To solve the equation [tex]\(\frac{y-2}{y+2} = \frac{y-5}{y+1} + 1\)[/tex], follow these steps:
1. Start with the given equation:
[tex]\[
\frac{y-2}{y+2} = \frac{y-5}{y+1} + 1
\][/tex]
2. Combine the terms on the right side of the equation into a single fraction:
[tex]\[
\frac{y-2}{y+2} = \frac{y-5 + (y+1)}{y+1}
\][/tex]
Simplify the numerator:
[tex]\[
y-5 + y+1 = 2y - 4
\][/tex]
So the equation now becomes:
[tex]\[
\frac{y-2}{y+2} = \frac{2y-4}{y+1}
\][/tex]
3. Clear the fractions by cross-multiplying:
[tex]\[
(y-2)(y+1) = (2y-4)(y+2)
\][/tex]
4. Expand both sides of the equation:
[tex]\[
y^2 - y - 2y - 2 = 2y^2 + 4y - 4y - 8
\][/tex]
Simplify each side:
[tex]\[
y^2 - y - 2 = 2y^2 - 8
\][/tex]
5. Move all terms to one side to set the equation to zero:
[tex]\[
y^2 - y - 2 - 2y^2 + 8 = 0
\][/tex]
Simplify:
[tex]\[
-y^2 - y + 6 = 0
\][/tex]
Multiply through by -1 to make calculation easier:
[tex]\[
y^2 + y - 6 = 0
\][/tex]
6. Factor the quadratic equation:
[tex]\[
y^2 + y - 6 = (y+3)(y-2) = 0
\][/tex]
7. Set each factor equal to zero to solve for [tex]\(y\)[/tex]:
[tex]\[
y+3 = 0 \quad \text{or} \quad y-2 = 0
\][/tex]
Solving these gives:
[tex]\[
y = -3 \quad \text{or} \quad y = 2
\][/tex]
Therefore, the solutions to the equation [tex]\(\frac{y-2}{y+2} = \frac{y-5}{y+1} + 1\)[/tex] are [tex]\(y = -3\)[/tex] and [tex]\(y = 2\)[/tex].
