Question

section 1.5 daily work questions
solve by graphing.

1. (\frac{1}{3}x - 4 = -2x + 3)
2. (|x + 5| = 2x + 4)
3. (\frac{3}{4}x + 3 < -3)
4. (-|x| + 5 geq -1)
5. (-\frac{1}{2}x + 2 > |x - 1|)
6. (2|x| + 3 leq -x + 6)
7. write an inequality & solve

cindy is longboarding 6 mi ahead of tamira.
cindy is traveling at an average rate of 2 mph.
tamira is traveling at a rate of 4 mph. let (x) represent the number of hours since tamira started longboarding. when will tamira be ahead of cindy? write an inequality to represent this situation.
(image of longboarding: cindy, 2 mph; tamira, 4 mph)

Explanation:

For Questions 1-6 (Solve by graphing, presented with algebraic solutions for verification):

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Question 1: $\frac{1}{3}x - 4 = -2x + 3$

Step1: Collect like terms

$\frac{1}{3}x + 2x = 3 + 4$

Step2: Combine terms

$\frac{7}{3}x = 7$

Step3: Solve for $x$

$x = 7 \times \frac{3}{7} = 3$

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Question 2: $|x + 5| = 2x + 4$

Step1: Split into two cases

Case 1: $x + 5 = 2x + 4$ (when $x+5\geq0$)
Case 2: $-(x + 5) = 2x + 4$ (when $x+5<0$)

Step2: Solve Case1

$5 - 4 = 2x - x \implies x=1$ (valid, $1+5=6\geq0$)

Step3: Solve Case2

$-x-5=2x+4 \implies -3x=9 \implies x=-3$ (invalid, $-3+5=2\geq0$ violates case condition)

Step4: Final valid solution

$x=1$

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Question 3: $\frac{3}{4}x + 3 < -3$

Step1: Isolate $x$ term

$\frac{3}{4}x < -3 - 3$

Step2: Simplify right side

$\frac{3}{4}x < -6$

Step3: Solve for $x$

$x < -6 \times \frac{4}{3} = -8$

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Question 4: $-|x| + 5 \geq -1$

Step1: Isolate absolute value

$-|x| \geq -1 - 5$

Step2: Simplify right side

$-|x| \geq -6$

Step3: Multiply by -1 (reverse inequality)

$|x| \leq 6$

Step4: Rewrite without absolute value

$-6 \leq x \leq 6$

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Question 5: $-\frac{1}{2}x + 2 > |x - 1|$

Step1: Split into two cases

Case1: $-\frac{1}{2}x + 2 > x - 1$ (when $x-1\geq0$)
Case2: $-\frac{1}{2}x + 2 > -(x - 1)$ (when $x-1<0$)

Step2: Solve Case1

$2+1 > x + \frac{1}{2}x \implies 3 > \frac{3}{2}x \implies x<2$; combined with $x\geq1$: $1\leq x<2$

Step3: Solve Case2

$-\frac{1}{2}x + 2 > -x + 1 \implies \frac{1}{2}x > -1 \implies x>-2$; combined with $x<1$: $-2

Step4: Combine solutions

$-2 < x < 2$

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Question 6: $2|x| + 3 \leq -x + 6$

Step1: Split into two cases

Case1: $2x + 3 \leq -x + 6$ (when $x\geq0$)
Case2: $-2x + 3 \leq -x + 6$ (when $x<0$)

Step2: Solve Case1

$2x + x \leq 6 - 3 \implies 3x \leq 3 \implies x\leq1$; combined with $x\geq0$: $0\leq x\leq1$

Step3: Solve Case2

$-2x + x \leq 6 - 3 \implies -x \leq 3 \implies x\geq-3$; combined with $x<0$: $-3\leq x<0$

Step4: Combine solutions

$-3 \leq x \leq 1$

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For Question 7:

Step1: Define distances

Cindy's distance: $6 + 2x$; Tamira's distance: $4x$

Step2: Write inequality (Tamira ahead)

$4x > 6 + 2x$

Step3: Solve for $x$

$4x - 2x > 6 \implies 2x > 6 \implies x>3$

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Answer:

1. $x=3$
2. $x=1$
3. $x < -8$
4. $-6 \leq x \leq 6$
5. $-2 < x < 2$
6. $-3 \leq x \leq 1$
7. Inequality: $4x > 2x + 6$; Tamira will be ahead after $x > 3$ hours (3 hours after she starts)