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solve for f. $-\frac{5}{2}(4 - f) = 3 + \frac{1}{3}(6 + f) - 2$

Question

solve for f.
$-\frac{5}{2}(4 - f) = 3 + \frac{1}{3}(6 + f) - 2$

Explanation:

Step1: Simplify both sides

Left side: Distribute $-\frac{5}{2}$: $-\frac{5}{2}\times4 - (-\frac{5}{2})f = -10 + \frac{5}{2}f$
Right side: Simplify constants and distribute $\frac{1}{3}$: $3 - 2 + \frac{1}{3}\times6 + \frac{1}{3}f = 1 + 2 + \frac{1}{3}f = 3 + \frac{1}{3}f$
Equation becomes: $-10 + \frac{5}{2}f = 3 + \frac{1}{3}f$

Step2: Subtract $\frac{1}{3}f$ from both sides

$-10 + \frac{5}{2}f - \frac{1}{3}f = 3$
Find common denominator for $f$ terms: $\frac{15}{6}f - \frac{2}{6}f = \frac{13}{6}f$
Equation: $-10 + \frac{13}{6}f = 3$

Step3: Add 10 to both sides

$\frac{13}{6}f = 3 + 10 = 13$

Step4: Solve for $f$

Multiply both sides by $\frac{6}{13}$: $f = 13\times\frac{6}{13} = 6$

Answer:

$f = 6$