Question

2 x(1 1/2-y)= 2 1/3+1/15Y=?

Answer 1

Hey there, lets solve for each side!

$2x\left(1\cdot \left(\frac{1}{2}\right)-y\right)=2\left(\frac{1}{3}\right)+\left(\frac{1}{15}\right)$

$2x\left(1\cdot \left(\frac{1}{2}\right)-y\right) \; (a) = a\ \textgreater \ Remove\;parenthesis \ \textgreater \ 2x\left(1\cdot \frac{1}{2}-y\right)$

$\mathrm{Multiply:}\:1\cdot \frac{1}{2}=\frac{1}{2} \ \textgreater \ 2x\left(\frac{1}{2}-y\right)$

$\mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac$
$Where\; a=2x,\:b=\frac{1}{2},\:c=-y$
$2x\cdot \frac{1}{2}+2x\left(-y\right)$

$\mathrm{Apply\:minus-plus\:rules} \ \textgreater \ +\left(-a\right)=-a \ \textgreater \ 2x\frac{1}{2}-2xy$

$2x\frac{1}{2} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:2x}{2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{2x}{2} \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ x \ \textgreater \ x-2xy$

Moving on
$2\left(\frac{1}{3}\right)+\left(\frac{1}{15}\right)$

Remove parenthesis again
$2\cdot \frac{1}{3}+\frac{1}{15} \ \textgreater \ 2\cdot \frac{1}{3} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:2}{3}$
$\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{2}{3} \ \textgreater \ \frac{2}{3}+\frac{1}{15}$

Now we want to find the LCD for $\frac{2}{3}+\frac{1}{15}$
$\mathrm{Factor\:each\:denominator\:into\:its\:primes} \ \textgreater \ 15=3\cdot \:5 \ \textgreater \ 15$

Now adjust the fractions based on the LCD
$\frac{2\cdot \:5}{15}+\frac{1}{15}$

Since the denominators are equal, you can combine the fractions
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{2\cdot \:5+1}{15}$

Of course no simply multiply 2 by 5 then add 1
$\frac{11}{15}$

Combine the two again
$x-2xy=\frac{11}{15}$

Hope this helps!