Yes if you tell us wha5 your question is.
Box 1: 6 Box 2: 16 Box 3: 20 Box 4: 24
Answer:
screenshot linked below,
i'm not sure if I wrote your equation right, but if this answer is not correct please correct me. I will be happy to fix it!
Oldest = 2x
Middle = x + 5
Youngest = x
2x + x + 5 + x = 57
combine like terms
4x + 5 = 57
subtract 5 from both sides
4x = 52
divide both sides by 4 to isolate x
x = 13
Oldest = 2x = 2(13) = 26
Middle = x + 5 = 13 + 5 = 18
Youngest = x = 13
Complete Question:
Is the value of the fraction 7−2y/6 greater than the value of the fraction 3y−7/12 ? For what values?(Make sure to use an inequality)
Answer:
y < 3
Step-by-step explanation:
The given two fractions are:
and ![\frac{3y-7}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B3y-7%7D%7B12%7D)
We have to tell for which range of values is the value of first fraction larger than the second fraction. This can be done by setting up an inequality as shown:
![\frac{7-2y}{6} >\frac{3y-7}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B7-2y%7D%7B6%7D%20%3E%5Cfrac%7B3y-7%7D%7B12%7D)
The range of y which will satisfy this inequality will result in first fraction of larger value as compared to the second fraction.
Multiplying both sides of inequality by 12, we get:
![12 \times \frac{7-2y}{6} > 12 \times \frac{3y-7}{12} \\\\ 2(7-2y)>3y-7\\\\ 14-4y>3y-7\\\\ 14+7>3y+4y\\\\ 21>7y\\\\ 3>y\\\\ y](https://tex.z-dn.net/?f=12%20%5Ctimes%20%5Cfrac%7B7-2y%7D%7B6%7D%20%3E%2012%20%5Ctimes%20%5Cfrac%7B3y-7%7D%7B12%7D%20%5C%5C%5C%5C%20%202%287-2y%29%3E3y-7%5C%5C%5C%5C%2014-4y%3E3y-7%5C%5C%5C%5C%2014%2B7%3E3y%2B4y%5C%5C%5C%5C%2021%3E7y%5C%5C%5C%5C%203%3Ey%5C%5C%5C%5C%20y%3C3)
This means, for y lesser than 3, the value of first fraction is larger than the second one.