Problem 3 - Algebra 2 and Trigonometry

Question

In \(3-20,\) solve each equation and check. $$ \frac{1}{4} a+8=\frac{1}{2} a $$

Step-by-Step Solution

Answer

The solution is \(a = 32\). It checks out when substituted back.

Step 1: Eliminate Fractions

To eliminate the fractions, multiply every term in the equation by 4, which is the least common denominator. This gives you: \[ 4 \cdot \left( \frac{1}{4}a \right) + 4 \cdot 8 = 4 \cdot \left( \frac{1}{2}a \right) \]Simplifying, we have:\[ a + 32 = 2a \]

Step 2: Isolate the Variable

Move all terms involving \(a\) to one side of the equation. Subtract \(a\) from both sides:\[ a + 32 - a = 2a - a \]This simplifies to:\[ 32 = a \]

Step 3: Check the Solution

To check the solution, substitute \(a = 32\) back into the original equation:\[ \frac{1}{4}(32) + 8 = \frac{1}{2}(32) \]Simplify both sides:Left: \( \frac{1}{4} \times 32 + 8 = 8 + 8 = 16 \)Right: \( \frac{1}{2} \times 32 = 16 \)Since both sides equal 16, our solution is correct.

Key Concepts

Fractions in Equations

Understanding fractions in equations is crucial, as they can often make solving equations seem more complex. Fractions are expressed as a division of numbers, usually in the form \( \frac{a}{b} \), where \(a\) is the numerator, and \(b\) is the denominator. In equations, fractions can complicate calculations, especially when solving for a variable.
To make things easier, it's helpful to eliminate fractions early on. This involves finding the least common denominator (LCD) of all the fractional terms involved. The LCD is the smallest number that each denominator divides into without leaving a remainder.
Once the LCD is found, multiply every term in the equation by this number. This step effectively removes the fractions, simplifying the equation and making it easier to solve. It's essential to remember to apply this multiplication to all parts of the equation to keep it balanced. For example, in the given equation \( \frac{1}{4}a + 8 = \frac{1}{2}a \), multiplying each term by 4 removes the fractions, resulting in the simpler linear equation \(a + 32 = 2a\).

Solving Linear Equations

Linear equations are equations of the first degree, meaning they have variables raised only to the power of one. They typically take the form \(ax + b = cx + d\). Solving a linear equation involves finding the value of the variable that makes the equation true.
After eliminating any fractions, the next step in solving the equation is to isolate the variable. This often involves moving all terms with the variable on one side of the equation and constants on the other.

• Begin by adding or subtracting terms on both sides to gather variable terms on one side.
• In the example \(a + 32 = 2a\), we subtract \(a\) from both sides. This gives \(32 = a\), effectively isolating \(a\).

By arranging the equation so that only one term contains the variable, it becomes straightforward to determine its value. The simplicity of linear equations is that once the variable is isolated, it leads directly to the solution.

Checking Solutions

Checking solutions is an often overlooked but vital step in solving equations. It confirms that the solution is correct. Once a solution is found, replace the variable in the original equation with this calculated value.
For the equation \( \frac{1}{4}a + 8 = \frac{1}{2}a \) with the solution \(a = 32\):

• Substitute \(32\) for \(a\) back into the original equation.
• This results in \( \frac{1}{4}(32) + 8 = \frac{1}{2}(32) \).
• Simplify both sides: \(8 + 8 = 16\) on the left side, and \(16\) on the right side.

Since both sides equal, the solution \(a = 32\) satisfies the original equation, confirming its correctness.
Always ensure both sides match when checking. If they don't, re-evaluate your steps to find potential errors. This meticulous check prevents mistakes and solidifies understanding.