# Algebra 2 Solver

Algebra 2 is a  more advanced level of algebra 1 where you will learn  about polynomials, quadratic expressions, trigonometry and more. Algebra 2 is  simpler if you've got a good grasp on algebra 1 and pre algebra basics but if you don't, then worry not. Algebra 2 solver is an online resource that you can use to upgrade your algebra skills right from home. Algebra 2 solvers assign math helpers to each student, ensuring individual attention and the freedom to learn at your own pace. Students schedule sessions themselves, whenever they want. So you can learn algebra 2 online once a month, once a week or every day, depending on how much help you want.

## Free Algebra 2 Solver

Algebra 2 solvers give you freedom to study from the comfort of your home, and to schedule sessions for when you want them. This saves you the hassle of having to go to tutoring centers after school and also guarantees that you have a math helper when you really need one. Students looking for more advanced math than what is done in class can use algebra 2 solver to find challenging algebra 2 problems which will expand your understanding of the subject. Algebra 2 solver has a lot of advantages to offer students. It ensures that students actually learn the subject and don't just copy answers blindly from online calculators or from services which give you the 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,

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

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

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

[ a2 - b2 = (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 - 6i2

[ i2 = - 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:

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:

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).