Equations With Solution Z=5 A Step-by-Step Guide

In mathematics, solving equations is a fundamental skill. When we are given an equation, we often need to find the value(s) of the variable(s) that make the equation true. These values are called solutions to the equation. In this article, we will explore how to determine if a given value is a solution to an equation. Specifically, we will focus on the case where z=5 and identify which equations from a given set have z=5 as a solution. Understanding how to substitute values into equations and verify solutions is crucial for success in algebra and beyond. It's a foundational concept that underpins more complex problem-solving strategies. By mastering this skill, you'll be well-equipped to tackle a wide range of mathematical challenges. The process involves substituting the given value for the variable in the equation and then simplifying both sides to check if they are equal. If the sides are equal after the substitution, the given value is a solution. If not, it is not a solution. This straightforward approach is applicable to various types of equations, including linear equations, quadratic equations, and more. Let’s delve into the specifics of how to apply this method and solve the given problem.

Problem Statement

We are presented with a set of equations and asked to select the ones where z=5 is a solution. This means we need to substitute 5 for z in each equation and check if the equation holds true. This problem reinforces the core concept of what it means for a value to be a solution of an equation – it must satisfy the equation. Each equation presents a unique context, and verifying the solution involves basic arithmetic operations. Therefore, our task is to meticulously examine each equation and perform the necessary calculations to ascertain whether z=5 fits the criteria. This exercise not only tests our ability to perform substitutions and arithmetic but also enhances our understanding of equation solving in general. Let's proceed by individually analyzing each equation to determine if z=5 is indeed a valid solution.

Analyzing Equation A: 8 = z + 3

The first equation we'll analyze is 8 = z + 3. To determine if z=5 is a solution, we substitute 5 for z in the equation. This gives us 8 = 5 + 3. Now, we simplify the right side of the equation: 5 + 3 = 8. So the equation becomes 8 = 8. Since this statement is true, z=5 is a solution for equation A. This process demonstrates the direct application of substitution to verify a solution. The simplicity of this equation allows us to clearly see the fundamental principle at play. We replaced the variable with the potential solution and checked if the resulting equation holds. The fact that both sides of the equation are equal after the substitution confirms that z=5 is indeed a solution. This type of verification is a cornerstone of algebra, enabling us to validate the correctness of our solutions. Let's move on to the next equation and apply the same methodical approach.

Analyzing Equation B: z - 1 = 6

Next, we consider the equation z - 1 = 6. Again, we substitute z=5 into the equation, resulting in 5 - 1 = 6. Simplifying the left side, we find 5 - 1 = 4. Therefore, the equation becomes 4 = 6. This statement is false, indicating that z=5 is not a solution for equation B. This example highlights the importance of accurate arithmetic. A small error in calculation can lead to an incorrect conclusion about whether a value is a solution. In this case, the difference between the two sides of the equation is evident, clearly demonstrating that substituting z=5 does not satisfy the equation. Consequently, we can confidently rule out z=5 as a solution for this equation. The process of elimination is a crucial strategy in problem-solving, allowing us to narrow down the possibilities and focus on the correct answers. Now, let’s move on to the next equation and continue our analysis.

Analyzing Equation C: 20 = 40 ÷ z

Now, let's examine equation C: 20 = 40 ÷ z. Substituting z=5, we get 20 = 40 ÷ 5. We simplify the right side by performing the division: 40 ÷ 5 = 8. Thus, the equation becomes 20 = 8. This statement is false, meaning z=5 is not a solution for equation C. This example introduces the operation of division, adding another layer to the arithmetic involved in verifying solutions. It’s important to remember the order of operations (PEMDAS/BODMAS) when simplifying equations. In this case, the division must be performed before any other operation. The result of the division clearly shows that the two sides of the equation are unequal when z=5, confirming that it is not a solution. This reinforces the understanding that a solution must make the equation true, which is not the case here. Moving forward, we continue to apply the same systematic approach to each equation, building our confidence in the process.

Analyzing Equation D: 12z = 60

Moving on to equation D, 12z = 60, we substitute z=5. This gives us 12 * 5 = 60. Multiplying 12 by 5, we get 60. Therefore, the equation becomes 60 = 60. This statement is true, indicating that z=5 is a solution for equation D. This equation involves multiplication, another fundamental arithmetic operation. The substitution and subsequent multiplication clearly demonstrate that both sides of the equation are equal when z=5, validating it as a solution. Equations of this form, where a variable is multiplied by a constant, are common in algebra, and understanding how to solve them is crucial. This example reinforces the concept that a solution is a value that satisfies the equation, making it a true statement. We are steadily building a repertoire of examples, each illustrating a different aspect of solution verification. Let’s proceed to the final equation and complete our analysis.

Analyzing Equation E: 12 + z = 17

Finally, we analyze equation E: 12 + z = 17. Substituting z=5, we get 12 + 5 = 17. Adding 12 and 5, we obtain 17. Therefore, the equation becomes 17 = 17. This statement is true, confirming that z=5 is a solution for equation E. This equation involves addition, another basic arithmetic operation. The result of the addition confirms that both sides of the equation are equal when z=5, thus validating it as a solution. This final example reinforces the concept of a solution as a value that makes the equation true. By systematically analyzing each equation, we have gained a comprehensive understanding of how to verify solutions. The consistency of our approach ensures accuracy and builds confidence in our problem-solving abilities. Now that we have examined all the equations, we can confidently identify the ones where z=5 is a solution.

Conclusion

In conclusion, we have successfully analyzed each of the given equations to determine whether z=5 is a solution. By substituting z=5 into each equation and simplifying, we found that the following equations hold true: 8 = z + 3, 12z = 60, and 12 + z = 17. Therefore, equations A, D, and E are the ones where z=5 is a solution. This exercise has reinforced the importance of substitution and simplification in verifying solutions to equations. The methodical approach we employed, analyzing each equation individually, is a valuable strategy for solving mathematical problems. It minimizes errors and ensures that we arrive at the correct answer. Understanding how to verify solutions is a fundamental skill in algebra and is essential for tackling more complex mathematical problems. By mastering this skill, students can confidently solve equations and gain a deeper understanding of mathematical concepts. The ability to accurately verify solutions not only enhances problem-solving skills but also builds a solid foundation for future mathematical endeavors.