No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2.
A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.
No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2.
A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.