how do i solve: 1/b-a + 1/a+b - 2b/a^2-b^2 and 3y-1/2y^2+y-3 - 2-y/y-1 - y/1-yalgebra

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beckden eNotes educator| Certified Educator

You cannot solve these, only simplify.  I see that you use spaces, but it is easier to undersand if you use parenthesis.

I am suspecting you mean

(a) 1/(b-a) + 1/(a+b) - 2b/(a^2-b^2)

(b) (3y-1)/(2y^2+y-3) - (2-y)/(y-1) - y/(1-y)

(a) First note that 1/(b-a) = -1/(a-b) since (b-a) = -(a-b)
Now we can factor (a^2-b^2) = (a-b)(a+b) and our expression is

-1/(a-b) + 1/(a+b) - 2b/((a+b)/(a-b))

Now like all fractions we can only add if we have a common denominator so I am going to multiply -1/(a-b) * (a+b)/(a+b).  I can do this because (a+b)/(a+b)=1 and I can always multiply by 1.  With the same idea I can multiply 1/(a+b) by (a-b)/(a-b) again because (a-b)/(a-b) = 1. so I get

(-1(a+b))/((a-b)(a+b)) + (1(a-b)/((a-b)(a+b)) - 2b((a+b)(a-b))

Please verify that if you simplify everything I get the original equation.  Now that I have the same denominator, I add the numerators, just like (1/4+2/4 = 3/4).

(-1(a+b) + 1(a-b) - 2b)/((a+b)(a-b))  Now use the distributive property to simplify the numerator.

(-a + -b + a + -b + -2b)/((a+b)(a-b)) = (-4b)/((a+b)(a-b)) = -4b/((a+b)(a-b)).

So 1/(b-a) + 1/(a+b) - 2b/(a^2-b^2) = -4b/((a+b)(a-b))

(b) (3y-1)/(2y^2+y-3) - (2-y)/(y-1) - y/(1-y)  first factor,
2y^2 + y - 3 = (2y+3)(y-1). and (1-y) = -(y-1) so y/(1-y) = -y/(y-1). this gives us

(3y-1)/((2y+3)(y-1)) - (2-y)/(y-1) - (-y/(y-1)) Now we need a common denominator which is (2y+3)(y-1) so multiply the last two rational expressions with (2y+3)/(2y+3) to get

(3y-1)/((2y+3)(y-1)) - ((2-y)(2y+3))/((y-1)(2y+3)) - (-y(2y+3)/((y-1)(2y+3))

Now we have the same denominator we can add the numerators again

((3y-1) - (2-y)(2y+3) - (-y(2y+3)))/((y-1)(2y+3))  Now distribute to get

(3y - 1 - (4y - 2y^2 + 6 - 3y) - (-2y^2 - 3y))/((y-1)(2y+3))  Now change subtractions to additions by changing the sign of what we are subtracting.

(3y +  -1 + -4y + 2y^2 + -6 + 3y + 2y^2 + 3y)/((y-1)(2y+3)) now add like terms to get

(4y^2 + 5y - 7)/((y-1)(2y+3))

So our answers are

(a) 1/(b-a) + 1/(a+b) - 2b(a^2-b^2) = -4b/(a^2-b^2) = 4b/(b^2-a^2)

(b) (3y-1)/(2y^2+y-3) - (2-y)/(y-1) - y/(1-y) = (4y^2 + 5y - 7)/((y-1)(2y+3))

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