Question

Can someone help me?

Answer 1

\[(4 - 1/3y) / (5/6y +1)\]

Answer 2

the top divided by the bottom

Answer 3

4 minus one third y divided by 5 over 6y plus 1

Answer 4

the first part is correct the second fraction is wrong

Answer 5

\[4-\frac{ 1 }{ 3y }/\frac{ 5 }{ 6y }+1\]

Answer 6

\[\frac{ 4*3y-1 }{ 3y }/\frac{ 5+1*6y }{ 6y }\]

Answer 7

\[\frac{ 12y-1 }{ 3y }*\frac{ 6y }{ 5+6y }\]

Answer 8

\[\frac{ (12y-1)*2 }{ 5+6y }\]

Answer 9

\[\frac{ 24y-2 }{ 5+6y }\]

Answer 10

\( \Large{\dfrac{ 4 - \frac{1} {3y} } { \frac{5}{6y} + 1} }\)

\( \Large{=\dfrac{ 4 - \frac{1} {3y} } { \frac{5}{6y} + 1} \cdot \dfrac{6y}{6y}}\)

\( \Large{=\dfrac{ 24y - 2} { 5 + 6y} }\)

Answer 11

I am still confused where the extra 6y came in

Answer 12

@aliciabennett
in which step u r confused?

Answer 13

the second

Answer 14

in second step , @mathstudent55 has multiplied and divide the whole term by the same number so that calculation can get more simpler
for example - \[\frac{ 2 }{ 2 } = 1\]
similarly,
\[\frac{ 6y }{ 6y } = 1\]

Answer 15

I still don't get it :/

Answer 16

@aliciabennett
The big fraction has a denominator of 3y in its numerator and a denominator of 6y in its denominator.

The LCD of 3y and 6y is 6y.

By multiplying the numerator and denominator of the big fraction by the LCD, 6y, you eliminate the fractions in the numerator and denominator. That helps you simplify the big fraction.

Answer 17

For example, look this problem:

\( \dfrac{ \frac{1}{2} + 1}{ 2 - \frac{1}{3} } \)

Once again you have a big fraction which has the fraction 1/2 in the numerator and the fraction 1/3 in the denominator.

One way to simplify this fraction is to find the LCD of the denominators of the small fractions, 2 and 3. The LCD of 2 and 3 is 6.

Now you multiply the entire fraction by 6/6.

\(= \dfrac{ \frac{1}{2} + 1}{ 2 - \frac{1}{3} } \cdot \dfrac{6}{6}\)

\( =\dfrac{ \frac{1}{2} \cdot 6+ 1 \cdot6}{ 2 \cdot 6- \frac{1}{3} \cdot 6 } \)

\( =\dfrac{ 3 + 6}{ 12 - 2} \)

\(= \dfrac{9}{10} \)

Answer 18

how do you do it without finding the LCD, how do you do it using the keep change flip method

Answer 19

@mathstudent55

Answer 20

Without simplifying with the LCD, you can use the method of dividing a fraction by a fraction:

Remember that to divide a fraction by a fraction, you multiply the first fraction by the reciprocal of the second fraction.

\(\Large{\dfrac{ 4 - \frac{1} {3y} } { \frac{5}{6y} + 1} }\)

\(= \left(4 - \dfrac{1} {3y} \right) \div \left(\dfrac{5}{6y} + 1 \right)\)

Now we need a common denominator for each factor.

Answer 21

but you can do the keep change flip method by flipping the 5/6y + 1 into 6y/5 +1

Answer 22

You can only do that when the denominator is written as a single fraction. That's what we're going to do now. We're going to use common denominators to make each factor in parentheses into a fraction. Then we can flip the second onend use multiplication.

Answer 23

\(= \left(4 \cdot \dfrac{3y}{3y} - \dfrac{1} {3y} \right) \div \left(\dfrac{5}{6y}
+ 1 \cdot \dfrac{6y}{6y}\right)\)

\(= \left( \dfrac{12y}{3y} - \dfrac{1} {3y} \right) \div \left(\dfrac{5}{6y}
+ \dfrac{6y}{6y}\right)\)

\(= \left( \dfrac{12y - 1}{3y} \right) \div \left(\dfrac{5 + 6y}
{6y}\right)\)

Now both quantities are fractions, so we change the division to a multiplication by flipping the second fraction.

Answer 24

\(= \dfrac{12y - 1}{3y} \cdot \dfrac {6y} {5 + 6y} \)

\(= \dfrac{12y - 1}{3y} \cdot \dfrac {2 \cdot 3y} {5 + 6y} \)

\(= \dfrac{12y - 1}{\cancel{3y}} \cdot \dfrac {2 \cdot \cancel{3y}} {5 + 6y} \)

\(= \dfrac{2(12y - 1)}{5 + 6y} \)

\(= \dfrac{24y - 2}{5 + 6y} \)

Answer 25

That's how you'd do your first problem without multiplying by the LCD.