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The values of x and y which satisfy the equations: 47x + 31y = 63 and 31x + 47y=15 are ____________​
  • a)
    -2 and -1
  • b)
    2 and -1
  • c)
    2 and 1
  • d)
    -2 and 1
Correct answer is 'B'. Can you explain this answer?
Most Upvoted Answer
The values of x and y which satisfy the equations:47x + 31y = 63 and 3...
Given pair of linear equations:
47x+31y =63  ---(1)
31x+47y =15  ---(2)
multiply equation (1) by 31 and equation (2) by 47
substract (2) from (1) we get
(961 - 2209)y = 63x31 - 15x47
-1248 y = 1953 - 705
-1248 y = 1248
Therefore, y = -1.
Substitute y = -1 in equation (1) we get
47x = 94
So, x = 2.

Hence, The values of x and y are 2, -1.
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Community Answer
The values of x and y which satisfy the equations:47x + 31y = 63 and 3...
Solution:

We can solve this system of equations by elimination method.

Multiplying the first equation by 47 and the second equation by 31, we get:

(47)(47x) + (47)(31y) = (47)(63)

(31)(31x) + (31)(47y) = (31)(15)

Simplifying, we get:

2209x + 1457y = 2961

961x + 1457y = 465

Subtracting the second equation from the first, we get:

1248x = 2496

x = 2

Substituting x = 2 in the second equation, we get:

(31)(2) + (47)y = 15

62 + 47y = 15

47y = -47

y = -1

Therefore, the values of x and y that satisfy the equations are:

x = 2

y = -1

Hence, the correct answer is option B.
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