Yes.
To show the conditions on a, b, c and d given that if a/b = c/d then a+b = c+d. Suppose b != d (and that both b and d are non-zero) then:
d = kb for some number k (!= 0), so
c/d = c/kb = (c/k)/b
so
a/b = (c/k)/b
=> a = c/k
=> c = ka
Thus:
c + d = ka + kb
= k(a + b)
Which means that c + d = a + b only if k = 1.
Thus if a/b = c/d then a + b = c + d only if a = c and b = d.
The condition on b and d both being non-zero prevents the possibility of division by zero. If either is zero, a division by zero will occur and at least one of the fractions is infinite.
If: b/5+b/3 = 10 then b works out as 18.75
Yes because if 1+0=1 than 0 plus b equals b
b = 14
a=24 b=16 c=18
Because there is no way to define the divisors, the equations cannot be evaluated.
If: b/5+b/3 = 10 then b works out as 18.75
the answer is a
4b
Yes because if 1+0=1 than 0 plus b equals b
b = 14
a=24 b=16 c=18
Because there is no way to define the divisors, the equations cannot be evaluated.
(a-2(ab)1/2+b)/(a-b)
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
A.
a+b=a+b
a = 20 b = 60 c = 100