Question

1. solve the equation \\(\frac{x + 3}{x - 8} = \frac{x + 8}{x + 9}\\).

a. \\(x = -3\\) c. \\(x = -\frac{91}{12}\\)
b. \\(x = \frac{91}{12}\\) d. \\(x = 8, x = -9\\)

1. what is the solution to the rational equation \\(\frac{x - 2}{x + 4} + \frac{x + 1}{x + 6} = \frac{11x + 32}{x^2 + 10x + 24}\\)?

Explanation:

Problem 2

Step1: Cross - multiply the equation

Given the equation \(\frac{x + 3}{x - 8}=\frac{x + 8}{x + 9}\), cross - multiplying (since if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\) for \(b
eq0\) and \(d
eq0\)) gives \((x + 3)(x + 9)=(x + 8)(x - 8)\).

Step2: Expand both sides

Expand the left - hand side: \((x + 3)(x + 9)=x^{2}+9x+3x + 27=x^{2}+12x + 27\).
Expand the right - hand side: \((x + 8)(x - 8)=x^{2}-64\) (using the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\) where \(a = x\) and \(b = 8\)).
So the equation becomes \(x^{2}+12x + 27=x^{2}-64\).

Step3: Simplify the equation

Subtract \(x^{2}\) from both sides of the equation: \(x^{2}-x^{2}+12x+27=x^{2}-x^{2}-64\), which simplifies to \(12x+27=-64\).
Then subtract 27 from both sides: \(12x=-64 - 27=-91\).

Step4: Solve for x

Divide both sides by 12: \(x =-\frac{91}{12}\). We also need to check the domain. The original equation has restrictions \(x
eq8\) and \(x
eq - 9\). When \(x =-\frac{91}{12}\), it does not make the denominators zero, so it is a valid solution.

Step1: Factor the denominator on the right - hand side

First, factor \(x^{2}+10x + 24\). We need two numbers that multiply to 24 and add up to 10. The numbers are 4 and 6. So \(x^{2}+10x + 24=(x + 4)(x + 6)\).
The equation \(\frac{x - 2}{x + 4}+\frac{x + 1}{x + 6}=\frac{11x+32}{(x + 4)(x + 6)}\) has the domain restriction \(x
eq - 4\) and \(x
eq - 6\).

Step2: Multiply through by the common denominator

Multiply each term by \((x + 4)(x + 6)\) to clear the fractions:
\((x - 2)(x + 6)+(x + 1)(x + 4)=11x + 32\)

Step3: Expand the left - hand side

Expand \((x - 2)(x + 6)\): \(x^{2}+6x-2x-12=x^{2}+4x-12\)
Expand \((x + 1)(x + 4)\): \(x^{2}+4x+x + 4=x^{2}+5x + 4\)
Add the two expanded expressions: \((x^{2}+4x-12)+(x^{2}+5x + 4)=2x^{2}+9x-8\)
So the equation becomes \(2x^{2}+9x-8=11x + 32\)

Step4: Rearrange the equation to standard quadratic form

Subtract \(11x\) and 32 from both sides: \(2x^{2}+9x-11x-8 - 32=0\)
Simplify to get \(2x^{2}-2x-40=0\)
Divide through by 2: \(x^{2}-x - 20=0\)

Step5: Factor the quadratic equation

We need two numbers that multiply to - 20 and add up to - 1. The numbers are - 5 and 4. So \(x^{2}-x - 20=(x - 5)(x + 4)=0\)

Step6: Solve for x

Set each factor equal to zero: \(x - 5=0\) or \(x + 4=0\). So \(x = 5\) or \(x=-4\). But \(x=-4\) is not in the domain (it makes the denominator of the original equation zero), so we discard \(x=-4\).

Answer:

C. \(x =-\frac{91}{12}\)

Problem 1