Question

Help with my algebra homework.

1.) what is the wuotient in simplified form? state any restrictions on the variable. (x^2-16/x^2+5x+6) / (x^2+5x+4/x^2-2x-8)

2.) simplify the sum
(7/a+8) + (7/a^2-64)


3.) simplify the difference
(a^2-2a-3/a^2-9a+18) - (a^2-5a-6/a^2+9a+8)


4.) simplify the complex fraction

(4/x+3) / ((1/x)+3)

5.) solve the equation. Check the solution

(-2/x+4)=(4/x+3)


* I have been trying to figure these out all weekend and I can't. Please help.

Answer 1

ok... so why you divide by a fraction, you find the reciprocal of the 2nd fraction and multiply

so you get

\[\frac{x^2 - 16}{x^2 + 5x + 6} \times \frac{x^2 -2x - 8}{x^2 + 5x + 4}\]

if you factorise the numerators and denominators the you will see common factors that will cancel

\[\frac{(x -4)(x+4)}{(x + 2)(x + 3)} \times \frac{(x -4)(x + 2)}{(x + 1)(x + 4)}\]

once you have eliminated the common factors distribute the numerator and denominator for the answer.

hope this make sense

Answer 2

I figured everyone except for one. I cannot get this one

(3x-7/x^2) / ((x^2/2)+(2/x))

Can anyone help?

Answer 3

Is that \[\frac{\frac{3x-7}{x^2}}{\frac{x^2}{2}+\frac{2}{x}}\]
Yechhhh!

Let's simplify the denominator first:
\[\frac{x^2}{2} + \frac{2}{x}=\frac{x*x^2}{x*2} + \frac{2*2}{x*2} = \frac{x^3+4}{2x}\]
Now remember that dividing fractions can be done by multiplying by the reciprocal giving us
\[\frac{3x-7}{x^2}*\frac{2x}{x^3+4} = \frac{2(3x-7)}{x(x^3+4)}\] after we cancel matching terms.

Answer 4

thanks guys. I sent it in lastnight and got two wrong. Im horrible at algebra

Answer 5

Why don't you show us what you did on the two you got wrong, and we'll tell you where the mistake happened?