Absolute Value in Equations and Inequalities
Absolute Value of a Number
The absolute value of a real number a, written $ |a| $ , was defined (Chapter 1) as follows:
\[ \left|a\right|=\left\{\begin{aligned}a&\quad if a\geq0\\ -a&\quad if a<0\end{aligned}\right. \]
Absolute Value, Interpreted Geometrically
Geometrically, the absolute value of a real number is the distance of that number from the origin (see Fig. 7-1).
Similarly, the distance between two real numbers a and b is the absolute value of their difference: $ |a - b| $ or $ |b - a| $ .
Properties of Absolute Values
\[ \left|-a\right|=\left|a\right| \]
\[ \left|a\right|=\sqrt{a^{2}} \]
\[ \left|a b\right|=\left|a\right|\left|b\right| \]
\[ \left|a+b\right|\leq\left|a\right|+\left|b\right|(Triangle inequality) \]
EXAMPLE 7.1 (a) $ |-5|=|5|=5 $ ; (b) $ |-6|=6 $ ; $ ==6 $ , thus, $ |-6|= $ .
EXAMPLE 7.2 (a) $ |-5x{2}|=|-5|x{2}|=5x^{2}| $ ; (b) $ |3y|=|3|y|=3|y|| $
EXAMPLE 7.3 Triangle inequality: $ |5+(-7)|=2|5|+|-7|=5+7=12 $
Absolute Value in Equations
Since $ |a| $ is the distance of a from the origin,
The equation $ |a| = b $ is equivalent to the two equations a = b and a = -b, for b > 0. (The distance of a from the origin will equal b precisely when a equals b or -b.)
The equation $ |a|=|b| $ is equivalent to the two equations a=b and a=-b.
EXAMPLE 7.4 Solve: $ |x + 3| = 5 $
Transform into equivalent equations that do not contain the absolute value symbol and solve:
\[ x+3=5 \]
\[ x+3=-5 \]
\[ x=2 \]
\[ x=-8 \]
EXAMPLE 7.5 Solve: $ |x - 4| = |3x + 1| $
Transform into equivalent equations that do not contain the absolute value symbol and solve:
\[ x-4=3x+1\qquad or\qquad x-4=-(3x+1) \]
\[ -2x=5 \]
\[ x-4=-3x-1 \]
\[ x=-\frac{5}{2} \]
\[ 4x=3 \]
\[ x=\frac{3}{4} \]
Absolute Value in Inequalities
For b > 0,
- The inequality $ |a| < b $ is equivalent to the double inequality -b < a < b. (Since the distance of a from the origin is less than b, a is closer to the origin than b; see Fig. 7-2.)
- The inequality $ |a| > b $ is equivalent to the two inequalities a > b and a < -b. (Since the distance of a from the origin is greater than b, a is farther from the origin than b; see Fig 7-3.)
EXAMPLE 7.6 Solve: $ |x-5|>3 $
Transform into equivalent inequalities that do not contain the absolute value symbol and solve:
\[ \begin{aligned}&x-5>3\quad&or\quad&x-5<-3\\&x>8\quad&x<2\end{aligned} \]
SOLVED PROBLEMS
7.1. Solve: $ |x - 7| = 2 $
Transform into equivalent equations that do not contain the absolute value symbol and solve:
\[ x-7=2 \]
or
\[ x-7=-2 \]
\[ x=9 \]
\[ x=5 \]
7.2. Solve: $ |x + 5| = 0.01 $
\[ x+5=0.01 \]
\[ \mathbf{o r} \]
\[ x+5=-0.01 \]
\[ x=-4.99 \]
\[ x=-5.01 \]
7.3. Solve: $ |6x + 7| = 10 $
\[ 6x+7=10\qquad or\qquad6x+7=-10 \]
\[ 6x=3 \]
\[ 6x=-17 \]
\[ x=\frac{1}{2} \]
\[ x=-\frac{17}{6} \]
7.4. Solve: $ 5|x| - 3 = 6 $
First isolate the absolute value expression, then write the two equivalent equations that do not contain the absolute value symbol.
\[ 5\left|x\right|=9 \]
\[ \left|x\right|=\frac{9}{5} \]
\[ x=\frac{9}{5}\quad or\quad x=-\frac{9}{5} \]
7.5. Solve: $ 3|5 - 2x| + 4 = 9 $
First isolate the absolute value expression.
\[ 3\left|5-2x\right|=5 \]
\[ \left|5-2x\right|=\frac{5}{3} \]
Now write and solve the two equivalent equations that do not contain the absolute value symbol.
\[ 5-2x=\frac{5}{3}\qquad or \]
\[ 5-2x=-\frac{5}{3} \]
\[ -2x=-\frac{10}{3} \]
\[ -2x=-\frac{20}{3} \]
\[ x=\frac{5}{3} \]
\[ x=\frac{10}{3} \]
7.6. Solve: $ |5x-3|=-8 $
Since the absolute value of a number is never negative, this equation has no solution.
7.7. Solve: $ |2x-5|=|8x+3| $
Transform into equivalent equations that do not contain the absolute value symbol and solve:
\[ 2x-5=8x+3 \]
\[ 2x-5=-(8x+3) \]
\[ -6x=8 \]
\[ 2x-5=-8x-3 \]
\[ x=-\frac{4}{3} \]
\[ 10x=2 \]
\[ x=\frac{1}{5} \]
7.8. Solve: $ lx + 5l > 3 $
Transform into equivalent inequalities that do not contain the absolute value symbol and solve:
\[ \begin{aligned}&x+5>3\quad&or\quad&x+5<-3\\&&&x<-2\end{aligned} \]
Solution: \((-\infty,-8)\cup(-2,\infty)\)
7.9. Solve: $ |x - 3| $
Transform into an equivalent double inequality and solve:
\[ \begin{aligned}-10&\leq x-3\leq10\\-7&\leq x\leq13\end{aligned} \]
Solution: $ [-7,13] $
7.10. Solve: $ 4|2x - 7| + 5 < 19 $
Isolate the absolute value symbol, then transform into an equivalent double inequality and solve:
\[ 4\left|2x-7\right|<14 \]
\[ \left|2x-7\right|<\frac{7}{2} \]
\[ -\frac{7}{2}<2x-7<\frac{7}{2} \]
\[ \frac{7}{2}<2x<\frac{21}{2} \]
\[ \frac{7}{4}<x<\frac{21}{4} \]
Solution: $ (, ) $
7.11. Solve: $ |5x-3|>-1 $
Since the absolute value of a real number is always positive or zero—hence, always greater than any negative number—all real numbers are solutions.
7.12. Write as an inequality statement with and without the absolute value symbol and graph the solutions on a number line: The distance between x and a is less than $ $ .
In terms of the absolute value symbol, this statement becomes $ |x - a| < $ . Rewrite as a double linequality and solve:
\[ -\delta<x-a<\delta \]
\[ a-\mathfrak{d}<x<a+\mathfrak{d} \]
The graph is shown in Fig. 7-4:
SUPPLEMENTARY PROBLEMS
7.13. Prove: $ |ab| = |a||b| $ . (Hint: Consider the cases separately for various signs of a and b.)
7.14. (a) Prove: for any real number x, $ -|x| x |x| $ . (b) Use part (a) to prove the triangle inequality.
7.15. Write as an equation or inequality and solve:
The distance between x and 3 is equal to 7. (b) 5 is twice the distance between x and 6.
The distance between x and -3 is more than 2.
\[ \mathrm{Ans.}\quad(\mathrm{a})\mathrm{l}x-3\mathrm{l}=7;\left\{-4,10\right\};(\mathrm{b})5=2\mathrm{l}x-6\mathrm{l};\left\{\frac{17}{2},\frac{7}{2}\right\};(\mathrm{c})\mathrm{l}x+3\mathrm{l}>2;(-\infty,-5)\cup(-1,\infty) \]
7.16. Solve: (a) $ |x + 8| = 5 $ ; (b) $ |x + 5| < 8 $ ; (c) $ |x - 3| $
Ans. (a) $ {-13,-3} $ ; (b) $ (-13,3) $ ; (c) $ (-,-1] $ ; (c) $ (-,-14)(-2,) $
7.18. Solve: $ |5-2x|=3|x+1| $ Ans. $ {-8,} $
7.19. Solve: $ |3-5x| $ Ans. $ (-,-]$ ; (c) $ [c - , c) (c, c + ] $