Key Concept: Solving linear equations, transposing terms, fractions
b) 2
[Solution Description]
To solve the equation $\frac{2x}{3} + \frac{1}{2} = \frac{x}{2} + \frac{5}{6}$, we first eliminate the fractions by finding a common denominator.
Step 1: Multiply every term in the equation by $6$ (the least common multiple of $2, 3,$ and $6$) to eliminate the denominators:
$6 \times \left(\frac{2x}{3}\right) + 6 \times \left(\frac{1}{2}\right) = 6 \times \left(\frac{x}{2}\right) + 6 \times \left(\frac{5}{6}\right)$
Calculating each term, we get:
$4x + 3 = 3x + 5$
Step 2: Subtract $3x$ from both sides to get all the $x$ terms on one side:
$4x - 3x + 3 = 3x - 3x + 5$
Simplifying this, we get:
$x + 3 = 5$
Step 3: Subtract $3$ from both sides to solve for $x$:
$x + 3 - 3 = 5 - 3$
Simplifying this, we get:
$x = 2$
Therefore, the solution is $x = 2$.
Your Answer is correct.
b) 2
[Solution Description]
To solve the equation $\frac{2x}{3} + \frac{1}{2} = \frac{x}{2} + \frac{5}{6}$, we first eliminate the fractions by finding a common denominator.
Step 1: Multiply every term in the equation by $6$ (the least common multiple of $2, 3,$ and $6$) to eliminate the denominators:
$6 \times \left(\frac{2x}{3}\right) + 6 \times \left(\frac{1}{2}\right) = 6 \times \left(\frac{x}{2}\right) + 6 \times \left(\frac{5}{6}\right)$
Calculating each term, we get:
$4x + 3 = 3x + 5$
Step 2: Subtract $3x$ from both sides to get all the $x$ terms on one side:
$4x - 3x + 3 = 3x - 3x + 5$
Simplifying this, we get:
$x + 3 = 5$
Step 3: Subtract $3$ from both sides to solve for $x$:
$x + 3 - 3 = 5 - 3$
Simplifying this, we get:
$x = 2$
Therefore, the solution is $x = 2$.