The First And The Last
I have 4 numbers,
the sum of the first two numbers is 2,
the sum of the middle two numbers is 3, and
the sum of the last two numbers is 4.
What is the sum of the first number and the last number?
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4 solutions
a+b = 2 b+c = 3 c+d= 4
Evaluating the equations, we have
a+2(b+c)+d= 9
Since b+c= 3
a+2(3)+d= 9 a+d+6= 9
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Your solution is also a good one, but Roger Erisman already did this solution below, in Sept. 12, 2016.
Relevant wiki: Systems of Linear Equations Word Problems - Basic
Let ’a’ be the first number, ’b’, ’c’, ’d’ be the second, third and fourth numbers respectively. Given a + b = 2 ... (i) b + c = 3 ...(ii) c + d = 4 ...(iii) From (i) a + b = 2 b = 2 − a From (ii) b + c = 3 c = 3 − b c = 3 − ( 2 − a ) = 3 − 2 + a = 1 + a Using above in (iii) c + d = 4 1 + a + d = 4 a + d = 4 − 1 a + d = 3
There is a much shorter way:
(i) + (iii) - (ii) :
(a + b) + (c + d) - (b + c) = a + d
2 + 4 − 3 = 3
Let a = 1st, b = 2nd, c = 3rd, d = 4th.
a+b = 2
b+c =3
c+d =4
Add left sides and right sides giving:
a + 2 b + 2 c + d = 9
but b+c = 3 so 2*(b+c) = 6
subtracting 6 from both sides
a + d = 3
I added the three equations together to get a+2b+2c+d=9. Since 2b + 2c is always even, then a+d has to be odd.Checking the possible answers,it had to be 3.
Let a , b , c , d be the four numbers. So a + b = 2 ( 1 ) b + c = 3 ( 2 ) c + d = 4 ( 3 ) Add equations (1) and (3): a + b + c + d = 6 ( 4 ) Subtract equation (2) from equation (4): a + d = 3