Hey there, lets solve for each side!
[tex]2x\left(1\cdot \left(\frac{1}{2}\right)-y\right)=2\left(\frac{1}{3}\right)+\left(\frac{1}{15}\right)[/tex]
[tex]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)[/tex]
[tex]\mathrm{Multiply:}\:1\cdot \frac{1}{2}=\frac{1}{2} \ \textgreater \ 2x\left(\frac{1}{2}-y\right)[/tex]
[tex]\mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac[/tex]
[tex]Where\; a=2x,\:b=\frac{1}{2},\:c=-y[/tex]
[tex]2x\cdot \frac{1}{2}+2x\left(-y\right)[/tex]
[tex]\mathrm{Apply\:minus-plus\:rules} \ \textgreater \ +\left(-a\right)=-a \ \textgreater \ 2x\frac{1}{2}-2xy[/tex]
[tex]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 [/tex]
[tex]\frac{2x}{2} \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ x \ \textgreater \ x-2xy[/tex]
Moving on
[tex]2\left(\frac{1}{3}\right)+\left(\frac{1}{15}\right)[/tex]
Remove parenthesis again
[tex]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}[/tex]
[tex]\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{2}{3} \ \textgreater \ \frac{2}{3}+\frac{1}{15}[/tex]
Now we want to find the LCD for [tex]\frac{2}{3}+\frac{1}{15}[/tex]
[tex]\mathrm{Factor\:each\:denominator\:into\:its\:primes} \ \textgreater \ 15=3\cdot \:5 \ \textgreater \ 15[/tex]
Now adjust the fractions based on the LCD
[tex]\frac{2\cdot \:5}{15}+\frac{1}{15}[/tex]
Since the denominators are equal, you can combine the fractions
[tex]\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{2\cdot \:5+1}{15} [/tex]
Of course no simply multiply 2 by 5 then add 1
[tex]\frac{11}{15}[/tex]
Combine the two again
[tex]x-2xy=\frac{11}{15}[/tex]
Hope this helps!
