Simplify the following algebraic expressions in modular arithmetic: 1) 2e - 3 + 6e + 2 mod 5, 2) (x+1) - (4x-10) mod 7, 3) 4f + e - 7f - 5e mod 7. Also, perform polynomial division: 4) (2x2 + 4)(2x -

Here are the step-by-step solutions for the problems:

1) Evaluate $2e - 3 + 6e + 2$ in mod 5

Step 1: Combine like terms.

Step 2: Reduce the coefficients modulo 5.
Since $8 \equiv 3 \pmod{5}$ and $-1 \equiv 4 \pmod{5}$:

The expression in mod 5 is $\boxed{\text{3e + 4}}$.

2) Evaluate $(x+1) - (4x-10)$ in mod 7

Step 1: Distribute the negative sign and combine like terms.

Step 2: Reduce the coefficients modulo 7.
Since $-3 \equiv 4 \pmod{7}$ and $11 \equiv 4 \pmod{7}$:

The expression in mod 7 is $\boxed{\text{4x + 4}}$.

3) Evaluate $4f + e - 7f - 5e$ in mod 7

Step 1: Combine like terms.

Step 2: Reduce the coefficients modulo 7.
Since $-3 \equiv 4 \pmod{7}$ and $-4 \equiv 3 \pmod{7}$:

The expression in mod 7 is $\boxed{\text{4f + 3e}}$.

4) Evaluate $(2x^2 + 4)(2x - x + 1) \div (x-3)$

Step 1: Simplify the second factor.

The expression becomes $(2x^2 + 4)(x + 1) \div (x-3)$.

Step 2: Perform the multiplication.

Step 3: Perform polynomial division $(2x^3 + 2x^2 + 4x + 4) \div (x-3)$.
Using polynomial long division:

        2x^2 + 8x + 28
      ________________
x - 3 | 2x^3 + 2x^2 + 4x + 4
        -(2x^3 - 6x^2)
        ________________
              8x^2 + 4x
            -(8x^2 - 24x)
            ________________
                    28x + 4
                  -(28x - 84)
                  ___________
                          88

The quotient is $2x^2 + 8x + 28$ with a remainder of 88.
So, the result is $2x^2 + 8x + 28 + \frac{88}{x-3}$.
The evaluated expression is $\boxed{2x^2 + 8x + 28 + \frac{88}{x-3}}$.

5) Evaluate $(2t^2+1) \div (2t-1)$ in mod 3

Step 1: Reduce the coefficients modulo 3 in the expressions.
For $2t^2+1$: $2 \equiv 2 \pmod{3}$, $1 \equiv 1 \pmod{3}$. So, $2t^2+1 \pmod{3}$.
For $2t-1$: $2 \equiv 2 \pmod{3}$, $-1 \equiv 2 \pmod{3}$. So, $2t+2 \pmod{3}$.
The problem becomes $(2t^2+1) \div (2t+2)$ in mod 3.

Step 2: Perform polynomial division modulo 3.
We need to find $Q(t)$ and $R$ such that $2t^2+1 = (2t+2)Q(t) + R \pmod{3}$.
We can multiply the divisor by the inverse of its leading coefficient to simplify. The inverse of $2 \pmod{3}$ is 2 (since $2 \times 2 = 4 \equiv 1 \pmod{3}$).
So, $2t+2 \equiv 2(t+1) \pmod{3}$.
Let's divide $2t^2+1$ by $2t+2$ directly.

        t + 2
      _________
2t+2 | 2t^2 + 0t + 1
       -(2t^2 + 2t)
       ___________
             -2t + 1
             -(2t + 2)
             _________
                   -1

All calculations are modulo 3.
$-2t \equiv t \pmod{3}$
$-1 \equiv 2 \pmod{3}$
So, the division steps are:

        t + 2
      _________
2t+2 | 2t^2 + 0t + 1
       -(2t^2 + 2t)  (since 2t * (2t+2) = 4t^2 + 4t = t^2 + t mod 3, this is wrong.
                      We need to find a term to multiply (2t+2) to get 2t^2.
                      (2t+2) * t = 2t^2 + 2t. So, subtract (2t^2 + 2t))
       ___________
             -2t + 1  (which is t + 1 mod 3)
             -(2t + 2)  (We need to find a term to multiply (2t+2) to get t.
                         (2t+2) * 2 = 4t + 4 = t + 1 mod 3. This is not t.
                         Let's try to multiply by 2. (2t+2) * 2 = t+1.
                         We need to get t.
                         Let's restart the division carefully modulo 3.

Let $P(t) = 2t^2+1$ and $D(t) = 2t+2$.
To get $2t^2$ from $2t$, we multiply by $t$.
$t(2t+2) = 2t^2 + 2t$.
Subtract this from $P(t)$:
$(2t^2+1) - (2t^2+2t) = -2t+1 \equiv t+1 \pmod{3}$.
Now we need to divide $t+1$ by $2t+2$.
To get $t$ from $2t$, we multiply by 2 (since $2 \times 2t = 4t \equiv t \pmod{3}$).
$2(2t+2) = 4t+4 \equiv t+1 \pmod{3}$.
Subtract this from $t+1$:
$(t+1) - (t+1) = 0 \pmod{3}$.
So the quotient is $t+2$ and the remainder is 0.
The evaluated expression in mod 3 is $\boxed{t+2}$.

6) Solve $3a = 6$ in mod 5

Step 1: Reduce the constant term modulo 5.
$6 \equiv 1 \pmod{5}$.
So the congruence is $3a \equiv 1 \pmod{5}$.

Step 2: Find the multiplicative inverse of 3 modulo 5.
We need a number $x$ such that $3x \equiv 1 \pmod{5}$.
By testing values:
$3 \times 1 = 3 \pmod{5}$
$3 \times 2 = 6 \equiv 1 \pmod{5}$
So, the inverse of 3 modulo 5 is 2.

Step 3: Multiply both sides of the congruence by the inverse.
$2 \times (3a) \equiv 2 \times 1 \pmod{5}$
$6a \equiv 2 \pmod{5}$
$a \equiv 2 \pmod{5}$
The solution is $a \equiv \boxed{2 \pmod{5}}$.

7) Solve $5r = 4$ in mod 8

Step 1: The congruence is $5r \equiv 4 \pmod{8}$.

Step 2: Find the multiplicative inverse of 5 modulo 8.
We need a number $x$ such that $5x \equiv 1 \pmod{8}$.
By testing values:
$5 \times 1 = 5 \pmod{8}$
$5 \times 2 = 10 \equiv 2 \pmod{8}$
$5 \times 3 = 15 \equiv 7 \pmod{8}$
$5 \times 4 = 20 \equiv 4 \pmod{8}$
$5 \times 5 = 25 \equiv 1 \pmod{8}$
So, the inverse of 5 modulo 8 is 5.

Step 3: Multiply both sides of the congruence by the inverse.
$5 \times (5r) \equiv 5 \times 4 \pmod{8}$
$25r \equiv 20 \pmod{8}$
$r \equiv 4 \pmod{8}$ (since $25 \equiv 1 \pmod{8}$ and $20 \equiv 4 \pmod{8}$)
The solution is $r \equiv \boxed{4 \pmod{8}}$.

8) Solve $35x+3 = 6x-1$

Step 1: Collect $x$ terms on one side and constant terms on the other side.
Subtract $6x$ from both sides:
$35x - 6x + 3 = -1$
$29x + 3 = -1$

Step 2: Subtract 3 from both sides.
$29x = -1 - 3$
$29x = -4$

Step 3: Divide by 29 to solve for $x$.

The solution is $x = \boxed{-\frac{4}{29}}$.

9) Evaluate $(3x+6)(3x-2)$ in mod 9

Step 1: Expand the product.

Step 2: Reduce the coefficients modulo 9.
Since $9 \equiv 0 \pmod{9}$:
$9x^2 \equiv 0x^2 \equiv 0 \pmod{9}$

Since $12 \equiv 3 \pmod{9}$:
$12x \equiv 3x \pmod{9}$

Since $-12 \equiv 6 \pmod{9}$ (because $-12 = -2 \times 9 + 6$ or $-12+18=6$):
$-12 \equiv 6 \pmod{9}$

Substitute these back into the expression: