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solutions but don't explain how to solve the problems.### Solved Examples

**Question 1:**Solve $\frac{x^3 - x}{2x}$

**Solution:**

Given $\frac{x^3 - x}{2x}$

Solve for x,

$\frac{x^3 - x}{2x}$ = $\frac{x(x^2 - 1)}{2x}$

= $\frac{x^2 - 1}{2}$

= $\frac{(x + 1)(x - 1)}{2}$

[ a

=> $\frac{x^3 - x}{2x}$ = $\frac{(x + 1)(x - 1)}{2}$

Solve for x,

$\frac{x^3 - x}{2x}$ = $\frac{x(x^2 - 1)}{2x}$

= $\frac{x^2 - 1}{2}$

= $\frac{(x + 1)(x - 1)}{2}$

[ a

^{2}- b^{2}= (a - b)(a + b) ]=> $\frac{x^3 - x}{2x}$ = $\frac{(x + 1)(x - 1)}{2}$

**Question 2:**Solve (4 + 2i)(5 - 3i)

**Solution:**

Given (4 + 2i)(5 - 3i)

(4 + 2i)(5 - 3i) = 4(5 - 3i) + 2i(5 - 3i)

= 4 * 5 - 4 * 3i + 2i * 5 - 2i * 3i

= 20 - 12i + 10i - 6i

[ i

= 20 - 2i + 6

= 26 - 2i

=> (4 + 2i)(5 - 3i) = 26 - 2i

(4 + 2i)(5 - 3i) = 4(5 - 3i) + 2i(5 - 3i)

= 4 * 5 - 4 * 3i + 2i * 5 - 2i * 3i

= 20 - 12i + 10i - 6i

^{2}[ i

^{2}= - 1 ]= 20 - 2i + 6

= 26 - 2i

=> (4 + 2i)(5 - 3i) = 26 - 2i

**Question 3:**Solve system of equations

2x + y = 5 and x - y = 1

**Solution:**

**Step 1:**

Given system of linear equations

2x + y = 5 ...................................(1)

x - y = 1 .....................................(2)

Step 2:

Step 2:

Solve equation (1) and equation (2)

Add (1) and (2), to eliminate the coefficient of y.

=>

2x + y = 5

+ x - y = 1

------------------

3x + 0 = 6

------------------

=> 3x = 6

Divide each side by 3

=> $\frac{3x}{3} = \frac{6}{3}$

=> x = 2

Step 3:

Step 3:

Put x = 2 in equation (2), to find the value of y.

=> 2 - y = 1

Subtract 2 from both sides

=> 2 - y - 2 = 1 - 2

=> - y = - 1

or y = 1

Hence, the solution of the system is (x, y) = (2, 1).