The sum of 3 numbers is 61. The second number is 5 times the first. While the 3rd is 2 less then the first. How do you find the numbers?

1 Answer
Oct 18, 2015

Answer:

#9#, #45#, and #7#

Explanation:

Let's say that your numbers are #x#, #y#, and #z#.

The first thing to write down is that their sum is equal to #61#

#x + y + z = 61" " " "color(purple)((1))#

Next, you know that the second number, #y#, is five times bigger than the first, #x#. This is equivalent to saying that you need to multiply #x# by #5# to get #y#

#y = 5 * x" " " "color(purple)((2))#

Finally, you know that if you subtract #2# from #x# you get the #z#

#z = x - 2" " " "color(purple)((3))#

Use equations #color(purple)((2))# and #color(purple)((3))# to find the value of #x# in equation #color(purple)((1))#

#x + 5x + x - 2 =61#

#7x = 63 implies x = 63/7 = color(green)(9)#

Now take this value of #x# and find #y# and #z#

#y = 5 * (9) = color(green)(45)#

#z = (9) - 2 = color(green)(7)#

The three numbers are #9#, #45#, and #7#.