Problem 17 - Prealgebra

Question

Solve equation. Check your solution. \(2 f-6=7 f+24\)

Step-by-Step Solution

Answer

The solution is \(f = -6\), verified by substituting back into the equation.

Step 1: Move Variables to One Side

First, we need to isolate the variable terms on one side of the equation. We can do this by subtracting \(7f\) from both sides:\(2f - 7f - 6 = 7f - 7f + 24\)This simplifies to:\(-5f - 6 = 24\)

Step 2: Move Constant to Other Side

Next, we want to move the constant term (-6) to the other side of the equation. We do this by adding 6 to both sides:\(-5f - 6 + 6 = 24 + 6\)This simplifies to:\(-5f = 30\)

Step 3: Solve for the Variable

Now, we need to solve for \(f\) by dividing both sides of the equation by -5:\(\frac{-5f}{-5} = \frac{30}{-5}\)This gives us:\(f = -6\)

Step 4: Check the Solution

To verify our solution, substitute \(f = -6\) back into the original equation:\(2(-6) - 6 = 7(-6) + 24\)Calculate each side:Left side: \(2(-6) - 6 = -12 - 6 = -18\)Right side: \(7(-6) + 24 = -42 + 24 = -18\)Both sides equal, confirming that \(f = -6\) is correct.

Key Concepts

Solving Equations

To grasp solving linear equations, it's essential to comprehend the primary goal: finding the value of the unknown that makes the equation true. Linear equations are often written in the form \(ax + b = c\), where \(x\) is the variable we need to solve. The process involves different steps to manipulate the equation while maintaining equality. The golden rule is: whatever you do to one side of the equation, do it to the other side as well. This maintains balance, just like a seesaw that's balanced at both ends.

In our exercise, the equation given is \(2f - 6 = 7f + 24\). Our task is to compute the value of \(f\) that makes this equation true. We begin by getting all variable terms on one side and all constants on the other. This sets us on the path of finding the value of the unknown.

• Subtraction is used to balance terms.
• All terms containing the variable \(f\) are gathered on one side.

These operations simplify the equation significantly, reducing the clutter and making it easier to identify what \(f\) must be.The art of solving equations is about following these steps carefully and ensuring the equation stays balanced with each action.

Variable Isolation

Variable isolation is at the core of solving equations. It refers to the process of rearranging the equation so that the variable we wish to solve for is by itself on one side of the equation. Doing this involves shifting other components of the equation to the opposite side.

Consider the step \(-5f - 6 = 24\). Here, our goal is to isolate \(f\). We proceed by adding 6 to both sides, aiming to move the constant term away from the variable. This yields \(-5f = 30\).

With the variable term by itself, the final step is to undo the multiplication of \(-5\) by dividing both sides of the equation. The division of both sides by \(-5\) results in \(f = -6\), confirming that \(f\) is isolated.

• Addition or subtraction helps move constant terms.
• Division or multiplication assists in isolating the variable fully.

Proper isolation is crucial because it allows us to clearly discern the value that satiates the equation, providing an accurate solution.

Checking Solutions

After finding a solution, it's vital to verify its correctness by substituting the solution back into the original equation. Checking solutions is the practice of confirming that both sides of the equation yield the same value when the found solution is inserted.

For our equation, the solution \(f = -6\) is checked by substituting it back into the original equation: \(2(-6) - 6 = 7(-6) + 24\). This involves calculating:

• The left side becomes: \(-12 - 6 = -18\).
• The right side becomes: \(-42 + 24 = -18\).

Both sides equate to \(-18\), confirming that our solution of \(f = -6\) is indeed correct.

Checking solutions prevents errors by ensuring that the proposed value truly balances the equation, reinforcing confidence in our algebraic work.