Add 5 in mod 8. (You can always subtract 3 in mod 8 as well, eg 5 + 3 = 0; 0 - 3 = 5.)
If: 6x-4+7 = 5 Then: x = 1/3
14x = 5 x = 5/14
6
(2, 3) is a solution only if it satisfies the equation. x + 5y = 4 (2) + 5(3) =? 4 2 + 15 =? 4 17 = 4 False Therefore, (2, 3) is not a solution to the given equation.
Yes. It is THE solution.
Yes, x = 2 is the solution.
the solution is supersaturated
x = 5
Yes
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
x = 3/5 and y = 1/5
Add 5 in mod 8. (You can always subtract 3 in mod 8 as well, eg 5 + 3 = 0; 0 - 3 = 5.)
The solution to 5 = 3x + 2 is found by first subtracting 2:3 = 3x.Then dividing by 3:x = 1.
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
It is: x = 5/3 and x = -3
They are: x = 5/3 and x = -3