An algebra problem by math man

After seeing all we can tell that: 2(A+B+C+D+E)=2+3+4+5+6=20 A+B+C+D+E=20/2=10 So, X=10

Adding up all the expressions given we obtain: $(a+b)+(b+c)+(c+d)+(d+e)+(e+a)=2+3+4+5+6$

$2(a+b+c+d+e)=20$

$2x=20$

$x=10$

I'm not a maths person, so I solved it through the process of elimination:

A+B = 2 might mean: A = 0 or 1 or 2 B = 2 or 1 or 0

B+C = 3 might mean: B = 0 or 1 or 2 or 3 C = 3 or 2 or 1 or 0

C+D = 4 might mean: C = 4 or 3 or 2 or 1 or 0 D = 0 or 1 or 2 or 3 or 4

D+E = 5 might mean: D = 5 or 4 or 3 or 2 or 1 or 0 E = 0 or 1 or 2 or 3 or 4 or 5

E + A = 6 might mean: E = 6 or 5 or 4 or 3 or 2 or 1 or 0

But we have already established that A might be 0 or 1 or 2, thus eliminating E's possible values of 0, 1, 2, and 3 to get a value of 6 for E + A.

From this we can infer: A = 2 B = 0 C = 3 D = 1 E = 4

This solution could be one way of looking at the problem or the answer I gave might be one hell of a coincidence! :D

2A+2B+2C+2D+2E = 2+3+4+5+6

A+B+C+D+E = 10

$A+B+B+C+C+D+D+E+E+A=20$ $2(A+B+C+D+E)=20$ $A+B+C+D+E=10$

$\boxed{\text{X}=10}$

A+B=2 => B=2-A B+C=3 => C=3-B C+D=4 => D=4-C D+E=5 => E=5-D E+A=6 => A=6-E THEN A+B+C+D+E=6-E+2-A+3-B+4-C+5-D A+B+C+D+E=20-(A+B+C+D+E) 2(A+B+C+D+E)=20 X=10

Create matrix A such that it is 5x5.

Then make the first row, 1, 1, 0, 0, 0. The second row, 0, 1, 1, 0, 0. The third row, 0, 0, 1, 1, 0. The fourth row, 0, 0, 0, 1, 1. And the fifth row, 1, 0, 0, 0, 1.

Then create matrix be such that it is 5x1.

Make the first column, 2, 3, 4, 5, 6.

Then multiple [A]^-1 by [B] to get your answers for each variable. Add rhe together and you will get 10, the final answer.

way overthought that one... I did a string of substitutions b=2-a; c=a+1; d=3-a; e=a+2 which mikes the last equation 2a+2=6 or a=2, b=0, c=3, d=1, e=4. could have added all the equations and been done with it in half the time.

b+c=3

but a+b=2 then b=2-a

then 2-a+c=3

but c+d=4 then c=4-d

then 2-a+4-d=3

but e+a=6 then a=6-e and d+e=5 then d=5-e

so 2-(6-e)+4-(5-e)=3

2-6+4-5+e+e=3

2e=8

e=4

so x=a+b+c+d+e=2+4+4=10

a+b=2 means that the numbers can either be 0,2; 2,0; or 1,1 and be substituted into a and b.

Next you figure out if a=0 and b=2 and you continue down the line adding the numbers so;

a+b=2 a=0, b=2 0+2=2

b+c=3 b=2, c=1 2+1=3

c+d=4 c=1, d=3 1+3=4

d+e=5 d=3, e=2 3+2=5

e+a=6 e=2, a=0

Therefore since a is not equal to 0 you substitute in a different pair of numbers; a=1, b=1

a+b=2 a=1, b=1 1+1=2

b+c=3 b=1, c=2 1+2=3

c+d=4 c=2, d=2 2+2=4

d+e=5 d=2, e=3 2+3=5

e+a=6 e=3, a=1

Therefore since a is not equal to 1, you substitute in the final pair

a+b=2 a=2, b=0 2+0=2

b+c=3 b=0, c=3 0+3=3

c+d=4 c=3, d=1 3+1=4

d+e=5 d=1, e=4 1+4=5

e+a=6 e=4, a=2

Therefore since a=2 that pair substitutes in to the equation properly and now you just need to add all the variables together.

a=2 b=0 c=3 d=1 3=4

2+0+3+1+4=10!

Add all equation given above. You will get 2(a+b+c+d+e)=20 or a+b+c+d+e=10 Since x =a+b+c+d+e Therefore x =10

(2+3+4+5+6)/2 =10 So X = 10

(B+C)-(A+B)=3-2 C-A=1 + E+A=6 C+E=7 X=2+7+D=9+D (E-B)-(D-B)=4-1 E-D=3 - (E-C)=1 C-D=2 C=D+2 D+2+D=4 2D=2 D=1 X=9+1=10

add all 2A+2B+2C+2D+2E=2+3+4+5+6 =>2(A+B+C+D+E)=20 =>A+B+C+D+E=10 X=10

2(A+B+C+D+E)=2+3+4+5+6=20

A+B+C+D+E = 10

A+B=2

                        B+C = 3

                        C+D = 4

                        D+E = 5

      +               E+A = 6

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2(A+B+C+D+E)=20

=> A+B+C+D+E=10

Adding

2(A + B + C + D + E) = 20

2 x = 20

x = 10

Write a solution. 2+0+3+1+4=10

B+C=3,D+E =5.ONLY FIND VALUE OF A.for this solve 2eqn one by one.and get the eqn A-E=2.soget value of a=2 from this and first equation

2(A+B+C+D+E) = 2+3+4+5+6 =20

A+B+C+D+E = 10 = x