a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
A=8 B=4 C=14 Yes, I know I'm a genius. Mak - the all knowing one. MWAHAHAH!
That's going to depend on the values of 'a', 'b', 'c', and 'f'.
Roughly speaking, to get a unique solution - or at least, a limited number of solutions - if you have 3 variables, you need 3 equations, not just 2. With the two equations, you can get a relationship between the three variables, but not a unique value for a, b, and c. To get the general relationship, solve both equations for "c", replace one in the other, and solve the resulting equation for "a" to get the relationship between the variables "a" and "b". Then, for any valid combination of values for "a" and "b", use the simpler of the original equations (a + b + c = 24) to get the corresponding value for "c".
If a + b + c + d + 80 + 90 = 100, then a + b + c + d = -70.
between A and B
if f :- a+b = ac then fd:- a.b = a+c
Difficult to tell when you cannot use parentheses. a*(b+c) or a(b+c) = ab + ac This is known as the distributive property of multiplication over addition.
the answer is a
Because there is no way to define the divisors, the equations cannot be evaluated.
a=24 b=16 c=18
A=0 b=0 c=0
b + b + b + c + c + c + c = 3b + 4c
a = 20 b = 60 c = 100
2a. (a, b and c are all equal.)
(a + b)/(a - b) = (c + d)/(c - d) cross multiply(a + b)(c - d) = (a - b)(c + d)ac - ad + bc - bd = ac + ad - bc - bd-ad + bc = -bc + ad-ad - ad = - bc - bc-2ad = -2bcad = bc that is the product of the means equals the product of the extremesa/b = b/c