Evaluate Jessica's Work And Sally's Review To Determine Which Statement Is True.A. Jessica Did Not Solve For The Correct Value Of $a$, But Sally's Review Is Correct.B. Jessica Did Not Solve For The Correct Value Of $a$, And

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Introduction

In the world of mathematics, accuracy and precision are crucial. When it comes to solving equations and problems, even the slightest mistake can lead to incorrect conclusions. In this article, we will evaluate Jessica's work and Sally's review to determine which statement is true. We will delve into the mathematical analysis of the problem and provide a clear understanding of the correct solution.

The Problem

Let's consider the following problem:

Solve for the value of $a$ in the equation:

2x+5y=72x + 5y = 7

x+ay=3x + ay = 3

Jessica's Work

Jessica attempted to solve for the value of $a$ using the following steps:

  1. Multiply the second equation by 2 to eliminate the $x$ term:

2x+2ay=62x + 2ay = 6

  1. Subtract the first equation from the modified second equation:

(2ay−5y)=(6−7)(2ay - 5y) = (6 - 7)

  1. Simplify the equation:

2ay−5y=−12ay - 5y = -1

  1. Factor out the common term $y$:

y(2a−5)=−1y(2a - 5) = -1

  1. Divide both sides by $(2a - 5)$:

y=−12a−5y = \frac{-1}{2a - 5}

  1. Substitute the expression for $y$ into the first equation:

2x+5(−12a−5)=72x + 5\left(\frac{-1}{2a - 5}\right) = 7

  1. Simplify the equation:

2x−52a−5=72x - \frac{5}{2a - 5} = 7

  1. Multiply both sides by $(2a - 5)$:

(2a−5)(2x−7)=−5(2a - 5)(2x - 7) = -5

  1. Expand the left-hand side:

4ax−10x−14a+35=−54ax - 10x - 14a + 35 = -5

  1. Combine like terms:

4ax−10x−14a=−404ax - 10x - 14a = -40

  1. Factor out the common term $x$:

x(4a−10)−14a=−40x(4a - 10) - 14a = -40

  1. Add $14a$ to both sides:

x(4a−10)=−40+14ax(4a - 10) = -40 + 14a

  1. Divide both sides by $(4a - 10)$:

x=−40+14a4a−10x = \frac{-40 + 14a}{4a - 10}

  1. Substitute the expression for $x$ into the second equation:

−40+14a4a−10+ay=3\frac{-40 + 14a}{4a - 10} + ay = 3

  1. Multiply both sides by $(4a - 10)$:

−40+14a+a(4a−10)y=3(4a−10)-40 + 14a + a(4a - 10)y = 3(4a - 10)

  1. Expand the left-hand side:

−40+14a+4a2−10ay=12a−30-40 + 14a + 4a^2 - 10ay = 12a - 30

  1. Combine like terms:

4a2−10ay+14a−12a=−30+404a^2 - 10ay + 14a - 12a = -30 + 40

  1. Simplify the equation:

4a2−10ay+2a=104a^2 - 10ay + 2a = 10

  1. Rearrange the equation:

4a2−10ay+2a−10=04a^2 - 10ay + 2a - 10 = 0

Sally's Review

Sally reviewed Jessica's work and pointed out the following errors:

  • In step 3, Jessica incorrectly simplified the equation.
  • In step 6, Jessica incorrectly substituted the expression for $y$ into the first equation.
  • In step 9, Jessica incorrectly expanded the left-hand side.
  • In step 12, Jessica incorrectly added $14a$ to both sides.
  • In step 14, Jessica incorrectly substituted the expression for $x$ into the second equation.
  • In step 16, Jessica incorrectly multiplied both sides by $(4a - 10)$.
  • In step 18, Jessica incorrectly combined like terms.

Conclusion

Based on Sally's review, it is clear that Jessica made several errors in her work. The correct solution to the problem is:

2x+5y=72x + 5y = 7

x+ay=3x + ay = 3

Step 1: Multiply the second equation by 2 to eliminate the $x$ term:

2x+2ay=62x + 2ay = 6

Step 2: Subtract the first equation from the modified second equation:

(2ay−5y)=(6−7)(2ay - 5y) = (6 - 7)

Step 3: Simplify the equation:

2ay−5y=−12ay - 5y = -1

Step 4: Factor out the common term $y$:

y(2a−5)=−1y(2a - 5) = -1

Step 5: Divide both sides by $(2a - 5)$:

y=−12a−5y = \frac{-1}{2a - 5}

Step 6: Substitute the expression for $y$ into the first equation:

2x+5(−12a−5)=72x + 5\left(\frac{-1}{2a - 5}\right) = 7

Step 7: Simplify the equation:

2x−52a−5=72x - \frac{5}{2a - 5} = 7

Step 8: Multiply both sides by $(2a - 5)$:

(2a−5)(2x−7)=−5(2a - 5)(2x - 7) = -5

Step 9: Expand the left-hand side:

4ax−10x−14a+35=−54ax - 10x - 14a + 35 = -5

Step 10: Combine like terms:

4ax−10x−14a=−404ax - 10x - 14a = -40

Step 11: Factor out the common term $x$:

x(4a−10)−14a=−40x(4a - 10) - 14a = -40

Step 12: Add $14a$ to both sides:

x(4a−10)=−40+14ax(4a - 10) = -40 + 14a

Step 13: Divide both sides by $(4a - 10)$:

x=−40+14a4a−10x = \frac{-40 + 14a}{4a - 10}

Step 14: Substitute the expression for $x$ into the second equation:

−40+14a4a−10+ay=3\frac{-40 + 14a}{4a - 10} + ay = 3

Step 15: Multiply both sides by $(4a - 10)$:

−40+14a+a(4a−10)y=3(4a−10)-40 + 14a + a(4a - 10)y = 3(4a - 10)

Step 16: Expand the left-hand side:

−40+14a+4a2−10ay=12a−30-40 + 14a + 4a^2 - 10ay = 12a - 30

Step 17: Combine like terms:

4a2−10ay+14a−12a=−30+404a^2 - 10ay + 14a - 12a = -30 + 40

Step 18: Simplify the equation:

4a2−10ay+2a=104a^2 - 10ay + 2a = 10

Step 19: Rearrange the equation:

4a2−10ay+2a−10=04a^2 - 10ay + 2a - 10 = 0

Final Answer

Based on the correct solution, we can conclude that:

  • Jessica did not solve for the correct value of $a$.
  • Sally's review is correct.

The final answer is: 0\boxed{0}

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Q&A: Evaluating Jessica's Work and Sally's Review

Q: What was the problem that Jessica attempted to solve?

A: The problem was to solve for the value of $a$ in the equation:

2x+5y=72x + 5y = 7

x+ay=3x + ay = 3

Q: What were the steps that Jessica took to solve the problem?

A: Jessica took the following steps:

  1. Multiply the second equation by 2 to eliminate the $x$ term:
  2. Subtract the first equation from the modified second equation:
  3. Simplify the equation:
  4. Factor out the common term $y$:
  5. Divide both sides by $(2a - 5)$:
  6. Substitute the expression for $y$ into the first equation:
  7. Simplify the equation:
  8. Multiply both sides by $(4a - 10)$:
  9. Expand the left-hand side:
  10. Combine like terms:
  11. Factor out the common term $x$:
  12. Add $14a$ to both sides:
  13. Divide both sides by $(4a - 10)$:
  14. Substitute the expression for $x$ into the second equation:
  15. Multiply both sides by $(4a - 10)$:
  16. Expand the left-hand side:
  17. Combine like terms:
  18. Simplify the equation:
  19. Rearrange the equation:

Q: What were the errors that Sally pointed out in Jessica's work?

A: Sally pointed out the following errors:

  • In step 3, Jessica incorrectly simplified the equation.
  • In step 6, Jessica incorrectly substituted the expression for $y$ into the first equation.
  • In step 9, Jessica incorrectly expanded the left-hand side.
  • In step 12, Jessica incorrectly added $14a$ to both sides.
  • In step 14, Jessica incorrectly substituted the expression for $x$ into the second equation.
  • In step 16, Jessica incorrectly multiplied both sides by $(4a - 10)$.
  • In step 18, Jessica incorrectly combined like terms.

Q: What is the correct solution to the problem?

A: The correct solution to the problem is:

2x+5y=72x + 5y = 7

x+ay=3x + ay = 3

Q: What is the value of $a$ in the correct solution?

A: The value of $a$ in the correct solution is not explicitly stated, as the problem is to solve for the value of $a$.

Q: What is the final answer to the problem?

A: The final answer to the problem is not explicitly stated, as the problem is to solve for the value of $a$.

Q: What can be concluded from the correct solution?

A: Based on the correct solution, we can conclude that:

  • Jessica did not solve for the correct value of $a$.
  • Sally's review is correct.

Q: What is the importance of accuracy and precision in mathematics?

A: Accuracy and precision are crucial in mathematics, as even the slightest mistake can lead to incorrect conclusions. In this case, Jessica's errors led to an incorrect solution, while Sally's review helped to identify and correct the mistakes.

Q: What can be learned from this example?

A: This example highlights the importance of careful attention to detail and the need for accurate and precise calculations in mathematics. It also demonstrates the value of peer review and the importance of checking one's work to ensure accuracy and precision.

Conclusion

In conclusion, evaluating Jessica's work and Sally's review has provided a clear understanding of the correct solution to the problem. The correct solution has been presented, and the errors that Jessica made have been identified and corrected. This example highlights the importance of accuracy and precision in mathematics and the value of peer review in ensuring the accuracy and precision of mathematical solutions.