Not sure how to write this out...
{(8/x+y)+1/4}/{(x/x+y)-1}
My converter says it should be written like
@DIV{@DIV{8;x+y}+@DIV{1;4};@DIV{x;x+y}-1} If that help. PLZ if you help show steps so I can figure out how to do this correctly.
Not sure how to write this out...
{(8/x+y)+1/4}/{(x/x+y)-1}
My converter says it should be written like
@DIV{@DIV{8;x+y}+@DIV{1;4};@DIV{x;x+y}-1} If that help. PLZ if you help show steps so I can figure out how to do this correctly.
First, it appears that you are not using parentheses to state your problem correctly. Second, I have no clue what you mean by a converter.
Is this your problem
$Simplify\ \left(\dfrac{8}{x + y} + \dfrac{1}{4}\right) \div \left(\dfrac{x}{x + y} - 1\right).$
If so, the first step is to change each expression in parentheses into a single fraction. What do you get for each one?
Now when you divide a fraction by a fraction, how do you proceed?
So now what do you have?
And that simplifies nicely to what?
My problem seems to be getting the 2 fractions to one, Which should be easy right..
32+x+y/4(x+y) the second I get -y/x+1
flip the 2nd and multiply x^2+33x+xy+y+32/-4y(x+y)
But It's not right of course.. I just can't put my finger on what I am doing wrong.
$Simplify\ \left(\dfrac{8}{x + y} + \dfrac{1}{4}\right) \div \left(\dfrac{x}{x + y} - 1\right).$
First step: get common denominators
$\left(\dfrac{8}{x + y} + \dfrac{1}{4}\right) \div \left(\dfrac{x}{x + y} - 1\right) = \left(\dfrac{4 * 8}{4 * (x + y)} + \dfrac{1 * (x + y)}{4 * (x + y) }\right) \div \left(\dfrac{x}{x + y} - \dfrac{x + y}{x + y}\right).$
Second step: add fractions
$\left(\dfrac{4 * 8}{4 * (x + y)} + \dfrac{1 * (x + y)}{4 * (x + y) }\right) \div \left(\dfrac{x}{x + y} - \dfrac{x + y}{x + y}\right) = \dfrac{32 + x + y}{4(x + y)} \div \left(-\ \dfrac{y}{x + y}\right).$
Third step: "flip" divisor
$\dfrac{32 + x + y}{4(x + y)} \div \left(-\ \dfrac{y}{x + y}\right) = \dfrac{32 + x + y}{4(x + y)} * \left(-\ \dfrac{x + y}{y}\right).$
Fourth step: Multiply and simplify
$\dfrac{32 + x + y}{4(x + y)} * \left(-\ \dfrac{x + y}{y}\right) = -\ \dfrac{(32 + x + y)(x + y)}{4y(x + y)} = -\ \dfrac{32 + x + y}{4y}.$
Hello, selfadmrier!
$\displaystyle \text{Simplify: }\: \dfrac{\dfrac{8}{x+y} + \dfrac{1}{4}}{\dfrac{x}{x+y} - 1}$
Multiply top and bottom by the LCD of all the denominators.
This would be: $\displaystyle 4(x+y)$
$\displaystyle \dfrac{4(x+y)\cdot \left(\dfrac{8}{x+y} + \dfrac{1}{4}\right)} {4(x+y)\cdot\left(\dfrac{x}{x+y} - 1\right)} \;\;=\;\;\dfrac{4(x+y)\cdot\dfrac{8}{x+y} + 4(x+y)\cdot\dfrac{1}{4}} {4(x+y)\cdot\dfrac{x}{x+y} - 4(x+y)\cdot 1} $
. . $\displaystyle =\;\;\frac{32 + x + y}{4x - 4x - 4y} \;\;=\;\;\frac{32+x+y}{-4y}$