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Solve using the addition principle. Don't forget to perform a check. \( x-\frac{3}{4}=\frac{5}{6} \)

Ask by Garrett Moran. in the United States
Feb 26,2025

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Answer

\( x = \frac{19}{12} \)

Solution

We are given the equation   x – (3/4) = 5/6. Step 1: Use the addition principle by adding 3/4 to both sides of the equation:   (x – 3/4) + 3/4 = 5/6 + 3/4. Step 2: Simplify both sides. On the left side, the –3/4 and +3/4 cancel:   x = 5/6 + 3/4. Step 3: To add the fractions on the right side, find a common denominator. The least common denominator (LCD) of 6 and 4 is 12:   5/6 = (5×2)/(6×2) = 10/12,   3/4 = (3×3)/(4×3) = 9/12. So, we have:   x = 10/12 + 9/12 = (10 + 9)/12 = 19/12. Step 4: Check the solution by substituting x = 19/12 back into the original equation:   x – 3/4 = 19/12 – 3/4. Convert 3/4 to a fraction with denominator 12:   3/4 = 9/12. Thus:   19/12 – 9/12 = 10/12 = 5/6. Since the left-hand side equals the right-hand side of the original equation, the solution is correct. Final Answer:   x = 19/12.

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To solve the equation \( x - \frac{3}{4} = \frac{5}{6} \) using the addition principle, you need to isolate \( x \). Start by adding \(\frac{3}{4}\) to both sides of the equation: \[ x - \frac{3}{4} + \frac{3}{4} = \frac{5}{6} + \frac{3}{4} \] This simplifies to: \[ x = \frac{5}{6} + \frac{3}{4} \] To add \(\frac{5}{6}\) and \(\frac{3}{4}\), you need a common denominator. The least common multiple of 6 and 4 is 12, so convert each fraction: - \(\frac{5}{6} = \frac{10}{12}\) - \(\frac{3}{4} = \frac{9}{12}\) Now add them together: \[ x = \frac{10}{12} + \frac{9}{12} = \frac{19}{12} \] Thus, the solution is: \[ x = \frac{19}{12} \] Now, perform a check by substituting \( x \) back into the original equation: \[ \frac{19}{12} - \frac{3}{4} \] Convert \(\frac{3}{4}\) to have a denominator of 12: \[ \frac{3}{4} = \frac{9}{12} \] So, \[ \frac{19}{12} - \frac{9}{12} = \frac{10}{12} = \frac{5}{6} \] The left-hand side matches the right-hand side of the original equation, confirming that: \[ x = \frac{19}{12} \] is indeed the correct solution!

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Algebra Mexico Feb 26, 2025
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