Question

1. \\(\frac{1}{r + 3} = \frac{r + 4}{r - 2} + \frac{6}{r - 2}\\)

Explanation:

Step1: Combine the right - hand side fractions

Since the two fractions on the right - hand side have the same denominator \(r - 2\), we can add them according to the rule \(\frac{a}{c}+\frac{b}{c}=\frac{a + b}{c}\). So \(\frac{r + 4}{r-2}+\frac{6}{r - 2}=\frac{(r + 4)+6}{r-2}=\frac{r+4 + 6}{r-2}=\frac{r + 10}{r-2}\). The equation becomes \(\frac{1}{r + 3}=\frac{r + 10}{r-2}\).

Step2: Cross - multiply

Cross - multiply the equation \(\frac{1}{r + 3}=\frac{r + 10}{r-2}\) to get rid of the fractions. We know that if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\). So \(1\times(r - 2)=(r + 3)\times(r + 10)\).

Step3: Expand both sides

Expand the right - hand side using the distributive property \((a + b)(c + d)=ac+ad+bc+bd\). So \(r-2=r^{2}+10r+3r + 30\).

Step4: Simplify the equation to standard quadratic form

Simplify the right - hand side: \(r-2=r^{2}+13r + 30\). Then move all terms to one side to get a quadratic equation. Subtract \(r\) and add 2 to both sides: \(0=r^{2}+13r - r+30 + 2\), which simplifies to \(r^{2}+12r+32 = 0\).

Step5: Factor the quadratic equation

We need to find two numbers that multiply to 32 and add up to 12. The numbers are 8 and 4. So \(r^{2}+12r + 32=(r + 8)(r+4)=0\).

Step6: Solve for r

Set each factor equal to zero: \(r+8 = 0\) or \(r + 4=0\). So \(r=-8\) or \(r=-4\).

Step7: Check for extraneous solutions

We need to check if these solutions make the original denominators zero. For the original equation, the denominators are \(r + 3\) and \(r-2\).

• When \(r=-8\): \(r + 3=-8 + 3=-5

eq0\) and \(r-2=-8-2=-10
eq0\).

• When \(r=-4\): \(r + 3=-4 + 3=-1

eq0\) and \(r-2=-4-2=-6
eq0\).

Answer:

\(r=-8\) or \(r = - 4\)