; with \ge or \le it is included and drawn as a filled dot.

The algebra answers in real numbers; the story may allow only some of them. A count must be a whole number, a time cannot be negative, money comes in cents. A story can even rule out every solution the algebra gives, as in Exercise 23.

Where this goes next

Multi-step inequalities put the unknown on both sides and add more brackets; the same allowed steps finish them, after collecting terms. A story often carries two limits at once: the coach may run only for at least 20 students, and the budget allows at most 38. Both together form a compound inequality, whose solution is the overlap of the two sets. Solution sets also have a compact written form, interval notation, which saves drawing a number line every time.

With two unknowns, say adults’ and students’ tickets at different prices, a limit becomes a linear inequality in two variables, whose solutions fill half of a plane, and several such limits together are a system of inequalities. Writing those limits from a situation, and saying what they mean, is writing and interpreting constraints, the same reading back as in Part 2, done with more than one quantity.

Exercises

Part 1: positive coefficients

Solve each inequality and describe its graph on a number line.

5x + 8 > 33
3y - 11 \le 10
29 \le 4k + 5
\displaystyle \frac{a}{3} + 2 < 7

Part 1: negative and fractional coefficients

Solve each inequality and describe its graph on a number line.

-2x + 9 \ge 1
14 - 5n < 39
\displaystyle -\frac{p}{4} - 3 > 2
-6 \ge -0.4t + 2
\displaystyle \frac{3}{5}w - 7 \le -10

Part 1: one bracket

3(x - 4) > 6
-4(m + 2) \le 12

Part 1: testing and building

Decide whether -3 is a solution of 7 - 2x > 12 by substituting it. Then solve the inequality and show that the solution set agrees.
The solution of 3x + r \le 20 is graphed as a filled dot at 4 with everything to its left shaded. Find r.

Part 2: from words to symbols

Write an inequality for each sentence, using n for the number, and solve it.

Five more than twice a number is at least 17.
Eight less than three times a number is less than 13.
When a number is multiplied by -4 and then 10 is added, the result is no more than 30.

Part 2: stories

Write an inequality, solve it, and answer the question in the story.

A parking garage charges 4 dollars to enter and 2.50 dollars for each hour, counting only whole hours. Ilaria has 21 dollars. For how many hours can she pay?
Jasna’s pottery studio has 20 kg of clay. She sets 3.5 kg aside for a large vase, and each mug uses 0.6 kg. How many mugs can she make?
Bogdan has 85 dollars and puts 15 dollars into savings every week. He wants at least 310 dollars for a bike. After how many weeks will he have enough?
Ludmila has 40 dollars and will make 12 equal weekly deposits. She wants at least 250 dollars at the end. How much must each deposit be?
Tamsin’s choir needs at least 600 dollars for a concert tour. A parent has donated 150 dollars, and raffle tickets sell for 7 dollars each. How many tickets must the choir sell?
Oskar’s school lunch card holds 60 dollars, and each day he buys lunch 7 dollars comes off it. He wants at least 5 dollars left on it for snacks. For how many days can he buy lunch?
A boat-hire stand charges 10 dollars to take a boat out and 3 dollars for each hour on the lake. Piet has 9 dollars. Write and solve an inequality for the number of hours h he can pay for, and say what the answer means.

Reasoning

When only whole numbers make sense, explain why the answer to 420 + 15s \le 1000 is found by going down from 38\tfrac{2}{3}, while the answer to 120 + 4.5c \ge 500 is found by going up from 84\tfrac{4}{9}, whatever the fraction parts are.
Solve 6 - 3x > 0 twice: once by dividing by -3, and once by first adding 3x to both sides. Explain why the two routes must agree.

Answers

x > 5: open circle at 5, shaded to the right. Subtract 8: 5x > 25; divide by 5.
y \le 7: filled dot at 7, shaded to the left. Add 11: 3y \le 21; divide by 3.
k \ge 6: filled dot at 6, shaded to the right. Subtract 5: 24 \le 4k; divide by 4 to get 6 \le k.
a < 15: open circle at 15, shaded to the left. Subtract 2: \displaystyle \frac{a}{3} < 5; multiply by 3.
x \le 4: filled dot at 4, shaded to the left. Subtract 9: -2x \ge -8; divide by -2 and turn the sign.
n > -5: open circle at -5, shaded to the right. Subtract 14: -5n < 25; divide by -5 and turn the sign.
p < -20: open circle at -20, shaded to the left. Add 3: \displaystyle -\frac{p}{4} > 5; multiply by -4 and turn the sign.
t \ge 20: filled dot at 20, shaded to the right. Subtract 2: -8 \ge -0.4t; divide by -0.4 and turn the sign, giving 20 \le t.
w \le -5: filled dot at -5, shaded to the left. Add 7: \displaystyle \frac{3}{5}w \le -3; multiply by \displaystyle \frac{5}{3}, which is positive, so the sign stays.
x > 6: open circle at 6, shaded to the right. Divide by 3: x - 4 > 2; add 4.
m \ge -5: filled dot at -5, shaded to the right. Divide by -4 and turn the sign: m + 2 \ge -3; subtract 2.
Yes: 7 - 2(-3) = 13, and 13 > 12. Solving, -2x > 5, and dividing by -2 turns the sign: \displaystyle x < -\frac{5}{2}. Since \displaystyle -3 < -\frac{5}{2}, the number -3 is in the solution set, which agrees.
r = 8. The boundary 4 makes the two sides equal: 3(4) + r = 20. Then 3x + 8 \le 20 gives x \le 4, shaded to the left as drawn.
2n + 5 \ge 17, so n \ge 6. Subtract 5: 2n \ge 12.
3n - 8 < 13, so n < 7. Add 8: 3n < 21.
-4n + 10 \le 30, so n \ge -5. Subtract 10: -4n \le 20; divide by -4 and turn the sign.
At most 6 hours. 4 + 2.5h \le 21 gives 2.5h \le 17, so h \le 6.8, and only whole hours count; with \le the answer goes down from the boundary. Six hours cost 19 dollars; seven would cost 21.50 dollars.
At most 27 mugs. 3.5 + 0.6m \le 20 gives 0.6m \le 16.5, so m \le 27.5, and mugs are whole. Twenty-seven mugs use 3.5 + 16.2 = 19.7 kg in all; 28 would need 20.3 kg.
After at least 15 weeks. 85 + 15w \ge 310 gives 15w \ge 225, so w \ge 15. The boundary is included: after 15 weeks he has exactly 310 dollars.
At least 17.50 dollars a week. 40 + 12d \ge 250 gives 12d \ge 210, so d \ge 17.5. Money comes in cents, and 17.50 dollars is a whole number of cents, so the boundary itself works.
At least 65 tickets. 150 + 7t \ge 600 gives 7t \ge 450, so t \ge 64\tfrac{2}{7}, and with \ge the answer goes up from the boundary. Sixty-four tickets raise 598 dollars in all, two short; 65 raise 605 dollars.
At most 7 days. 60 - 7d \ge 5 gives -7d \ge -55, and dividing by -7 turns the sign: d \le 7\tfrac{6}{7}. After 7 days 11 dollars are left; after 8 only 4.
10 + 3h \le 9 gives 3h \le -1, so \displaystyle h \le -\frac{1}{3}. No number of hours, not even 0, is that small, so he cannot hire a boat at all: the fixed charge alone is more than he has.
With \le the solutions are the numbers at or below the boundary, so the allowed whole numbers are those below 38\tfrac{2}{3}, and the largest is the whole number just under it, 38. With \ge the solutions are at or above the boundary, so the smallest allowed whole number is the one just over it, 85. How near the boundary a whole number is does not matter; which side it is on does, and the nearest whole number may be on the wrong side.
Both give x < 2. Dividing: -3x > -6, then divide by -3 and turn the sign. Adding 3x: 6 > 3x, then divide by 3 to get 2 > x. Every step on both routes leaves the solution set unchanged, so both end at the solution set of the original inequality, and there is only one.