Solve for z 3z 5 2z 25 5z – Embark on a mathematical journey to solve for z in the equation 3z + 5 = 2z + 25 + 5z. Join us as we unravel the mysteries of variables, operations, and the art of isolating the unknown.
We will delve into the intricacies of the problem, exploring the mathematical operations at play and the step-by-step process of isolating z. Along the way, we will discover the practical applications of solving for variables and how it empowers us to tackle real-world challenges.
Variable Identification
In mathematics, variables represent unknown values or quantities that can vary or change. They are typically represented by letters, such as x, y, or z. In the expression ‘3z 5 2z 25 5z’, the variable present is z.
Variables are essential in mathematical expressions because they allow us to represent unknown or changing quantities and manipulate them to solve equations or inequalities. For example, in the expression ‘3z 5 2z 25 5z’, we can use the variable zto represent an unknown number and solve for its value by simplifying the expression.
Mathematical Operations
The expression ‘3z 5 2z 25 5z’ involves several mathematical operations. These include addition, subtraction, and multiplication.
The order of operations is crucial in evaluating the expression correctly. According to the PEMDAS rule, parentheses, exponents, multiplication, division, addition, and subtraction are performed in that order.
Multiplication
Multiplication is represented by the symbol ‘×’ or simply by placing two terms adjacent to each other without an operator. In the given expression, there are two multiplication operations:
- 3z × 5 = 15z
- 2z × 25 = 50z
Addition
Addition is represented by the ‘+’ symbol. In the given expression, there are two addition operations:
- 15z + 50z = 65z
- 65z + 5z = 70z
Solving for Z
Now that we’ve identified the variables and operations involved, let’s dive into the steps of solving for ‘z’ in the expression ‘3z 5 2z 25 5z’.
Solving the equation 3z + 5 = 2z + 25 + 5z requires a bit of algebraic manipulation. If you’re looking for a distraction while you work through the steps, you might enjoy reading about Kade, who manages a hotel . Kade’s story is both inspiring and educational, and it will surely give you a new perspective on the hospitality industry.
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Step-by-Step Solution
- Combine like terms:Add or subtract the coefficients of like terms (terms with the same variable and exponent). In this case, we have two terms with ‘z’: 3z and 2z. Combining them gives us 5z.
- Simplify the expression:The expression now becomes ‘5z 5 25’.
- Isolate the variable term:To isolate the variable term (5z), we need to get rid of the constant term (25) on the same side of the equation. We do this by subtracting 25 from both sides.
- Solve for ‘z’:The expression becomes ‘5z =
- 20′. To solve for ‘z’, we divide both sides by 5, which gives us ‘z =
- 4′.
Therefore, the value of ‘z’ in the expression ‘3z 5 2z 25 5z’ is -4.
Solution Verification: Solve For Z 3z 5 2z 25 5z
Once we have found a potential solution for ‘z’, it is essential to verify if it indeed satisfies the original expression. This process ensures the accuracy of our solution and minimizes the chances of errors.
To verify the solution, we substitute the value of ‘z’ back into the original expression and check if both sides of the equation are equal.
Let’s consider our example: 3z + 5 = 2z + 25. Suppose we found a potential solution z = 5. To verify this solution, we substitute z = 5 back into the original expression:
3(5) + 5 = 2(5) + 25
15 + 5 = 10 + 25
20 = 35
Since both sides of the equation do not match, we can conclude that z = 5 is not a valid solution for the given expression.
Real-World Applications
Solving for variables is a fundamental skill used in various real-world scenarios. The process of solving for ‘z’ in the expression ‘3z 5 2z 25 5z’ is a practical example of how variable manipulation can be applied to solve problems.
Practical Problems, Solve for z 3z 5 2z 25 5z
- Balancing Chemical Equations:In chemistry, balancing chemical equations involves adjusting the coefficients of reactants and products to ensure that the number of atoms of each element is the same on both sides of the equation. Solving for the coefficients (variables) allows for the correct representation of chemical reactions.
- Motion Analysis:In physics, solving for variables like velocity, acceleration, and time is crucial in analyzing motion. By manipulating equations like v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time, physicists can determine unknown values and understand the motion of objects.
- Financial Planning:In finance, solving for variables like interest rates, loan amounts, and repayment periods is essential for budgeting, investment planning, and debt management. By manipulating equations like I = P – r – t, where I is interest, P is principal, r is interest rate, and t is time, individuals can make informed financial decisions.
- Construction and Engineering:In construction and engineering, solving for variables like dimensions, angles, and forces is vital for designing and building structures. By manipulating equations like F = ma, where F is force, m is mass, and a is acceleration, engineers can ensure the stability and safety of buildings and infrastructure.
Top FAQs
What is the first step in solving for z?
Combine like terms on both sides of the equation.
How do I isolate z on one side of the equation?
Add or subtract the same value from both sides of the equation to move z to one side and the constant to the other.
What is the importance of checking my solution?
Verifying your solution ensures that you have solved the equation correctly and that your answer is valid.
