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solve for f. 2 - \\frac{1}{4}f = f - \\frac{3}{4}f + 3 f = \\boxed{}

Question

solve for f.
2 - \frac{1}{4}f = f - \frac{3}{4}f + 3
f = \boxed{}

Explanation:

Step1: Simplify right - hand side

First, combine the \(f\) terms on the right - hand side. We have \(f-\frac{3}{4}f\). Since \(f = \frac{4}{4}f\), then \(\frac{4}{4}f-\frac{3}{4}f=\frac{4 - 3}{4}f=\frac{1}{4}f\). So the equation \(2-\frac{1}{4}f=f-\frac{3}{4}f + 3\) becomes \(2-\frac{1}{4}f=\frac{1}{4}f+3\).

Step2: Move \(f\) terms to one side

Add \(\frac{1}{4}f\) to both sides of the equation. \((2-\frac{1}{4}f)+\frac{1}{4}f=(\frac{1}{4}f + 3)+\frac{1}{4}f\). The left - hand side simplifies to \(2\), and the right - hand side simplifies to \(\frac{1 + 1}{4}f+3=\frac{2}{4}f+3=\frac{1}{2}f+3\). So now we have the equation \(2=\frac{1}{2}f+3\).

Step3: Solve for \(f\)

Subtract \(3\) from both sides: \(2 - 3=\frac{1}{2}f+3 - 3\). The left - hand side is \(- 1\), and the right - hand side is \(\frac{1}{2}f\). Then multiply both sides by \(2\) to solve for \(f\): \(-1\times2=\frac{1}{2}f\times2\). So \(f=-2\).

Answer:

\(-2\)