Question

If 4a - 13 = 6b + 35 = 8c -17 = d, find the smallest possible value of a + b + c + d.

Answer 1

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 $b=\dfrac{2}{3}a-8$

on using the first and third equality we have:

4a-13=8c-17

$c=\dfrac{1}{2}a+\dfrac{1}{2}$

also from the first and last equality we have:

d=4a-13

Hence,

$a+b+c+d=a+\dfrac{2}{3}a-8+\dfrac{1}{2}a+\dfrac{1}{2}+4a-13\\\\a+b+c+d=\dfrac{37a}{6}-\dfrac{41}{2}$

$a+b+c+d=\dfrac{37a-123}{6}$

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 $\dfrac{37a-123}{6}>0$

$a>3.324$

But as a is an integer, hence the smallest such value is 4.