the answer is a
a/b=c/d =>ad=bc =>a =bc/d b =ad/c c =ad/b d =bc/a so if a+b=c+d is true => (bc/d)+(ad/c)=(ad/b)+(bc/a) => (bc2+ad2)/dc=(da2+cb2)/ab => ab(bc2+ad2)=dc(da2+cb2) and since ad=bc, => ab(adc+add) =dc(ada+adc) => abadc+abadd =dcada + dcadc => abadc-dcadc =dcada-abadd => (ab-dc)adc =(dc-ab)add ad cancels out => (ab-dc)c =(dc-ab)d => -(dc-ab)c =(dc-ab)d => -c = d so there's your answer :)
If: a = b+c+d Then: c = a-b-d
c = 6 and d = 7
A plus b plus c equals d. A is the largest answer b is the smallest answer and d is less than 6?''
If a, b, c and d are all non-zero then ab = CD if and only if a/c = d/b or (equivalently) a/d = b/c
And how does this relate to coins?
z remains undefined.
(a - c)(b - d)
a+b+c+d+e = 30 (i) c+e = 14 (ii) d+b = 1 (iii) a = 2b-1 (iv) a+c = 10 (v) By (i)-(ii)-(iii): a = 15 then, by (iv): b = 8 and by (v): c = -5 Also, b = 8 so by (iii): d = -7 and then by (i), e = 19
If a=b and c=d then (a+c)=(b+d) ? This is proved very simply by the direct application of perhaps the most fundamental statement in all of Algebra: "If equals are added to equals, the sums are equal."