Problem 57 - Elementary Algebra

Question

For exercises 43-58, (a) solve. (b) check. $$ \frac{d+1}{3}=\frac{d-3}{6} $$

Step-by-Step Solution

Answer

The solution is \(d = -5\). Verification confirms it is correct.

Step 1 - Eliminate the Fractions

Multiply both sides of the equation by 6 to eliminate the fractions. This is done because 6 is the least common multiple (LCM) of 3 and 6: \[6 \times \frac{d+1}{3} = 6 \times \frac{d-3}{6}\]

Step 2 - Simplify the Equation

Simplify both sides of the equation after multiplying: \[2(d+1) = d-3\]This is obtained because \(6 \times \frac{d+1}{3} = 2(d+1)\) and \(6 \times \frac{d-3}{6} = d-3\).

Step 3 - Distribute and Isolate d

Distribute the 2 on the left side and then isolate the variable d: \[2d + 2 = d - 3\]Subtract d from both sides: \[d + 2 = -3\]Subtract 2 from both sides: \[d = -5\]

Step 4 - Check the Solution

Substitute \(d = -5\) back into the original equation to verify: \[\frac{-5+1}{3} = \frac{-5-3}{6}\]Simplify both sides: \[\frac{-4}{3} = \frac{-8}{6}\]Since \(\frac{-8}{6}\) simplifies to \(\frac{-4}{3}\), the solution \(d = -5\) is correct.

Key Concepts

Least Common Multiple (LCM)

When solving equations with fractions, finding the Least Common Multiple (LCM) is essential. Here, the LCM helps us eliminate fractions to simplify the equation. In the problem \(\frac{d+1}{3}=\frac{d-3}{6}\), the denominators are 3 and 6. The LCM of 3 and 6 is 6.

To clear the fractions, we multiply both sides by 6.

This gives us:
\[6 \times \frac{d+1}{3} = 6 \times \frac{d-3}{6}\]
By multiplying, we remove the denominators, making the equation easier to solve.

Isolating Variables

Isolating variables means rearranging the equation to get the unknown variable on one side. After clearing fractions, we have the simplified equation:

\[2(d+1) = d-3\]

We distribute the 2 to remove parentheses:
\[2d + 2 = d - 3\]

Next, we need to get all terms with the variable d on one side. Subtract d from both sides to keep d on one side:
\[2d + 2 - d = -3\]
\[d + 2 = -3\]

Finally, subtract 2 from both sides to get d alone:
\[d + 2 - 2 = -3 - 2\]
\[d = -5\]

Fraction Elimination

Eliminating fractions makes equations simpler. Multiplying both sides of an equation by the LCM removes fractions. In our example:

\(\frac{d+1}{3} = \frac{d-3}{6}\)

We chose 6 (the LCM) to multiply both sides:
\[6 \times \frac{d+1}{3} = 6 \times \frac{d-3}{6}\]

This multiplication gives:
\[2(d+1) = d-3\]

By clearing fractions, the equation simplifies to whole numbers, making it easier to solve for d.

Verifying Solutions

Verification ensures the solution is correct. Substitute the solution back into the original equation and check both sides. For \(d=-5\), put it in the original equation:

\(\frac{-5+1}{3} = \frac{-5-3}{6}\)

Simplifying both sides gives:
\(\frac{-4}{3} = \frac{-8}{6}\)

Since \(\frac{-8}{6}\) simplifies to \(\frac{-4}{3}\), our solution \(d=-5\) is correct.

Always check your solution to confirm it satisfies the original equation.