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solve for f. \\(\\frac{f - 8}{f - 10} = \\frac{f + 7}{f + 10}\\)

Question

solve for f.
\\(\frac{f - 8}{f - 10} = \frac{f + 7}{f + 10}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{f - 8}{f - 10}=\frac{f + 7}{f + 10}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=c\times b\). So we get \((f - 8)(f + 10)=(f + 7)(f - 10)\).

Step2: Expand both sides

Expand the left - hand side using the FOIL method: \((f - 8)(f + 10)=f\times f+f\times10-8\times f - 8\times10=f^{2}+10f - 8f-80=f^{2}+2f - 80\).
Expand the right - hand side using the FOIL method: \((f + 7)(f - 10)=f\times f+f\times(- 10)+7\times f+7\times(-10)=f^{2}-10f + 7f-70=f^{2}-3f - 70\).

Step3: Subtract \(f^{2}\) from both sides

Subtract \(f^{2}\) from both sides of the equation \(f^{2}+2f - 80=f^{2}-3f - 70\). We have \(f^{2}+2f - 80-f^{2}=f^{2}-3f - 70 - f^{2}\), which simplifies to \(2f-80=-3f - 70\).

Step4: Add \(3f\) to both sides

Add \(3f\) to both sides of the equation \(2f-80=-3f - 70\). So \(2f + 3f-80=-3f+3f - 70\), which gives \(5f-80=-70\).

Step5: Add 80 to both sides

Add 80 to both sides of the equation \(5f-80=-70\). We get \(5f-80 + 80=-70 + 80\), so \(5f = 10\).

Step6: Divide both sides by 5

Divide both sides of the equation \(5f = 10\) by 5. \(\frac{5f}{5}=\frac{10}{5}\), which gives \(f = 2\).

We need to check for extraneous solutions. When \(f = 2\), \(f-10=2 - 10=-8
eq0\) and \(f + 10=2+10 = 12
eq0\), so \(f = 2\) is a valid solution.

Answer:

\(f = 2\)