Algebra 1 · A.REI.B.3

Multi-step inequalities

When the unknown sits on both sides of an inequality, tangled in brackets, how do you find every number that makes it true?

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Equations with variables on both sidesMath 8 Two-step inequalitiesMath 7

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Writing and interpreting constraintsAlgebra 1

Where you are 3 in Solving linear inequalities

Algebra9

Inequalities7

Solving linear inequalities3

One-step inequalities and the sign flip Two-step inequalities

Multi-step inequalities

When you reach for it

Two-step inequalities had the unknown in one place: px + q > r. Many limits are not that tidy. Two companies quote for a class museum tour, each with a booking charge and a price per student, and you want to know when the first is no dearer than the second. Now the unknown appears on both sides:

250 + 6s \le 80 + 10s

These are multi-step inequalities: the unknown may sit on both sides, inside brackets, with fractions or decimals, and the sign is one of <, >, \le or \ge. The equation versions, multi-step equations and equations with variables on both sides, used the same moves. The one change is the one one-step inequalities found: multiplying or dividing both sides by a negative number turns the sign round. The answer is a whole solution set.

The procedure

Every step below must leave the solution set exactly as it was, so each one comes with the reason it is allowed.

Make each side as simple as it can be

Multiply out brackets with the distributive property and combine like terms on each side. In 5 - 2(x + 3) \ge 4x + 11, the left side becomes 5 - 2x - 6, which is -2x - 1:

-2x - 1 \ge 4x + 11

For every x the new left side is the same number as the old one, so the solution set is unchanged. Fractions are cleared by multiplying both sides by a common denominator, a positive number, so the direction stays.

Gather the variable on one side

Now add or subtract a variable term on both sides, as in equations with variables on both sides. For any particular x, 4x is one particular number, and adding or subtracting the same number on both sides slides both the same distance along the number line, which never changes which is larger. So this step never turns the sign, whatever the sign of x.

Finish with two steps, and the one place the sign turns

What remains is a two-step inequality. Move the number term across, then divide by the coefficient of the variable. Dividing by a positive number keeps the direction. Dividing by a negative number reverses it: < becomes >, and \le becomes \ge.

Why the flip. Multiplying by -1 sends every number to its mirror image across 0 on the number line. Two numbers in order from left to right come out in order from right to left. Take 2 < 5: after multiplying by -1, the numbers are -2 and -5, and now -2 is the one further right, so -2 > -5.

Figure 1Multiplying by -1 mirrors every number across 0. On top, 2 lies left of 5; after the mirror, -2 lies right of -5, so the order has turned round.

Any negative number does this, since dividing by -6 is the same as multiplying by -1 and then by \displaystyle \frac{1}{6}. The positive part only stretches or shrinks the line, which keeps the order; the -1 mirrors it, which turns the order round. So the sign turns exactly when both sides are multiplied or divided by a negative number, and at no other step.

Choosing the side: two routes, one answer

Back to -2x - 1 \ge 4x + 11. You may gather x on either side.

Gather on the left. Subtract 4x from both sides, then add 1:

\begin{aligned} -6x - 1 &\ge 11 \\ -6x &\ge 12 \end{aligned}

Divide both sides by -6. It is negative, so the sign turns:

x \le -2

Gather on the right. Add 2x to both sides instead, which leaves 6x with a positive coefficient. Then subtract 11 and divide by the positive number 6:

\begin{aligned} -1 &\ge 6x + 11 \\ -12 &\ge 6x \\ -2 &\ge x \end{aligned}

“-2 is at least x” says the same as “x is at most -2”, so both routes agree, as they must. Keeping the coefficient positive is a choice, not a rule: it avoids the flip, at the price of reading the answer from the variable’s side. The pointed end of the symbol always faces the smaller quantity.

The solution is every number at or to the left of -2: a filled dot at -2, because \le includes it, with the line shaded to the left.

Figure 2The solution of 5 - 2(x + 3) \ge 4x + 11. The filled dot says that -2 is included.

Checking with the boundary and a test value

The boundary -2 should make the two sides equal. Left: 5 - 2(-2 + 3) = 3. Right: 4(-2) + 11 = 3. They match, and 3 \ge 3 is true, so -2 belongs. Then a test value: x = 0 gives -1 on the left and 11 on the right, and -1 \ge 11 is false. Since 0 lies to the right of -2, the shading belongs on the left. The boundary check catches a wrong number; the test value catches a wrong direction.

When the variable vanishes

Sometimes the variable terms on the two sides are equal, and gathering them makes the variable disappear. What is left is a statement about numbers alone, true or false, as in equations with no solution or every number.

If it is true, it was true whatever x was, so every real number is a solution. The graph shades the whole line.
If it is false, there is no solution. Nothing is shaded.

Figure 3When the variable cancels, the answer is all or nothing. A true statement left behind shades the whole line; a false one shades none of it.

Whether the leftover statement is strict matters: 3 \ge 3 is true, but 3 > 3 is false.

Writing the answer

Write the solution with the variable first (x \le -2, not -2 \ge x), and graph it: an open circle at the boundary for < or >, a filled dot for \le or \ge, as in inequalities on the number line. Interval notation writes x \le -2 as (-\infty, -2]; it has its own lesson.

In a story, “at least” is \ge and “at most” is \le. After solving, read the answer back in the story, which may allow only whole numbers.

Worked examples

Example 1

Variables on both sides

Solve each inequality and describe its graph.

(a) 7y - 5 > 3y + 11 \qquad (b) 6a + 1 \le 9a - 14

Solution

(a) Subtract 3y from both sides, leaving 4y with a positive coefficient. Then add 5 and divide by the positive number 4:

\begin{aligned} 4y - 5 &> 11 \\ 4y &> 16 \\ y &> 4 \end{aligned}

An open circle at 4, shaded to the right. Check: y = 4 gives 23 on both sides, and 23 > 23 is false, so 4 is left out. The test value y = 5 gives 30 > 26, true.

(b) The larger coefficient, 9a, is on the right. Subtracting 6a from both sides leaves 3a there, positive, so no flip is needed:

\begin{aligned} 1 &\le 3a - 14 \\ 15 &\le 3a \\ 5 &\le a \end{aligned}

Read from the variable’s side, a \ge 5: a filled dot at 5, shaded to the right. Check: a = 5 gives 31 \le 31, true; a = 0 gives 1 \le -14, false, and 0 is to the left.

Example 2

Brackets on both sides

Solve 3(2k - 1) - 4 < 2(5k + 3) + 1.

Solution

Distribute on each side and combine like terms:

\begin{aligned} 6k - 3 - 4 &< 10k + 6 + 1 \\ 6k - 7 &< 10k + 7 \end{aligned}

Subtract 6k from both sides, keeping the variable term positive on the right. Then subtract 7 and divide by 4:

\begin{aligned} -7 &< 4k + 7 \\ -14 &< 4k \\ -\tfrac{7}{2} &< k \end{aligned}

So \displaystyle k > -\frac{7}{2}, an open circle at -3.5 shaded to the right.

Check: \displaystyle k = -\frac{7}{2} makes the left side 3(-8) - 4 = -28 and the right side \displaystyle 2(-\frac{29}{2}) + 1 = -28. Equal, and -28 < -28 is false, so the circle is open. The test value k = 0 gives -7 < 7, true.

Example 3

Fractions on both sides

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

Solution

The denominators are 3, 4 and 2, and 12 is a multiple of all three. Multiply both sides by 12. It is positive, so the sign stays:

8x - 12 > 9x + 6

The larger coefficient is 9x, on the right. Subtract 8x from both sides, then subtract 6:

\begin{aligned} -12 &> x + 6 \\ -18 &> x \end{aligned}

So x < -18: an open circle at -18, shaded to the left. Check: x = -18 gives \displaystyle \frac{2}{3}(-18) - 1 = -13 and \displaystyle \frac{3}{4}(-18) + \frac{1}{2} = -13, equal, so the boundary is right and, with >, left out. The test value x = -24 gives -17 > -17.5, true.

Example 4

When the variable vanishes

Solve each inequality.

(a) 4(x - 2) + x \le 5x + 3 \qquad (b) 3(2x + 1) > 6x + 5 \qquad (c) 2(x + 4) > 2x + 8

Solution

(a) The left side simplifies to 5x - 8, so the inequality is 5x - 8 \le 5x + 3. Subtract 5x from both sides:

-8 \le 3

The variable is gone and the statement left is true. It was reached by allowed steps for any x, so every real number is a solution. Test values agree: x = 10 gives 42 \le 53.

(b) The left side is 6x + 3, so the inequality is 6x + 3 > 6x + 5. Subtract 6x:

3 > 5

False, whatever x is: no solution. The left side is always 2 less than the right.

(c) The left side is 2x + 8, the same as the right. Subtracting 2x leaves 8 > 8, false, so there is no solution. With \ge instead, 8 \ge 8 would be true and every real number would be a solution: the sides are always equal, and the only question is whether “equal” is allowed.

Example 5

Two tour companies

Ines is booking a guided museum tour for her class. Company A charges 250 dollars to book plus 6 dollars for each student. Company B charges 80 dollars to book plus 10 dollars for each student. For how many students does Company A cost at most what Company B costs?

Solution

Let s be the number of students. “At most” is \le:

250 + 6s \le 80 + 10s

The larger coefficient is on the right, so subtract 6s from both sides, then subtract 80, then divide by the positive number 4:

\begin{aligned} 250 &\le 80 + 4s \\ 170 &\le 4s \\ 42.5 &\le s \end{aligned}

So s \ge 42.5. A number of students is a whole number at or above 42.5, so the smallest allowed is 43, not the nearer 42. Company A is no dearer for 43 students or more.

Figure 4Example 5. The inequality allows every number from 42.5 up. Only whole numbers of students make sense, and they carry on past 47 off the picture, so 43 is the fewest.

Check: for 42 students, A costs 250 + 252 = 502 dollars and B costs 80 + 420 = 500 dollars, so A is dearer. For 43, A costs 508 dollars and B costs 510 dollars. The question said “at most”, yet the answer is a lower limit on s: A’s larger booking charge is a head start that its smaller price per student wears down only once enough students come.

Example 6

A bake sale breaking even

Thibault runs a bake sale stall. The stall costs 36 dollars, and the ingredients for each muffin cost 0.75 dollars. He gives 5 of the muffins he bakes to the helpers and sells the rest at 2.25 dollars each. To break even, the money taken must be at least the money spent. How many muffins must he bake?

Solution

Let m be the number of muffins baked. He sells m - 5, so he takes 2.25(m - 5) dollars, and he spends 36 + 0.75m dollars, since every muffin baked costs its ingredients. “At least” is \ge:

2.25(m - 5) \ge 36 + 0.75m

Distribute, then subtract 0.75m from both sides, which leaves a positive 1.5m on the left:

\begin{aligned} 2.25m - 11.25 &\ge 36 + 0.75m \\ 1.5m - 11.25 &\ge 36 \\ 1.5m &\ge 47.25 \\ m &\ge 31.5 \end{aligned}

To avoid decimals, multiply the original by the positive number 4 first: 9(m - 5) \ge 144 + 3m, then 6m \ge 189, the same m \ge 31.5.

Muffins come whole, and the answer goes up from 31.5 because the sign is \ge: he must bake at least 32 muffins. Check: with 31 he takes 2.25 \times 26 = 58.50 dollars and spends 36 + 23.25 = 59.25 dollars. With 32 he takes 2.25 \times 27 = 60.75 dollars and spends 60 dollars.

Boundaries

The inequality is linear. The variable appears only to the first power, never squared, and never in a denominator. With \displaystyle \frac{3}{x} < 1 you cannot simply multiply both sides by x, because whether that flips the sign depends on whether x is negative, which is the very thing you do not know.

The sign turns exactly when both sides are multiplied or divided by a negative number. Adding or subtracting anything, a number or a variable term, never turns it, and neither does a negative number that merely appears on one side. Multiplying both sides by 0 is never allowed: it turns the true 2 < 5 into the false 0 < 0.

A vanishing variable gives all or nothing. If the variable terms cancel, the solution set is every real number or no number at all, decided by whether the statement left over is true.

The algebra answers in real numbers; the story may allow only some of them. A count must be a whole number, so a non-whole boundary is moved in the direction the inequality points. A length can be any positive number.

Where this goes next

Interval notation gives every solution set here a compact written form. A story often holds two limits at once, as when the cheaper tour must also fit within a budget, and the two together make a compound inequality, whose solution is the overlap of two sets.

With two unknowns a limit becomes a linear inequality in two variables, whose solutions fill half a plane. Writing limits from a situation and deciding which solutions make sense is writing and interpreting constraints.

Exercises

Variables on both sides

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

9x - 4 > 5x + 20
3m + 8 \le 7m - 20
2 - 5p \ge 3p + 26
11 - 4t < 2t - 1

Brackets first

Solve each inequality.

4(x + 2) > 2x - 6
3(y - 5) + 2y \le 7y - 9
-2(3k - 1) \ge 4(k + 3)
5 - (2n - 7) < 3(n - 1)

Fractions and decimals

\displaystyle \frac{x}{2} + 3 \le \frac{x}{3} + 5
0.6w - 1.4 > 0.2w + 1

When the variable vanishes

Solve each inequality, or say that every real number is a solution, or that there is none.

2(3x - 4) \ge 6x - 10
5x - (x - 3) < 4x + 1
(a) 4(x - 1) \ge 4x - 4 \qquad (b) 4(x - 1) > 4x - 4

Checking and graphing

Wystan solves 8 - 3x \le x + 20 and writes x \le -3. Use the test value x = 0 to show his answer cannot be right, then solve the inequality correctly.
Which of these inequalities have the solution graphed as an open circle at 2 with everything to its left shaded? (a) 3x + 1 < x + 5 \qquad (b) 1 - 3x > 5 - x \qquad (c) 2(x + 1) > 4x - 2

Stories

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

Uzoma’s phone has 128 GB of storage, and apps take up 46 GB of it. Each photo album he keeps takes 3 GB. He wants the free space left over to be at least twice the space his albums use. How many albums can he keep?
Verena is fencing a rectangular vegetable bed whose length is 3 m more than twice its width. The fence runs all the way round and must be at least 40 m long. What widths are possible?
Yevette’s class grows plant cuttings to sell at the school fair. Each tray costs 6 dollars to plant and holds 12 pots; 2 from every tray fail to root, and the rest sell for 0.75 dollars each. The stall costs 15 dollars. To break even, how many trays must they plant?

Reasoning

Explain why subtracting 3x from both sides of an inequality never turns the sign, even when x stands for a negative number, while dividing both sides by -3 always does.
For which values of c is 2(x + c) \ge 2x + 6 true for every real number x, and for which values does it have no solution? Explain.

Answers

x > 6: open circle at 6, shaded to the right. Subtract 5x and add 4: 4x > 24.
m \ge 7: filled dot at 7, shaded to the right. Subtract 3m and add 20: 28 \le 4m.
p \le -3: filled dot at -3, shaded to the left. Add 5p: 2 \ge 8p + 26, so -24 \ge 8p. (Subtracting 3p instead gives -8p \ge 24, and dividing by -8 turns the sign: the same answer.)
t > 2: open circle at 2, shaded to the right. Add 4t and 1: 12 < 6t.
x > -7. Distribute: 4x + 8 > 2x - 6; subtract 2x and 8: 2x > -14.
y \ge -3. The left side is 5y - 15; subtract 5y and add 9: -6 \le 2y.
k \le -1. Distribute: -6k + 2 \ge 4k + 12; add 6k and subtract 12: -10 \ge 10k.
n > 3. The left side is 12 - 2n and the right is 3n - 3. Add 2n and 3: 15 < 5n.
x \le 12. Multiply by the positive number 6: 3x + 18 \le 2x + 30; subtract 2x and 18.
w > 6. Subtract 0.2w and add 1.4: 0.4w > 2.4; divide by 0.4.
Every real number. The left side is 6x - 8; subtracting 6x leaves -8 \ge -10, true.
No solution. The left side is 4x + 3; subtracting 4x leaves 3 < 1, false.
(a) Every real number: both sides are 4x - 4, and subtracting 4x leaves -4 \ge -4, true. (b) No solution: the same steps leave -4 > -4, false.
At x = 0 the inequality reads 8 \le 20, true, so 0 is a solution; but 0 is not in x \le -3. Correctly: add 3x to get 8 \le 4x + 20, so -12 \le 4x, so x \ge -3. The boundary was right and the direction was not.
(a) and (c). (a) 2x < 4, so x < 2. (b) Add 3x: 1 > 5 + 2x, so -4 > 2x and x < -2, a different set. (c) 2x + 2 > 4x - 2, so 4 > 2x and x < 2.
At most 9 albums. With a albums, 128 - 46 - 3a \ge 2(3a), so 82 \ge 9a and a \le 9\tfrac{1}{9}; albums are whole. With 9 albums 55 GB are free against 54 GB needed; with 10, 52 GB against 60 GB.
Any width of at least 5\tfrac{2}{3} m. The length is 2w + 3, so 2(w + 2w + 3) \ge 40, that is 6w + 6 \ge 40 and \displaystyle w \ge \frac{17}{3}. A width need not be whole, so the boundary itself works.
At least 10 trays. With t trays they sell 10t pots, so 0.75(10t) \ge 6t + 15, that is 7.5t \ge 6t + 15, so 1.5t \ge 15 and t \ge 10. Ten trays take 75 dollars and cost 75 dollars, exactly even; nine take 67.50 dollars and cost 69 dollars.
For any particular x, the term 3x is one particular number, and subtracting the same number from both sides moves both the same distance along the number line, so the larger side stays larger, negative x or not. Dividing by -3 is multiplying by -1 and then by \displaystyle \frac{1}{3}; the \displaystyle \frac{1}{3} keeps the order, but the -1 mirrors every number across 0, which reverses the order of any two numbers.
Every real number when c \ge 3; no solution when c < 3. Distributing gives 2x + 2c \ge 2x + 6, and subtracting 2x leaves 2c \ge 6, a statement with no x in it. It is true exactly when c \ge 3, and then it holds for every x; otherwise it holds for none.