Solutions to Problem Set 2

Math 101-06

9-6-2017

[Linear inequalities]

1. Solve the inequality $17 x + 3
   > 6 - 11 x$ for x. Write your answer using either inequality notation or interval notation.

$$\eqalign{ 17 x + 3 & > 6 - 11 x \cr 17 x + 3 + 11 x - 3 & > 6 - 11 x + 11 x - 3 \cr 28 x & > 3 \cr \noalign{\vskip2pt} x & > \dfrac{3}{28} \cr}$$

The solution is $x >
   \dfrac{3}{28}$ or $\left(\dfrac{3}{28}, \infty\right)$ .


2. Solve the inequality $22 x \ge
   3(5 x - 2) + 17$ for x. Write your answer using either inequality notation or interval notation.

$$\eqalign{ 22 x & \ge 3(5 x - 2) + 17 \cr 22 x & \ge 15 x - 6 + 17 \cr 22 x & \ge 15 x + 11 \cr 22 x - 15 x & \ge 15 x + 11 - 15 x \cr 7 x & \ge 11 \cr \noalign{\vskip2pt} \dfrac{7 x}{7} & \ge \dfrac{11}{7} \cr \noalign{\vskip2pt} x & \ge \dfrac{11}{7} \quad\halmos \cr}$$


3. Solve the inequality $17 - 8 x
   < 4(3.5 - 2 x)$ for x. Write your answer using either inequality notation or interval notation.

$$\eqalign{ 17 - 8 x & < 4(3.5 - 2 x) \cr 17 - 8 x & < 14 - 8 x \cr 17 - 8 x + 8 x & < 14 - 8 x + 8 x \cr 17 & < 14 \cr}$$

The last inequality is false, because 17 is not less than 14. Hence, the original inequality has no solutions.

You may write your answer as $\emptyset$ , the symbol for the empty set. Don't write things like "DNE".


4. Solve the inequality $12 x -
   17 < 3 (2 + 4 x)$ for x. Write your answer using either inequality notation or interval notation.

$$\eqalign{ 12 x - 17 & < 3 (2 + 4 x) \cr 12 x - 17 & < 6 + 12 x \cr 12 x - 17 - 12 x & < 6 + 12 x - 12 x \cr -17 & < 6 \cr}$$

The last inequality is true, because -17 is less than 6. Hence, the original inequality is true for all x.

You may write your answer as $\real$ , or as $(-\infty, \infty)$ .


5. Solve the inequality $3 x + 5
   \le 8(3 - x)$ for x, writing your answer using either inequality notation or interval notation.

$$\eqalign{ 3 x + 5 & \le 8(3 - x) \cr 3 x + 5 & \le 24 - 8 x \cr 3 x + 5 - 5 + 8 x & \le 24 - 8 x - 5 + 8 x \cr 11 x & \le 19 \cr \noalign{\vskip2pt} x & \le \dfrac{19}{11} \cr}$$

The solution is $x \le
   \dfrac{19}{11}$ , or $\left(-\infty,
   \dfrac{19}{11}\right]$ .


6. Solve the inequality $\dfrac{7}{10} x - 3 > \dfrac{5}{4} + \dfrac{1}{2} x$ for x. Write your answer using either inequality notation or interval notation.

The denominators of the fractions are 10, 4, and 2. The least common multiple of 10, 4, and 2 is 20. So I'll multiply both sides by 20 to clear the fractions.

$$\eqalign{ \dfrac{7}{10} x - 3 & > \dfrac{5}{4} + \dfrac{1}{2} x \cr \noalign{\vskip2pt} 20 \cdot \left(\dfrac{7}{10} x - 3\right) & > 20 \cdot \left(\dfrac{5}{4} + \dfrac{1}{2} x\right) \cr \noalign{\vskip2pt} 20 \cdot \dfrac{7}{10} x - 20 \cdot 3 & > 20 \cdot \dfrac{5}{4} + 20 \cdot \dfrac{1}{2} x \cr \noalign{\vskip2pt} 14 x - 60 & > 25 + 10 x \cr 14 x - 60 + 60 - 10 x & > 25 + 10 x + 60 - 10 x \cr 4 x & > 85 \cr \noalign{\vskip2pt} x & > \dfrac{85}{4} \quad\halmos \cr}$$


7. Solve the inequality $\dfrac{4}{3} (5 x - 7) > \dfrac{5}{6} - \dfrac{3}{4} x$ for x. Write your answer using either inequality notation or interval notation.

The denominators of the fractions are 3, 6, and 4. The least common multiple of 3, 6, and 4 is 12. So I'll multiply both sides by 12 to clear the denominators.

$$\eqalign{ \dfrac{4}{3} (5 x - 7) & > \dfrac{5}{6} - \dfrac{3}{4} x \cr \noalign{\vskip2pt} 12 \cdot \dfrac{4}{3} (5 x - 7) & > 12 \cdot \left(\dfrac{5}{6} - \dfrac{3}{4} x\right) \cr \noalign{\vskip2pt} 12 \cdot \dfrac{4}{3} (5 x - 7) & > 12 \cdot \dfrac{5}{6} - 12 \cdot \dfrac{3}{4} x \cr \noalign{\vskip2pt} 12 \cdot \dfrac{4}{3} (5 x - 7) & > 12 \cdot \dfrac{5}{6} - 12 \cdot \dfrac{3}{4} x \cr \noalign{\vskip2pt} 16 (5 x - 7) & > 10 - 9 x \cr 80 x - 112 & > 10 - 9 x \cr 80 x - 112 + 112 + 9 x & > 10 - 9 x + 112 + 9 x \cr 89 x & > 122 \cr \noalign{\vskip2pt} x & > \dfrac{122}{89} \quad\halmos \cr}$$


8. Solve the inequality $1.4 x +
   1.2 > 0.02 (5 x + 3)$ for x. Write your answer using either inequality notation or interval notation.

$$\eqalign{ 1.4 x + 1.2 & > 0.02 (5 x + 3) \cr 100 \cdot (1.4 x + 1.2) & > 100 \cdot 0.02 (5 x + 3) \cr 100 \cdot 1.4 x + 100 \cdot 1.2 & > 100 \cdot 0.02 (5 x + 3) \cr 140 x + 120 & > 2 (5 x + 3) \cr 140 x + 120 & > 10 x + 6 \cr 130 x & > -114 \cr \noalign{\vskip2pt} x & > -\dfrac{114}{13} = -\dfrac{57}{65} \cr} \quad\halmos$$


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