QUESTION IMAGE
Question
section 1.5 daily wo
solve by graphing.
- ⅓x − 4 = −2x + 3
- |x + 5| = 2x + 4
- ¾x + 3 < −3
- −|x| + 5 ≥ −1
- −½x + 2 > |x − 1|
- 2|x| + 3 ≤ −x + 6
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Problem 1: $\frac{1}{3}x - 4 = -2x + 3$
Step1: Isolate x terms
$\frac{1}{3}x + 2x = 3 + 4$
Step2: Combine like terms
$\frac{1}{3}x + \frac{6}{3}x = 7$
$\frac{7}{3}x = 7$
Step3: Solve for x
$x = 7 \times \frac{3}{7}$
$x = 3$
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Problem 2: $|x + 5| = 2x + 4$
Step1: Case 1: $x + 5 \geq 0$ (i.e., $x \geq -5$)
$x + 5 = 2x + 4$
$5 - 4 = 2x - x$
$x = 1$ (valid, since $1 \geq -5$)
Step2: Case 2: $x + 5 < 0$ (i.e., $x < -5$)
$-(x + 5) = 2x + 4$
$-x - 5 = 2x + 4$
$-5 - 4 = 2x + x$
$3x = -9$
$x = -3$ (invalid, since $-3
less -5$)
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Problem 3: $\frac{3}{4}x + 3 < -3$
Step1: Isolate x term
$\frac{3}{4}x < -3 - 3$
$\frac{3}{4}x < -6$
Step2: Solve for x
$x < -6 \times \frac{4}{3}$
$x < -8$
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Problem 4: $-|x| + 5 \geq -1$
Step1: Isolate absolute value
$-|x| \geq -1 - 5$
$-|x| \geq -6$
Step2: Reverse inequality (multiply by -1)
$|x| \leq 6$
Step3: Solve absolute value inequality
$-6 \leq x \leq 6$
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Problem 5: $-\frac{1}{2}x + 2 > |x - 1|$
Step1: Case 1: $x - 1 \geq 0$ (i.e., $x \geq 1$)
$-\frac{1}{2}x + 2 > x - 1$
$2 + 1 > x + \frac{1}{2}x$
$3 > \frac{3}{2}x$
$x < 2$ (valid, since $1 \leq x < 2$)
Step2: Case 2: $x - 1 < 0$ (i.e., $x < 1$)
$-\frac{1}{2}x + 2 > -(x - 1)$
$-\frac{1}{2}x + 2 > -x + 1$
$-\frac{1}{2}x + x > 1 - 2$
$\frac{1}{2}x > -1$
$x > -2$ (valid, since $-2 < x < 1$)
Step3: Combine valid intervals
$-2 < x < 2$
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Problem 6: $2|x| + 3 \leq -x + 6$
Step1: Isolate absolute value
$2|x| \leq -x + 6 - 3$
$2|x| \leq -x + 3$
Step2: Case 1: $x \geq 0$
$2x \leq -x + 3$
$2x + x \leq 3$
$3x \leq 3$
$x \leq 1$ (valid, since $0 \leq x \leq 1$)
Step3: Case 2: $x < 0$
$2(-x) \leq -x + 3$
$-2x \leq -x + 3$
$-2x + x \leq 3$
$-x \leq 3$
$x \geq -3$ (valid, since $-3 \leq x < 0$)
Step4: Combine valid intervals
$-3 \leq x \leq 1$
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