Problem 31 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve equation and check your proposed solution. Begin your work by rewriting each equation without fractions. \(\frac{x}{5}-4=-6\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = -10\).

Step 1: Eliminate the Fraction

First, we want to rewrite the equation without the fraction. The denominator of the fraction is '+5+', so we can multiply the whole equation by '5' to get rid of the fraction. So, we get \(5(\frac{x}{5}) - 5*4 = 5*(-6)\), which simplifies to \(x - 20 = -30\)

Step 2: Solve for '+x+'

Next, the goal is to isolate '+x+' on one side of the equation. For that, we should add '+20+' to both sides of the equation. This results in \(x = -30 + 20\), which further simplifies to \(x = -10\)

Step 3: Validate the Solution

Finally, we want to validate the obtained solution by substituting '+x+' with '-10' in the original equation, which is \(\frac{x}{5}-4=-6\). After substitution, we get \(\frac{-10}{5}-4=-6\), which further simplifies to \(-2-4=-6\). As this equation holds true, our solution is correct.

Key Concepts

Eliminating Fractions in Equations

When solving equations in algebra, fractions can often complicate the process. To simplify the operation, one effective method is to eliminate fractions altogether before proceeding with solving the equation.

Take for example the equation \(\frac{x}{5}-4=-6\). Our goal is to clear the fraction to make the equation easier to solve. We do this by finding the least common denominator (LCD) of all fractions present which, in this case, is 5. We then multiply every term in the equation by this number to cancel out the denominator. Here's how that looks: multiply both sides by 5 to get \(5(\frac{x}{5}) - 5\cdot4 = 5\cdot(-6)\), which simplifies to \(x - 20 = -30\).

By eliminating the fraction early on, we avoid potential mistakes when dealing with decimal or fractional answers and streamline the solution process. Remember, always find the LCD and use it to clear the fractions from your equation.

Isolating Variables in Algebra

Once fractions are out of the way, the next step in solving linear equations is to isolate the variable. This means getting the variable on one side of the equation and everything else on the other.

Continuing with our example, we now have \(x - 20 = -30\). To isolate \(x\), we need to eliminate any numbers attached to it. In this case, we add 20 to both sides of the equation to cancel the -20 that's being subtracted from \(x\). This gives us \(x = -30 + 20\), which simplifies further to \(x = -10\).

The goal is always to get the variable alone and clearly defined. When doing this, remember to perform the inverse operation to both sides of the equation to maintain equality. Addition cancels subtraction, multiplication cancels division, and vice versa.

Validating Algebraic Solutions

The final and often overlooked step is to validate your solution. Just because we've solved for the variable doesn't guarantee its accuracy; we must check our solution to ensure it satisfies the original equation.

In our exercise, we found that \(x = -10\). To validate, we substitute \(x\) back into the original equation: \(\frac{-10}{5}-4=-6\). When simplified, this results in \(\frac{-10}{5}-4=-6\), which further simplifies to \(\frac{-10}{5}-4=-2-4=-6\). Since both sides equal -6, our solution is validated.

Whenever you arrive at a solution, always plug it back into the original equation to ensure you arrive at a true statement. Validation is the key to confirming that you have solved the equation correctly, and it helps avoid errors that may have occurred during the solving process.