An old problem.
If 3 x 4 + 2 x 5 = 1 8 1 and x 1 , x 2 , x 3 , x 4 , and x 5 satisfy the system of equations below.
2 x 1 + x 2 + x 3 + x 4 + x 5 = m
x 1 + 2 x 2 + x 3 + x 4 + x 5 = 2 m
x 1 + x 2 + 2 x 3 + x 4 + x 5 = 4 m
x 1 + x 2 + x 3 + 2 x 4 + x 5 = 8 m
x 1 + x 2 + x 3 + x 4 + 2 x 5 = 1 6 m
find m
The answer is 6.
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3 solutions
Yep! That easy and rated a whooping 190 points!
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I am also surprised about its rating.
Its so simple !!!!!!!!!The equation is clearly symmetrical along the left leading diagonal, using gaussian elimination would be too long. Adding eq 1---> eq5 we see 6x1 + 6x2 + 6x3 + 6x4 + 6x5 = 31m hence sum of x s = 31/6m from equation 4 we see x4 + 31/6 = 8m this implies x4 = 8m - 31/6m , similarly x5 = 16 m - 31/6m. Subbing these values into the above equation ( 3x4 + 2x5 = 181) yields 24m - 31/2m + 32m - 31/3m = 181 . This will give m = 6
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We can arrange the whole systems as matrix A times matrix X, and we can evaluate the values of all x’s with the inverse matrix A:
Then we can calculate:
x4 = (-1/6)(m+2m+4m+16m)+(5/6)(8m) = (17/6)m
x5 = (-1/6)(m+2m+4m+8m)+(5/6)(16m) = (65/6)m
3x4+2x5 = 181
(17/2)m+(65/3)m = 181
(181/6)m = 181
So m=6.
Adding all 5 equations of the system, we obtain:
6 ( x 1 + x 2 + x 3 + x 4 + x 5 ) = 3 1 m ⇒ x 1 + x 2 + x 3 + x 4 + x 5 = 6 3 1 m
giving us
1 8 1 = 3 x 4 + 2 x 5 = 3 ( 8 m − 6 3 1 m ) + 2 ( 1 6 m − 6 3 1 m ) = 6 1 8 1 m
hence m = 6 .