and explain them?
THANKS
http://img149.imageshack.us/img149/5044/806m.gif
and explain them?
THANKS
http://img149.imageshack.us/img149/5044/806m.gif
Note that if two fractions have the same denominator they may be combined.
Mathematically: $\displaystyle \frac{a}{c} \pm \frac{b}{c} = \frac{a\pm b}{c}\: ,\: c \neq 0$
Difference of two squares: $\displaystyle a^2-b^2 = (a-b)(a+b)$
1. $\displaystyle \frac{5}{y-2} = y+2$
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. $\displaystyle \frac{x}{3} - \frac{2}{3} = \frac{1}{x}$
You can then combine the left side and multiply through by 3x to give
$\displaystyle x^2-2x = 3$ and solve the quadratic using your favourite method
3. $\displaystyle \frac{y+2}{y} = \frac{1}{y-5}$
Cross multiply to give $\displaystyle (y+2)(y+5) = y$. Expand and solve.
4. $\displaystyle \frac{x^2}{x-4} - \frac{7}{x-4} = 0$
Combine denominator to give: $\displaystyle \frac{x^2-7}{x-4} = 0$
In this case x-4 will cancel (but remember that x=4 cannot be a solution). $\displaystyle x^2-7=0$ can be solved using the difference of two squares.
5. $\displaystyle \frac{x^2}{x+3} - \frac{5}{x+3} = 0$
Do the same as in question 4 but solve $\displaystyle x^2-5=0$ instead.
$\displaystyle \frac {y+2}{y} = \frac {1}{y-5}$
$\displaystyle y+2 = y(\frac {1}{y-5})$
$\displaystyle y+2 = \frac {y}{y-5}$
$\displaystyle (y-5)(y+2) = y$
$\displaystyle y^2 +2y - 5y -10 = y$
$\displaystyle y^2 - 3y -10 = y$
$\displaystyle y^2 - 4y -10 = 0$
Find y using the quadratic formula : $\displaystyle y = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$