Question

solve the rational equation by following the instructions: (\frac{y - 4}{y + 5}=\frac{2}{y + 1}+\frac{-5y - 13}{y^{2}+6y + 5}) (1) clear the denominators using the least common multiple. do not yet balance or solve the resulting polynomial equation. (2) write the polynomial equation in factored form, with one side equal to zero. do not yet solve for y. (3) solve for y. if multiple solutions exist, list them using commas. if no solution exists, enter dne. no decimal answers! y = question help: video

Explanation:

Step1: Factor the denominators

Factor \(y^{2}+6y + 5=(y + 1)(y+5)\). The least - common multiple of \(y + 5\), \(y + 1\) and \(y^{2}+6y + 5=(y + 1)(y + 5)\) is \((y + 1)(y + 5)\). Multiply each term of the equation \(\frac{y - 4}{y + 5}=\frac{2}{y + 1}+\frac{-5y-13}{y^{2}+6y + 5}\) by \((y + 1)(y + 5)\):
\((y - 4)(y + 1)=2(y + 5)+(-5y - 13)\)

Step2: Expand and simplify to get the polynomial equation in standard form

Expand the left - hand side: \((y - 4)(y + 1)=y^{2}+y-4y-4=y^{2}-3y - 4\).
Expand the right - hand side: \(2(y + 5)+(-5y - 13)=2y+10-5y - 13=-3y - 3\).
So, \(y^{2}-3y - 4=-3y - 3\).
Move all terms to one side: \(y^{2}-3y - 4+3y + 3 = 0\), which simplifies to \(y^{2}-1=0\).

Step3: Factor the polynomial equation

Factor \(y^{2}-1\) using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here \(a = y\) and \(b = 1\), so \(y^{2}-1=(y + 1)(y - 1)=0\).

Step4: Solve for \(y\)

Set each factor equal to zero:
If \(y+1 = 0\), then \(y=-1\). But if \(y=-1\), the original denominators \(y + 1\) and \(y^{2}+6y + 5\) are zero, so \(y=-1\) is an extraneous solution.
If \(y - 1=0\), then \(y = 1\).

Answer:

(1) \((y - 4)(y + 1)=2(y + 5)+(-5y - 13)\)
(2) \((y + 1)(y - 1)=0\)
(3) \(1\)