Problem 40 - Introductory Algebra for College Students

Question

Solve each equation and check your proposed solution in Exercises. Begin your work by rewriting each equation without fractions. $$\frac{z}{5}-\frac{1}{2}=\frac{z}{6}$$

Step-by-Step Solution

Answer

The solution to the equation \(\frac{z}{5}-\frac{1}{2}=\frac{z}{6}\) is \(z = 15\).

Step 1: Rewrite the equation without fractions

To first remove fractions, find a common denominator. In this case, it's 30 (multiple of 5, 2 and 6). Therefore, multiply every term with 30. It would look like \(30*\frac{z}{5} - 30*\frac{1}{2} = 30*\frac{z}{6}\), which simplifies to \(6z - 15 = 5z\).

Step 2: Solve for z

Now, get the variables on one side and constants on the other, to figure out the value of z. It would look like this: \(6z - 5z = 15\), which simplifies to \(z = 15\).

Step 3: Check the solution

Substitute the solution found in step 2 back into the original equation to check if it is correct. Substituting \(z = 15\) into the original equation \(\frac{z}{5}-\frac{1}{2}=\frac{z}{6}\) yields \(\frac{15}{5}-\frac{1}{2}=\frac{15}{6}\), which simplifies to \(3 - 0.5 = 2.5\), confirming that the proposed solution is indeed correct.

Key Concepts

Fractions in Equations

Fractions often appear in algebraic equations, and they can initially look intimidating. However, dealing with them step-by-step simplifies the process. To start, recognize each fraction's components: the numerator and the denominator. The denominator informs us of the number of equal parts the whole is divided into. In an equation such as \( \frac{z}{5}-\frac{1}{2}=\frac{z}{6} \), each term includes a fraction involving variables or constants. When fractions are part of the equation, the main goal is to eliminate them, usually by finding a common denominator. This makes the equation easier to manage and solve by converting it into a simpler, fraction-free form.

Common Denominator

Finding a common denominator is a crucial step in solving equations with fractions. A common denominator is a shared multiple of the denominators of the fractions involved. It allows you to transform each fraction into an equivalent one with the same denominator, making it easier to eliminate fractions from the equation.

In the equation \( \frac{z}{5}-\frac{1}{2}=\frac{z}{6} \), the denominators are 5, 2, and 6. The smallest common multiple of these numbers is 30. By multiplying every term by 30, you can cancel out the denominators. Here's what it looks like:

• Multiplying \( \frac{z}{5} \) by 30 results in \( 6z \).
• Multiplying \( \frac{1}{2} \) by 30 results in \( 15 \).
• Multiplying \( \frac{z}{6} \) by 30 results in \( 5z \).

After this step, you're left with a much simpler equation: \( 6z - 15 = 5z \).

Algebra Concepts

Once you have a fraction-free equation, you can solve it using fundamental algebra concepts. The objective is to isolate the variable on one side of the equation, making it explicit. For instance, with \( 6z - 15 = 5z \), you need to get all the terms involving \( z \) on one side and the constant terms on the other.

To do this:

• Subtract \( 5z \) from both sides, resulting in \( 6z - 5z = 15 \).
• This simplifies to \( z = 15 \).

By applying these basic operations—such as addition, subtraction, multiplication, or division—you rearrange the equation to solve for the unknown variable effectively.

Solution Verification

After solving an equation, it's always important to verify the proposed solution to ensure it's correct. Verification involves substituting the found value back into the original equation to check if both sides equal.

For the equation \( \frac{z}{5}-\frac{1}{2}=\frac{z}{6} \), after determining that \( z = 15 \), substitute 15 back into the original fractions:

• \( \frac{15}{5} = 3 \)
• \( \frac{15}{6} = 2.5 \)
• Thus, \( 3 - 0.5 = 2.5 \)

The left side equals the right side after the substitution, confirming the proposed solution is indeed correct. Solution verification is an essential step—it helps catch any potential errors made during simplification or calculation.