Find the solution of the following system of congruences:
(circ)
3x+4y= 1(mod11) 2x+5y= 7(mod11)

Tahmid Knox
2020-11-12
Answered

Find the solution of the following system of congruences:
(circ)
3x+4y= 1(mod11) 2x+5y= 7(mod11)

You can still ask an expert for help

Aamina Herring

Answered 2020-11-13
Author has **85** answers

Step 1

The system of congruence are,

3x+4y=1(mod 11)...(1)

2x+5y=7(mod 11)...(2)

Step 2

Multiply equation (1) by 5 and multiply equation (2)by 4.

15x+20y=5(mod 11)

8x+20y=28(mod 11)

7x=-23(mod 11)

7x=10(mod 11)

21x=30(mod 11)

-x=8(mod 11)

x=-8(mod 11)

Step 3

Substitute x=-8 in equation (1).

3(-8)+4y=1(mod 11)

-24+4y=1(mod 11)

4y=25(mod 11)

4y=3(mod 11)

12y=9(mod 11)

y=9(mod 11)

The system of congruence are,

3x+4y=1(mod 11)...(1)

2x+5y=7(mod 11)...(2)

Step 2

Multiply equation (1) by 5 and multiply equation (2)by 4.

15x+20y=5(mod 11)

8x+20y=28(mod 11)

7x=-23(mod 11)

7x=10(mod 11)

21x=30(mod 11)

-x=8(mod 11)

x=-8(mod 11)

Step 3

Substitute x=-8 in equation (1).

3(-8)+4y=1(mod 11)

-24+4y=1(mod 11)

4y=25(mod 11)

4y=3(mod 11)

12y=9(mod 11)

y=9(mod 11)

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