For what values of y:
is the value of the fraction (7−2y)/6 greater than the value of the fraction (y−7)/12 ?
(7−2y)/6 > (y−7)/12
2(7-2y) > y-7
14-4y > y-7
21 > 5y
21/5 > y
(7−2y)/6 > (y−7)/12
multiplying by 12 ... 14 - 4y > y - 7
adding 4y and 7 ... 21 > 5y
To compare the values of these fractions, we need to find the range of values of y for which (7 - 2y)/6 is greater than (y - 7)/12.
Let's first simplify both fractions:
(7 - 2y)/6 > (y - 7)/12
Multiplying both sides by 12 to eliminate the denominators, we get:
12(7 - 2y) > 6(y - 7)
Now, expand and simplify the equation:
84 - 24y > 6y - 42
Next, let's gather the y terms on one side and the constant terms on the other side:
-24y - 6y > -42 - 84
Combining like terms:
-30y > -126
To isolate the variable y, divide both sides by -30 (and flip the inequality sign since we are dividing by a negative number):
y < 126/30
Reducing the right side:
y < 21/5
Thus, the solution is y < 21/5. The fraction (7 - 2y)/6 is greater than (y - 7)/12 for any y-values less than 21/5.