Question

Find the sum of the equation

Answer 1

Answer:

$\frac{{y}^{2}+13y-6}{{(y-1)}^{2}(y+7)}$

Step-by-step explanation:

1)  Rewrite ${y}^{2}-2y+1y$  in the form${a}^{2}-2ab+{b}^{2}$, where a = y and b = 1.

$\frac{y}{{y}^{2}-2(y)(1)+{1}^{2}}+\frac{6}{{y}^{2}+6y-7}$

2)  Use Square of Difference: ${(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$.

$\frac{y}{{(y-1)}^{2}}+\frac{6}{{y}^{2}+6y-7}$

3)  Factor ${y}^{2}+6y-7$.

1 - Ask: Which two numbers add up to 6 and multiply to -7?

-1 and 7

2 - Rewrite the expression using the above.

$(y-1)(y-7)$

Outcome/Result: $\frac{y}{(y-1)^2} +\frac{6}{(y-1)(y+7)}$

4) Rewrite the expression with a common denominator.

$\frac{y(y+7)+6(y-1)}{{(y-1)}^{2}(y+7)}$

5)  Expand.

$\frac{{y}^{2}+7y+6y-6}{{(y-1)}^{2}(y+7)}$

6) Collect like terms.

$\frac{{y}^{2}+(7y+6y)-6}{{(y-1)}^{2}(y+7)}$

7) Simplify  ${y}^{2}+(7y+6y)-6y$  to  ${y}^{2}+13y-6y$

$\frac{{y}^{2}+13y-6}{{(y-1)}^{2}(y+7)}$

Answer 2

$\boldsymbol{\dfrac{y}{y^{2}-2y+1 }+\dfrac{6}{y^{2}+6y-7 } \ \ \to \ \ \ Exercise \ to \ solve. }$

Factor y² - 2y + 1. Factor y² + 6y -7.

$\bf{\dfrac{y}{(y-1){2} }+\dfrac{6}{(y-1)(y+7)} }$

To add or subtract expressions, expand them so their denominators are the same. The least common multiple of (y - 1)² and (y - 1)(y + 2) it is (x + y)(y - 1)². Multiply $\bf{\frac{y}{(y-1)^{2} } }$ by $\bf{\frac{y+7}{y+7}. }$ Multiply $\bf{\frac{6}{(y-1)(y+7)} \ by \ \frac{y-1}{y-1}. }$

$\bf{\dfrac{y(y+7)}{(y+7)(y-1)^{2} }+\dfrac{6(y-1)}{(y+7)(y-1)^{2} } }$

Since $\bf{\frac{y(y+7)}{(y+7)(y-1)^{2} }\ and \ \frac{6(y-1)}{(y+7)(y-1)^{2} } }$ have the same denominator, add their numerators to add them together.

$\bf{\dfrac{y(y+7)+6(y-1)}{(y+7)(y-1)^{2} } }$

Do the multiplications on y(y + 7) + 6(y - 1).

$\bf{\dfrac{y^{2} +7+6y-1}{(y+7)(y-1)^{2} } }$

Combine like terms in y² + 7y + 6y - 6.

$\bf{\dfrac{13-6y+1}{(y+7)(y-1)^{2} } }$

xpande (y+7)(y−1)².

$\bf{\dfrac{13y-6+y^{2} }{y^{3}+5y^{2}-13y+7 } \ \ \to \ \ \ Answer }$

$\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}$