Find the number such that when 3/4 of it is added to 3 1/2 the sum is the same as when 2/3 of it is subtracted from 6 1/2.
To find the number, let's solve the problem step by step.
Let's say the number we are looking for is "x".
According to the problem, when 3/4 of the number is added to 3 1/2, we get the same result as when 2/3 of the number is subtracted from 6 1/2.
Mathematically, we can represent this as an equation:
(3/4)*x + 3 1/2 = 6 1/2 - (2/3)*x
Now, let's simplify the equation by converting the mixed fractions into improper fractions:
(3/4)*x + 7/2 = 13/2 - (2/3)*x
Next, let's get rid of the fractions by multiplying everything by the least common multiple (LCM) of the denominators, which is 12:
12 * [(3/4)*x + 7/2] = 12 * (13/2 - (2/3)*x)
Simplifying both sides of the equation:
9x + 42 = 78 - 8x
Now, let's collect like terms:
9x + 8x = 78 - 42
17x = 36
Finally, divide both sides of the equation by 17 to solve for x:
x = 36/17
So, the number we are looking for is approximately 2.12 (rounded to two decimal places).
Therefore, the number such that when 3/4 of it is added to 3 1/2, the sum is the same as when 2/3 of it is subtracted from 6 1/2, is approximately 2.12.
what's the trouble? Just write all that as math:
(3/4)x + 3 1/2 = 6 1/2 - (2/3)x
Now just solve for x.