Problem 33 - Introductory Algebra for College Students

Question

Solve each equation using the addition property of equality. Be sure to check your proposed solutions. $$x+\frac{3}{4}=-\frac{9}{2}$$

Step-by-Step Solution

Answer

Based on the solution steps, the solution to the equation \(x+\frac{3}{4}=-\frac{9}{2}\) is \(x=-6.25\).

Step 1: Isolate the Variable

To isolate the variable \(x\), subtract \(\frac{3}{4}\) from both sides of the equation: \(x+\frac{3}{4}-\frac{3}{4}=-\frac{9}{2}-\frac{3}{4}\)

Step 2: Simplify Both Sides

This simplifies to: \(x=-\frac{9}{2}-\frac{3}{4}\). Now, convert \(\frac{3}{4}\) to equivalent fractions of \(\frac{9}{2}\) to simplify the process. You could convert \(\frac{3}{4}\) to \(\frac{3.5}{2}\) to match the denominator of \(-\frac{9}{2}\). Hence, \(x=-\frac{9}{2}-\frac{3.5}{2}\)

Step 3: Perform the Subtraction

Subtracting these values results in: \(x=-\frac{12.5}{2}\)

Step 4: Simplify the Right Hand Side

Simplify \(x=-\frac{12.5}{2}\) to: \(x=-6.25\)

Step 5: Validate the Solution

Substitute the solution, -6.25, back into the original equation to validate it: if \(-6.25+\frac{3}{4}=-\frac{9}{2}\), then the solution is valid.

Key Concepts

Addition Property of Equality

Understanding the addition property of equality is crucial when learning to solve linear equations. It states that if the same amount is added to both sides of an equation, the equality is maintained. This is because equations represent a balance, and altering this balance equally does not change the truth of the equation.

For example, if you have an equation like \(x + 3 = 7\), you can subtract 3 from both sides to keep the balance and isolate \(x\). This gives you \(x + 3 - 3 = 7 - 3\), simplifying to \(x = 4\). This fundamental principle allows you to manipulate equations to find the value of the unknown variable while ensuring that the solutions are accurate. Always remember when using this property, whatever operation you perform on one side of the equation must be mirrored on the other side to maintain the balance.

Isolate the Variable

Isolating the variable is a technique used in algebra to find the value of the unknown in an equation. To isolate a variable means to get the variable alone on one side of the equation so that the other side contains all the constants or known values. The goal of this process is to make the value of the variable clear and straightforward.

Take the equation from the exercise \(x + \frac{3}{4} = -\frac{9}{2}\). The first step is indeed to isolate the variable \(x\). By subtracting \(\frac{3}{4}\) from both sides of the equation, you remove the fraction from the left side, giving \(x\) its own side of the equation. This is a crucial step toward finding the value of \(x\). After subtraction, the equation simplifies to just one variable on one side; the rest of the equation-solving process involves dealing with numbers to find the exact value of \(x\).

Equivalent Fractions

Equivalent fractions are different fractions that represent the same part of a whole. They're incredibly useful in algebra because they allow us to add, subtract, multiply, or divide fractions by ensuring we work with a common denominator. When the denominators are the same, it's easy to perform these operations on the numerators directly, without complex fraction manipulation.

In the context of our example, converting \(\frac{3}{4}\) into a fraction with a denominator of 2 makes it much simpler to perform the subtraction from \(-\frac{9}{2}\). However, there was an inaccuracy in the provided solution; the equivalent fraction for \(\frac{3}{4}\) with denominator 2 is \(\frac{6}{8}\), which simplifies to \(\frac{3}{4}\). The correct conversion should be \(\frac{3}{4}\) to \(\frac{6}{8} = \frac{3 \times 2}{4 \times 2} = \frac{3 \times 2}{8} = \frac{6}{8}\), which is the same as \(\frac{3}{4}\) but with a denominator of 8. This fraction can then be multiplied by 2/2 to get the denominator to 2, resulting in an equivalent fraction of \(\frac{12}{8}\) which simplifies to \(\frac{3}{2}\), and finally, the operation can be carried out with common denominators to find the value of \(x\).