Directions: Please Show All Of Your Work For Each Problem ✓ Solved

Directions : Please show all of your work for each problem.

1. Is 12 a solution to the equation 7 – x = 5?

To check if 12 is a solution, substitute x with 12 in the equation:

7 - 12 = 5

Calculating:

-5 ≠ 5, thus 12 is not a solution.

2. Is –9 a solution to the equation 9 – 8x = 81?

Substituting x with -9:

9 - 8(-9) = 81

Calculating:

9 + 72 = 81.

Hence, -9 is a solution.

3. Solve -2x + 7 ≥ 0:

-2x ≥ -7

x ≤ 3.5

4. Solve 3(x - 5)

3x - 15

Combining like terms yields:

-15 + 2

-13

x > -13.

5. Solve 8x + 2 = 7x:

8x - 7x = -2

x = -2.

6. Solve 7x - 0.96 = 6(x - 0.5):

7x - 0.96 = 6x - 3

x = -2.04.

7. Solve for x in 4x = -8:

x = -2.

8. A company estimates that 5% of the parts they manufacture are defective. If 8 defective parts are found one week by the quality assurance testers, how many parts were manufactured that week?

Let P be the total parts, then:

0.05P = 8.

P = 8 / 0.05 = 160.

9. Solve for x in 35 - 7x = 0:

7x = 35

x = 5.

10. Solve for x in -5 = 2x - 6:

2x = 1

x = 0.5.

11. Solve for x in 4(x - 2) + 4x = 5x + 6:

4x - 8 + 4x = 5x + 6

8x - 5x = 14

x = 14 / 3.

12. Solve for x in 7x - 3 + 3x = 10x - 7:

7x + 3x + 3 = 10x - 7

10x + 3 = 10x - 7

3 = -7 (which has no solution).

13. Solve for x in -10x + 1 - 7x = -17x + 5:

-17x + 1 = -17x + 5

1 = 5 (which has no solution).

14. Solve for y in the equation x + 5y = 20:

5y = 20 - x

y = (20 - x) / 5.

15. A rectangular solid has a base with length 5 cm and width 2 cm. If the volume of the solid is 100 cm3, find the height of the solid.

Volume V = LWH → 100 = 52H → H = 10 cm.

16. Translate the statement "1 less than 15 times a number is 9 times that same number" into an algebraic equation:

15x - 1 = 9x.

17. The sum of three consecutive odd integers is 201. Let the integers be x, x+2, x+4:

x + (x + 2) + (x + 4) = 201;

3x + 6 = 201 → 3x = 195 → x = 65.

18. At 9:00 a.m., a truck leaves traveling west at 35 mi/hr. A second truck leaves two hours later, traveling at 70 mi/hr. When will the second truck catch up?

Let t be the time traveled by second truck, and d = 35(2 + t):

70t = 35(2 + t) → 70t = 70 + 35t → 35t = 70 → t = 2 hours.

19. The base of an isosceles triangle is 1 in. less than the length of one of the equal sides. Let the equal side be x:

Perimeter = 2x + (x - 1) = 20 → 3x - 1 = 20 → 3x = 21 → x = 7.

20. Identify the amount in the statement "318 is 53% of 600":

0.53*600 = 318.

21. Elaine was charged $126 interest for 1 month on a $1800 credit card balance. What was the monthly interest rate?

Interest rate = (126/1800)*100 = 7%.

22. A broach was marked up $150 from cost, which amounts to a 50% increase. Find the original cost:

Let c be the cost. Then 150 = 0.5c → c = 300.

23. Solve the solution set 8x + 3

4x

24. Solve the solution set 5x + 12 > 10x - 8:

-5x > -20 → x

25. An arithmetic student needs an average of 70 or more. She scored 76, 69, and 84 on the exams:

(76 + 69 + 84 + x)/4 ≥ 70 → 229 + x ≥ 280 → x ≥ 51.

26. The length of a rectangle is 2 in. more than twice its width. If the perimeter is 28 in., find the width:

Let width = w, length = 2w + 2. Then:

2(length + width) = 28 → 4w + 4 = 28 → 4w = 24 → w = 6 in.

27. Solve and check: 6x = 4(x - 1):

6x = 4x - 4 → 2x = -4 → x = -2.

Paper For Above Instructions

This paper provides solutions to various algebraic problems ranging from determining if a number is a solution to an equation, solving inequalities, and translating contextual statements into algebraic equations. The problems cover topics like solving for variables, working with percentages, analyzing geometric properties, and understanding mathematical relationships. Each problem is addressed individually, presenting step-by-step workings to confirm the understanding of the algebraic principles involved.

The first few problems dealt with checking solutions by substituting values into the equations to verify their correctness. It starts with straightforward substitutions and evolves into understanding how to manipulate inequalities and literal equations. Special emphasis was placed on sequences when addressing problems involving consecutive integers and arithmetic averages, showcasing the application of algebra in real-life scenarios.

For instance, when calculating the average score required to pass a course, the formula incorporates a student's previous exam scores to establish a benchmark for success in future assessments. This approach is critical to understanding cumulative effects in many mathematical applications.

In the problem regarding the rectangular solid, the derived height was calculated using the formula for volume, showcasing how geometric interpretations can be tied back to algebraic expressions. The examination of perimeters in the context of triangles and rectangles demonstrated how algebra can be used to solve for unknown dimensions based on given conditions. This is especially useful in fields such as architecture and engineering, where measurements must be precise.

Moreover, problems pertaining to interest rates and markup calculations provided insight into financial mathematics, reflecting the essential role of algebra in economics and commerce. Understanding how percentages correlate to real monetary values is vital for budgeting and financial planning.

When it comes to inequalities, the paper navigated through solving linear inequalities and presenting solution sets. Analyzing inequalities enhances critical thinking and lays the foundation for understanding functions and limits, which are applicable in higher-level mathematics.

In summary, this paper elaborates on the resolution of various mathematical problems by adhering to standard procedures while showcasing the logical flow of mathematical reasoning. Each solution is approached systematically to validate the principles of algebra and its applications in diverse scenarios.

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