Clowns to the left of me, jokers to the right, here I am (sing along if you know the words)

Let's call the numbers in this order A, B, C, D, E. 1. E + C = 14 2. D = B + 1 3. A = 2 * B - 1 4. B + C = 10 5. A + B + C + D + E = 30 Substract (1.) from (5.): A + B + D = 16 Replace there D from (2.): A + 2 * B + 1 = 16 => A = 15 - 2 * B Replace there A from (3.): 2 * B - 1 = 15 - 2 * B => 4 * B = 16 => B = 4 Replace B in (2.): D = B + 1 = 5 Replace B in (3.): A = 2 * B - 1 = 2 * 4 - 1 = 7 Replace B in modified (4.): C = 10 - B = 10 - 4 = 6 Replace C in modified (1.): E = 14 - C = 14 - 6 = 8 Verification: 1. 8 + 6 = 14 2. 5 = 4 + 1 3. 7 = 2* 4 - 1 4. 4 + 6 = 10 5. 7 + 4 + 6 + 5 + 8 = 30

Ok... I'm going to try this... Been a LONG time, so this is MY way, not necessarily the RIGHT way! ;-) Let's assume the first number is "x" The 2nd number a calced by x = 2a - 1 or a = 1/2x + 1/2 The 3rd number b calced by 10 = b + (1/2x + 1/2) or b = 10 - (1/2x + 1/2) or b = 9.5 - 1/2x The 4th number c calced by = 1 + (1/2x + 1/2) or 1.5 + 1/2x The 5th number d calced by 14 = d + 1.5 + 1/2x or 12.5 - 1/2x The sum of all numbers is 30. x + (1/2x + 1/2) + (9.5 - 1/2x) + (1.5 + 1/2x) + (12.5 - 1/2x) = 30 It's now all addition: Solve for x. Put all like numbers together (whole + halfs + those with x in them) x + (0.5 + 9.5 + 1.5 + 12.5) or 24 + (2*1/2x - 2*1/2x) or 0 = 30 x + 23 + 0 = 30 or x = 6 Test it: #1 = x = 6 #2 = 1/2x + 1/2 = 3 + 1/2 = 3.5 #3 = 9.5 - 1/2x = 9.5 - 3 = 6.5 #4 = 1.5 + 1/2x = 1.5 + 3 = 4.5 #5 = 12.5 - 1/2x = 12.5 - 3 = 9.5 Add them: 6 + 3.5 + 6.5 + 4.5 + 9.5 = 30 Whew! I THINK that's correct, though I can't be sure how your teacher wanted it calced. ALWAYS test it. Took me a few tries to get the right numbers in there. Because they didn't add up right, I knew I'd missed something above.

See http://www.answerbag.com/q_view/612880

ANSWER IS 74658

1 st number is a 2 nd number is b 3 rd number is c 4 th number is d 5 th number is e therefore a = 2b-1 b + c = 10 or b = 10 - c c + e = 14 d = b + 1 e = 14 - c i know i need to get the common letter to replace all the terms in the a + b + c + d + e = 30 b can be replaced in all of the equations by 10-c therefore the equation should read like a = 2(10-c) -1 or a = 19 - 2c d = 10 - c + 1 or d = 11 - c therefore (19 - 2c) + (10 - c) + c + (11 - c) + (14 - c) = 30 54 - 4c = 30 4c = 24 c = 6 therefore a = 2(4) -1 = 7 b = 10 - 6 = 4 c = 6 d = 4 + 1 = 5 e = 8 7 + 4 + 6 + 5 + 8 = 30

Sorry the answer for this is 7, 4, 6, 5, 8.

Let the 1st number be a, the 2nd number be b, the 3rd number be c, the 4th number be d and the 5th be e. Add these algebraic terms to the clues given. Clues: 1. e + c = 14 2. d = b + 1 3. a = 2b - 1 4. b + c = 10 5. a + b + c + d + e = 30 Refer to the first clue, e + c = 14. Since all 5 numbers are equal to 30, the other 3 numbers must add up to 16, as 30 - 14 (which is c + e) equals 16. Now you know a + b + d = 16. Now refer to the fourth clue, b + c = 10. Since all 5 numbers are equal to 30, the other 3 numbers must add up to 20, as 30 - 10 (which is b + c) equals 20. This is the same procedure as the step above. Divide 16 by 4. Why? Well.. Clues 2 and 3 both use b, and have all the terms included in "a + b + d = 16". Therefore you find a way to use b in your calculation as well. d is b + 1 and a is 2b - 1. Get rid of the 1s for now. Since there is a 'plus 1' and a 'minus 1', add them to get 0. Now you have taken care of the 1s, you realise there are altogether 4 values of b in 16, since a = b (ignore the 1 for now), b is b, and d = 2b (ignore -1 here also). So to find out the value of b, divide 16 by 4. You get 4. Which is b. So... the rest is simple. You know a is 2b - 1, so a is 7. d is b + 1, so it is 5. Now that you have the values of a and d, use them in "a + d + e = 20". 20 - (5 + 7) = 8. e is 8. e + c = 14, so subtract e from 14, which is '14 - 8', and you get 6. c = 6. Finally.. a = 7 b = 4 c = 6 d = 5 e = 8 The answer is 74658! I hope you found this useful.

Whenever you have a problem like this, start by identifying that which you do not know. Here you do not know 5 numbers. I'll call them {A,B,C,D,E}, for lack of any more creative ideas. Next, write down the relations between these numbers. Take each statement, and convert it into mathematics. This is not hard! The fifth plus the third is 14. Ok, so the fifth number is E. The third is C. Their sum is 14. WRITE IT DOWN! E + C = 14 Do the next few too. One more than a number is that number plus 1! D = 1 + B Likewise, we find that A = 2B - 1 B + C = 10 And the sum of all five numbers? A + B + C + D + E = 30 We now have five equations in the five unknowns. These are LINEAR equations. The solution will generally be unique, except for some degenerate cases. Can we solve this system? E + C = 14 D - B = 1 A - 2B = -1 B + C = 10 A + B + C + D + E = 30 Now, can we eliminate one of these variables? Start with A. If we knew the value of B, then we would know A, because A = 2B - 1. And A appears in only one other place. So replace A in the last equation with 2B - 1. E + C = 14 D - B = 1 B + C = 10 3B + C + D + E = 31 Likewise, we see that E appears only in two places. Since we see that E = 14 - C, the problem reduces to D - B = 1 B + C = 10 3B + D = 17 See that C disappeared in that last equation! Next, look at D. D = 1 + B. Eliminate D next. B + C = 10 3B + 1 + B = 17 The last equation reduces to 4B = 16 Therefore B = 4. Once we know one of the numbers, recover the rest of them. If B + C = 10, and B = 4, then C is 6. If D = 1 + B, and B = 4, then D = 5. If E = 14 - C, and C = 6, then we have E = 8. If A = 2B - 1, and B = 4, then A = 7. We now have all 5 numbers, {A,B,C,D,E} = {7,4,6,5,8}. Is this the correct solution? ALWAYS check your results!!! The sum of the numbers must have been 30. Is it? Yes. So we have solved the problem.