Answer:
The smallest possible value of a+b+c+d is: 4
Step-by-step explanation:
since we are given that:
4a - 13 = 6b + 35 = 8c -17 = d
on taking the first two equality i.e. 4a-13=6b+35
we get [tex]b=\dfrac{2}{3}a-8[/tex]
on using the first and third equality we have:
4a-13=8c-17
[tex]c=\dfrac{1}{2}a+\dfrac{1}{2}[/tex]
also from the first and last equality we have:
d=4a-13
Hence,
[tex]a+b+c+d=a+\dfrac{2}{3}a-8+\dfrac{1}{2}a+\dfrac{1}{2}+4a-13\\\\a+b+c+d=\dfrac{37a}{6}-\dfrac{41}{2}[/tex]
[tex]a+b+c+d=\dfrac{37a-123}{6}[/tex]
the smallest possible value such that the expression a+b+c+d is positive will be claculated as:
a+b+c+d>0
that means [tex]\dfrac{37a-123}{6}>0[/tex]
[tex]a>3.324[/tex]
But as a is an integer, hence the smallest such value is 4.
