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

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

## Want to know more about Congruence?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Aamina Herring
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)