Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

In thinking about what number each term may be multiplied by to eliminate the fractions, we can take one of two courses:

1) Find the lowest common multiple, ie. the lowest value that each of the fraction denominators have in common as a multiple.

Now a multiple is basically that number multiplied by an integer - for example, multiples of 2 are 2, 4, 6, 8, 10, etc.

So, let us write out the first few multiples for each of the denominator values:

4 (from 3/4): 4, 8, 12, 16, 20

3 (from 1/3): 3, 6, 9, 12, 15, 18

2 (from 1/2): 2, 4, 6, 8, 10, 12, 14, 16

Looking at the values above, we can see that the lowest value that occurs in all three sets of numbers is 12, thus 12 is the lowest common multiple.

Therefor, in order to eliminate the fractions before solving, each term must be multiplied by 12.

2) You could alternatively try multiplying the equation (or simply each fraction) by each of the possible answers and seeing if that will eliminate all of the fractions - this may seem quicker at first but it is always worthwhile understanding how to calculate this question without having possible answers, and as you complete more questions, the process of finding the lowest common multiple will become more natural and even quicker in the end. Nonetheless, let us try this method:

a) Multiplying each fraction by 2

(3/4)*2 = 3/2

This does not eliminate the fraction, therefor 2 is not the answer.

b) Multiplying each fraction by 3

(3/4)*3 = 9/4

This does not eliminate the fraction, thus 3 is not the answer.

c) Multiplying each fraction by 6

(3/4)*6 = 9/2

This does not eliminate the fraction, therefor 6 is not the answer.

d) Multiplying each fraction by 12

(3/4)*12 = 9 (this works so far)

(1/3)*12 = 4 (this also works so far)

(1/2)*12 = 6 (this also works)

Since multiplying each fraction by 12 will eliminate the fractions, 12 is the answer.

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

add the table and i will answer correctly

step-by-step explanation: