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Simplify (y2+y-6)/(y2-4) + (y+3)/(2y2+y-1)

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Here are the solutions to Question 56.

Question 56:

  • Part (i): Simplify y2+y−6y2−4+y+32y2+y−1\frac{y^2+y-6}{y^2-4} + \frac{y+3}{2y^2+y-1}

  1. Factorize the numerators and denominators of both fractions.

    For the first fraction:
    Numerator: y2+y−6=(y+3)(y−2)y^2+y-6 = (y+3)(y-2)
    Denominator: y2−4=(y−2)(y+2)y^2-4 = (y-2)(y+2) (difference of squares)

    For the second fraction:
    Numerator: y+3y+3 (already in simplest form)
    Denominator: 2y2+y−1=(2y−1)(y+1)2y^2+y-1 = (2y-1)(y+1)

  2. Rewrite the expression with the factored terms.

    (y+3)(y−2)(y−2)(y+2)+y+3(2y−1)(y+1)\frac{(y+3)(y-2)}{(y-2)(y+2)} + \frac{y+3}{(2y-1)(y+1)}

  3. Simplify the first fraction by canceling the common factor (y−2)(y-2).

    y+3y+2+y+3(2y−1)(y+1)\frac{y+3}{y+2} + \frac{y+3}{(2y-1)(y+1)}

  4. Find a common denominator to add the fractions.

    The common denominator is (y+2)(2y−1)(y+1)(y+2)(2y-1)(y+1).

    (y+3)(2y−1)(y+1)(y+2)(2y−1)(y+1)+(y+3)(y+2)(y+2)(2y−1)(y+1)\frac{(y+3)(2y-1)(y+1)}{(y+2)(2y-1)(y+1)} + \frac{(y+3)(y+2)}{(y+2)(2y-1)(y+1)}

  5. Combine the numerators over the common denominator.

    (y+3)(2y−1)(y+1)+(y+3)(y+2)(y+2)(2y−1)(y+1)\frac{(y+3)(2y-1)(y+1) + (y+3)(y+2)}{(y+2)(2y-1)(y+1)}

  6. Factor out the common term (y+3)(y+3) from the numerator.

    (y+3)[(2y−1)(y+1)+(y+2)](y+2)(2y−1)(y+1)\frac{(y+3)[(2y-1)(y+1) + (y+2)]}{(y+2)(2y-1)(y+1)}

  7. Expand and simplify the expression inside the square brackets in the numerator.

    (2y−1)(y+1)+(y+2)=(2y2+2y−y−1)+(y+2)(2y-1)(y+1) + (y+2) = (2y^2 + 2y - y - 1) + (y+2) =(2y2+y−1)+(y+2)= (2y^2 + y - 1) + (y+2) =2y2+2y+1= 2y^2 + 2y + 1

  8. Substitute this back into the expression.

    (y+3)(2y2+2y+1)(y+2)(2y−1)(y+1)\frac{(y+3)(2y^2+2y+1)}{(y+2)(2y-1)(y+1)}

    The final answer is (y+3)(2y2+2y+1)(y+2)(2y−1)(y+1)\boxed{\frac{(y+3)(2y^2+2y+1)}{(y+2)(2y-1)(y+1)}}.

    • Part (ii): What must be added to x2+6xx^2+6x to make it a perfect square?

  9. Recall the formula for a perfect square trinomial.

    A perfect square trinomial is of the form (x+a)2=x2+2ax+a2(x+a)^2 = x^2 + 2ax + a^2.

  10. Compare the given expression x2+6xx^2+6x with the first two terms of the perfect square form.

    We have x2+6xx^2+6x. Comparing this to x2+2axx^2+2ax, we can equate the coefficients of xx:

    2a=62a = 6

  11. Solve for aa.

    a=62a = \frac{6}{2} a=3a = 3

  12. Determine the term a2a^2 that completes the square.

    a2=32a^2 = 3^2 a2=9a^2 = 9

  13. The value that must be added to x2+6xx^2+6x to make it a perfect square is 9.

    This results in x2+6x+9x^2+6x+9, which is (x+3)2(x+3)^2.

    The final answer is 9\boxed{9}.

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