1A) Solve by substitution.
2x − y + 3 = 0 x2 +y2 −4x=0
1B)
Solving system by Gaussian Elimination and then by back
substitution
3x − 3y + 6z = 6
x + 2y − z = 5 5x −8y +13z = 7
Solve a system of equations:
1-
2x = 5 mod 15
3x = 1 mod 4
2-
x = 5 mod 15
x = 2 mod 12
(Hint: Note that 15 and 12 are not relatively prime. Use the
Chinese remainder
theorem to split the last equation into equations modulo 4 and
modulo 3)