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anzhelika [568]
2 years ago
5

We can find a system of congruences where x≡a (mod 4) x≡b (mod 5) where the solution is x≡7 (mod 20).

Mathematics
1 answer:
Nonamiya [84]2 years ago
8 0

If x ≡ 7 (mod 20), then

x ≡ 7 ≡ 27 ≡ 47 ≡ 67 ≡ 87 ≡ … (mod 20)

or equivalently, x = 7 + 20k for some integer k.

Taken mod 4, we have

x ≡ 7 + 20k ≡ 3 (mod 4)

and mod 5,

x ≡ 7 + 20k ≡ 2 (mod 5)

so that a = 3 and b = 2.

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Inverse of y equals 12 to the x
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Math Questions :3 thank uu <3
coldgirl [10]
14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
                y = 10

Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
8 0
3 years ago
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