1. ## Can you please solve these for me?

2. Originally Posted by TheNotoriousVAG
Note that if two fractions have the same denominator they may be combined.

Mathematically: $\frac{a}{c} \pm \frac{b}{c} = \frac{a\pm b}{c}\: ,\: c \neq 0$

Difference of two squares: $a^2-b^2 = (a-b)(a+b)$

1. $\frac{5}{y-2} = y+2$

Multiply each side by y-2 to clear the denominator (note that the right side will be the difference of two squares so it'd be easier to expand). Then take 5 from each side and solve the quadratic (this will also be the difference of two squares)

2. $\frac{x}{3} - \frac{2}{3} = \frac{1}{x}$

You can then combine the left side and multiply through by 3x to give

$x^2-2x = 3$ and solve the quadratic using your favourite method

3. $\frac{y+2}{y} = \frac{1}{y-5}$

Cross multiply to give $(y+2)(y+5) = y$. Expand and solve.

4. $\frac{x^2}{x-4} - \frac{7}{x-4} = 0$

Combine denominator to give: $\frac{x^2-7}{x-4} = 0$

In this case x-4 will cancel (but remember that x=4 cannot be a solution). $x^2-7=0$ can be solved using the difference of two squares.

5. $\frac{x^2}{x+3} - \frac{5}{x+3} = 0$

Do the same as in question 4 but solve $x^2-5=0$ instead.

3. thanks man

Can you give me the answer to #3 I can't seem to get it right.

Can you give me the answer to #4 and 5 too. I don't get what u mean

4. Originally Posted by TheNotoriousVAG
Can you give me the answer to #3 I can't seem to get it right.
$\frac {y+2}{y} = \frac {1}{y-5}$

$y+2 = y(\frac {1}{y-5})$

$y+2 = \frac {y}{y-5}$

$(y-5)(y+2) = y$

$y^2 +2y - 5y -10 = y$

$y^2 - 3y -10 = y$

$y^2 - 4y -10 = 0$

Find y using the quadratic formula : $y = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

5. Originally Posted by TheNotoriousVAG
Can you give me the answer to #4 and 5 too. I don't get what u mean
e^(i*pi) explained exactly what you had to do, unfortunately, we are not here to provide answers but tips and paths for you to find your own.

6. Do i have to use the quadratic formula for 4 and 5?

7. Originally Posted by TheNotoriousVAG
Do i have to use the quadratic formula for 4 and 5?
If you can transform the equation into $ax^2 + bx + c = 0$ form, then, yes.