looks good so far
Problem:
The numbers a,b,c and d are such that the sum of any one of them and the product of the other three is equal to 2. if a>b , a>c, and a>d, find a.
i have here my partial answer/solution:
please help me correct on this.
a+bcd =2
b+acd =2
c+abd =2
d+abc =2
by subtracting first and second equations
then,
a-b + cd(a-b) = 0 <<=--- is this correct?
(a-b) (1+cd) = 0
a -b = 0 or 1 + cd = 0
i believe a = b is rejected because a is not equal to b.
and cd is equal to - 1.
is my solution above correct.. please help.
There is an error in your first step, but the basic idea is good. Subtracting b+ acd= 2 from a+ bcd= 2 gives b- a+ (a- b)cd= 0, not b- a+ (b- a)cd= 0. But you can still factor: that is (b- a)(1- cd)= 0. Yes, since you are told that a> b so b- a cannot be 0. That leads to 1- cd= 0 and cd= 1, not -1.
And you can do the same kind of thing for other pairs of equations. If you subtract the third equation from the first you get (a- c)+ (c- a)bd= 0 which leads to (a- c)(1- bd)= 0 and so bd= 1.
What do you get if you subtract the fourth equation from the first?
You are told that a is larger than b, c, or d but of b, c, and d may be the same.