Problem 12
Question
Solve each equation and check each solution. See Examples 1 through 3. $$ \frac{a+5}{4}+\frac{a+5}{2}=\frac{a}{8} $$
Step-by-Step Solution
Verified Answer
The solution is \( a = -6 \).
1Step 1: Find a Common Denominator
The first step in solving the equation \( \frac{a+5}{4} + \frac{a+5}{2} = \frac{a}{8} \) is to find a common denominator for the fractions. The denominators in the equation are 4, 2, and 8. The least common denominator (LCD) for these numbers is 8.
2Step 2: Rewrite Each Fraction with the Common Denominator
Rewrite each term in the equation with the common denominator of 8:- \( \frac{a+5}{4} \) becomes \( \frac{2(a+5)}{8} \) because multiplying the numerator and denominator by 2 gives this.- \( \frac{a+5}{2} \) becomes \( \frac{4(a+5)}{8} \) because multiplying the numerator and denominator by 4 gives this.The equation is now: \[ \frac{2(a+5)}{8} + \frac{4(a+5)}{8} = \frac{a}{8} \]
3Step 3: Combine Like Terms
Now that the fractions have the same denominator, we can combine them:\( \frac{2(a+5) + 4(a+5)}{8} = \frac{a}{8} \)Simplify the numerator:\( 2(a+5) + 4(a+5) = 2a + 10 + 4a + 20 = 6a + 30 \)Thus, we have:\( \frac{6a + 30}{8} = \frac{a}{8} \)
4Step 4: Simplify the Equation
Since the denominators are the same on both sides of the equation, we can equate the numerators:\[ 6a + 30 = a \]
5Step 5: Solve for a
Subtract \( a \) from both sides of the equation:\[ 6a - a + 30 = a - a \] This simplifies to:\[ 5a + 30 = 0 \]Subtract 30 from both sides:\[ 5a = -30 \]Divide both sides by 5:\[ a = -6 \]
6Step 6: Check the Solution
Substitute \( a = -6 \) back into the original equation to verify the solution:\[ \frac{-6+5}{4} + \frac{-6+5}{2} = \frac{-6}{8} \]Simplify each fraction:- \( \frac{-1}{4} \)- \( \frac{-1}{2} \) becomes \( \frac{-2}{4} \)Combine these on the left side:\( \frac{-1-2}{4} = \frac{-3}{4} \)The right side simplifies to \( \frac{-3}{4} \) as well.Both sides of the equation are equal, confirming that \( a = -6 \) is a correct solution.
Key Concepts
Common DenominatorSimplifying FractionsCombining Like TermsChecking Solutions
Common Denominator
When dealing with fractions in an equation, finding a common denominator is vital. This step ensures that each fraction is expressed in terms that can be easily combined. In our example, we have denominators 4, 2, and 8. The smallest number they all divide into without leaving a remainder is 8. This number is our common denominator (or least common denominator). Here's why it's useful:
- It makes addition and subtraction of fractions straightforward.
- Unifies different fractions, making the equation easier to handle.
Simplifying Fractions
After finding the common denominator, the next step is simplifying the fractions, if possible. In our problem, two fractions with different denominators, \( \frac{a+5}{4} ext{ and } \frac{a+5}{2} \) are rewritten with 8 as the denominator. Here's how simplifying works:
- Convert \( \frac{a+5}{4} \) to \( \frac{2(a+5)}{8} \) by multiplying both the numerator and denominator by 2.
- Convert \( \frac{a+5}{2} \) to \( \frac{4(a+5)}{8} \) by multiplying both the numerator and denominator by 4.
Combining Like Terms
Once the fractions have been rewritten with the same denominator, it's time to combine them. The equation still stays true to its form, but now it's more workable. Consider combining the terms:
- Begin with \( \frac{2(a+5) + 4(a+5)}{8} \).
- Simplify the expression in the numerator: distribute and add \( 2(a+5) + 4(a+5) = 2a + 10 + 4a + 20 = 6a + 30 \).
Checking Solutions
After determining a solution like \( a = -6 \), it's crucial to verify it by checking if it satisfies the original equation. This confirms accuracy and correctness of the solution.
- Replace \( a \) in the original equation: \( \frac{-6+5}{4} + \frac{-6+5}{2} = \frac{-6}{8} \).
- Simplify all terms: \( \frac{-1}{4} + \frac{-2}{4} = \frac{-3}{4} \).
- See if both sides equal: \[ \frac{-3}{4} = \frac{-3}{4} \] is true.
Other exercises in this chapter
Problem 12
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Perform each indicated operation. Simplify if possible. \(\frac{5}{y^{2}}-\frac{y}{2 y+1}\)
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