Answer:
Explanation:
Let n, m, and p be the 3 numbers such that when added together, they have a sum of 100. n is the smallest, m is the middle one, and p is the largest.
from question a. it's known that 2n+2=m. I'll just call this equation a
from question b. it's know that 5n-6=p. this is equation b
in question c, it asks to give an equation to show that n plus m plus p is equal to 100. The equation would be n+m+p = 100. (this is equation c)
Now that there are 3 equations, the 3 variables can be solved.
given that m and p are both known in terms of n, m and p can be substituted in the equation c
after substitution, the equation becomes
n+(2n+2)+(5n-6) = 100
expand and solve
8n-4=100
8n=104
n=13.
so n equals 13
since n has been solved for, it can be plugged back into equation a and b
m=2*(13)+2=28, so m equals 28
p = 5n-6 = 5(13)-6=65-6=59, so p equals 59
check these values by plugging it back into equation c
13+28+59 = 100, so the values work