Problem 15 - Prealgebra

Question

Solve equation. Check your solution. \(-s+4=7 s-3\)

Step-by-Step Solution

Answer

The solution is \(s = \frac{7}{8}\).

Step 1: Move all Terms Involving s to One Side

To isolate the variable, we should move all terms involving \(s\) to one side of the equation. We start by adding \(s\) to both sides to get rid of the \(-s\) on the left side. This gives us:\[ 4 = 8s - 3 \]

Step 2: Move Constant Terms to Opposite Side

Now, we need to solve for \(s\) by first moving the constant \(-3\) from the right side to the left side. We add 3 to both sides of the equation:\[ 4 + 3 = 8s \] \[ 7 = 8s \]

Step 3: Solve for s

To isolate \(s\), divide both sides by 8:\[ s = \frac{7}{8} \]

Step 4: Check the Solution

Substitute \(s = \frac{7}{8}\) back into the original equation to verify the solution is correct:Original equation: \(-s + 4 = 7s - 3\) Substitute \(s\): \[-\left(\frac{7}{8}\right) + 4 = 7\left(\frac{7}{8}\right) - 3\]Calculate both sides:Left side: \(-\frac{7}{8} + 4 = \frac{25}{8}\)Right side: \(\frac{49}{8} - 3 = \frac{25}{8}\)Since both sides are equal, the solution \(s = \frac{7}{8}\) is verified.

Key Concepts

Isolating the Variable

Isolating the variable is a fundamental step when solving equations. It means getting the variable, in this case, \(s\), all by itself on one side of the equation. This allows us to find the value of the variable.

In the given equation \(-s+4=7s-3\), we have terms involving \(s\) on both sides. To isolate \(s\), you start by bringing all \(s\)-related terms to one side. Follow these simple steps to effectively isolate variables:

• Add or subtract terms involving \(s\) to both sides. This helps eliminate \(s\) from one side, making the equation simpler.
• When moving terms, always perform the operation to both sides of the equation to keep it balanced.

In our case, adding \(s\) to both sides brings us to \(4 = 8s - 3\). Notice how the left side now consists solely of numbers, while the right focuses on the variable \(s\). This is the essence of isolating variables: simplifying your equation to a form where \(s\) can easily be solved.

Checking Solutions

Once you've found a potential solution, it's crucial to check whether it indeed satisfies the original equation. This ensures that no errors were made during algebraic manipulations. Here's how you can confidently verify your solution:

Insert the value of \(s\) back into the original equation:

• Original equation: \(-s+4 = 7s -3\)
• Substitute \(s = \frac{7}{8}\).

This means you'll plug the value into both sides of the equation and perform the arithmetic:

• For the left side: Compute \(-\left(\frac{7}{8}\right) + 4\).
• For the right side: Compute \(7\left(\frac{7}{8}\right) - 3\).

After solving both sides, make sure they equate. In this instance, both sides of the original equation resolve to \(\frac{25}{8}\). Hence, our solution \(s = \frac{7}{8}\) is indeed correct. Checking solutions confirms the integrity of your work and boosts confidence in your results.

Algebraic Manipulation

Algebraic manipulation involves using operations like addition, subtraction, multiplication, and division to rearrange and simplify equations. This skill is essential when solving equations like our original example.

Here's a breakdown of essential techniques:

• Adding or subtracting the same amount from both sides helps in shifting terms without changing the equality.
• Use multiplication or division for isolating the variable when it is being multiplied or divided by another number.
• Be mindful of performing each operation across the entire equation to maintain balance.

For example, after isolating \(s\) in \(7 = 8s\) by adding all constants to one side, we solve for \(s\) by dividing both sides by 8, resulting in \(s = \frac{7}{8}\). Remember: These manipulations are essential tools for discovering solutions and truths within algebraic expressions. Mastering them is fundamental for progressing in algebra.