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puteri [66]
2 years ago
6

Find the sum of the equation

Mathematics
2 answers:
stealth61 [152]2 years ago
5 0

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)}

Zina [86]2 years ago
4 0

\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}}}}}}

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