Problem 34 - Intermediate Algebra

Question

Solve each equation. Check each result. See Example 3. $$ -\frac{9}{8} s=3 $$

Step-by-Step Solution

Answer

The solution is \(s = -\frac{8}{3}\).

Step 1: Understand the Problem

The equation is given as \(-\frac{9}{8} s = 3\). The goal is to find the value of \(s\) that satisfies this equation.

Step 2: Isolate the Variable

To solve for \(s\), divide both sides of the equation by \(-\frac{9}{8}\). This will give you: \[ s = \frac{3}{-\frac{9}{8}} \].

Step 3: Simplify the Fraction

To divide by a fraction, multiply by its reciprocal. Therefore, we have: \[ s = 3 \times \frac{8}{-9} \] Multiply: \[ s = \frac{24}{-9} \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 3: \[ s = \frac{8}{-3} \] or simply \[ s = -\frac{8}{3} \].

Step 4: Check the Solution

Substitute \(s = -\frac{8}{3}\) back into the original equation to verify: \[ -\frac{9}{8} \times \left(-\frac{8}{3}\right) = 3 \] Calculate this: \[ \frac{9 \times 8}{8 \times 3} = 3 \] Simplifies to: \[ \frac{72}{24} = 3 \] Which is true, confirming that \(s = -\frac{8}{3}\) is correct.

Key Concepts

Fractions in Algebra

Working with fractions in algebra is a skill that often requires patience and precision. Fractions appear in equations as ratios of numbers.

• A fraction consists of two parts: the numerator (top part) and the denominator (bottom part).
• In algebra, fractions are used to express divisions that cannot be completed into whole numbers.

For example: In the equation \(-\frac{9}{8} s = 3\), \(-\frac{9}{8}\) is a fraction. When solving equations, you often need to manipulate these fractions, either by multiplying, dividing, or simplifying them.
To solve equations with fractions, one helpful strategy is to find the reciprocal. The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\). This is particularly useful for getting rid of fractions by multiplying them by their reciprocal, effectively turning them into the number 1. For instance, when solving the equation, \(-\frac{9}{8}\) becomes \(\frac{8}{-9}\) when finding the reciprocal, allowing us to find \(s\) easily.

Checking Solutions

Checking your work after solving an equation is an important step that help ensures accuracy. It's like double-checking your roadmap to make sure you're still heading in the right direction.

• This step involves substituting the solution back into the original equation to see if the equation holds true.
• If the left and right sides of the equation balance, then your solution is likely correct.

For example, after solving \(s = -\frac{8}{3}\), substitute it back into the original equation. When you replace \(s\) with \(-\frac{8}{3}\), you perform the calculations:
\(-\frac{9}{8} \times \left(-\frac{8}{3}\right)\) simplifies to \(3\), which exactly matches the other side of the equation. This confirms the answer \(s = -\frac{8}{3}\) is correct. This process increases confidence in the correctness of the solution.

Inverse Operations

Inverse operations are fundamental in solving equations. They act as the mathematical undo button, allowing you to isolate the variable and solve for it.

• Common inverse operations include addition/subtraction and multiplication/division.
• They are useful for rearranging equations by performing the opposite operation to isolate the variable.

In the equation \(-\frac{9}{8} s = 3\), we used inverse operations to solve for \(s\). Dividing both sides by \(-\frac{9}{8}\) is how we started, which is essentially multiplying by its reciprocal. This technique cancels out the fraction on one side, turning it into 1, effectively isolating \(s\). Inverse operations simplify complex equations, turning them into straightforward expressions where the variable easily stands alone.