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Varvara68 [4.7K]
3 years ago
14

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

Mathematics
1 answer:
Monica [59]3 years ago
8 0
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!
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