Algebra 2 is the next level in unraveling the mysteries of balancing equations and calculating trigonometry ratios. While a sound base in Algebra 1 goes a long way in making things easier, like all things mathematical, the key to getting it right is practice. The good news is that algebra 2 help is at the tip of your fingers. Whether you're looking for algebra 2 problem solvers, algebra 2 homework solvers, or algebra 2 word problem solvers, online Math solvers can be a great help and in fact save your grades!

Free Algebra 2 Problem Solver

Free algebra 2 solvers available online are a handy way to improve on your algebra skills. Online resources have the advantage of being quick, easily accessible and available 24x7. Algebra 2 problem solvers, are free to use and quick to calculate, giving you all the answers you need in a minute. Most sites explain the fundamental rules and give examples to illustrate the steps involved in the solving process. If you're looking for a more detailed tutorial, online tutoring services offer algebra 2 problem solvers which explain the math in a way you would understand. You can clear your doubts and get all the clarifications you need.

Solved Example

Question: Father is five years more than two times the age of his son. Five years ago, father was 4 times as old as his son. Find the present age of the father.
Solution:
Let the present age of the son = x years

Therefore present age of the father = 2x + 5

Step 1:

5 years ago:
Son's age = x - 5

and father's age = 2x + 5 - 5 = 2x

Step 2:

Age of father = 4(son's age)

=> 2x = 4(x - 5)

=> 2x = 4x - 20

Subtract 2x from both sides

=> 2x - 2x = 4x - 2x - 20

=> 0 = 2x - 20

=> 2x = 20

Divide each side by 2

=> x = $\frac{20}{2}$

=> x = 10

The present age of the father = 2x + 5 = 2 * 10 + 5

= 20 + 5

= 25

Thus the present age of the father is 25.
 

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Solved Examples

Question 1: The length of a rectangular garden is twice its width. The garden is surrounded by a rectangular wall having uniform width of 4 feet. If the area of the garden and the wall is 330 $feet^2$. Find the dimensions of the garden.
Solution:
Let width of the garden = x

Therefore length of the garden = 2x

Step 1:

Since wall has uniform width

=> Width of the largest rectangle = 4 + x + 4 = x + 8

and its length = 4 + 2x + 4 = 2x + 8

Area of the largest rectangle = 330 $feet^2$

Step 2:

Area of rectangle = length * width

=> 330 = (2x + 8)(x + 8)

=> 330 = 2x(x + 8) + 8(x + 8)

=> 330 = 2$x^2$ + 16x + 8x + 64

=> 330 = 2$x^2$ + 24x + 64

=> 2$x^2$ + 24x + 64 - 330 = 0

=> 2$x^2$ + 24x - 266 = 0

or $x^2$ + 12x - 133 = 0

Step 3:

Solve for x, $x^2$ + 12x - 133 = 0

=> $x^2$ + 12x - 133 = 0

=> $x^2$ + 19x - 7x - 133 = 0

=> x(x + 19) - 7(x + 19) = 0

=> (x - 7)(x + 19) = 0

Step 4:

either x - 7 = 0   or x + 19 = 0

=> x = 7    or    x = - 19

x = - 19, neglected since the width must be positive number.

=> x = 7 is only the value

therefore length of the garden = 2x = 2 * 7 = 14

Hence the dimensions of the garden are length = 14 feet and width = 7 feet. answer
 

Question 2: Find the three consecutive odd integers if the product of the first and the third integers is 6 more than three times the second integer.

Solution:
Let three odd consecutive integers

First integer = x

Second integer = x + 2

Third integer = x + 4

Step 1:
Product of the first and the third integers = 6 + three times the second integer

=> x(x + 4) = 3(x + 2) + 6

=> $x^2$ + 4x = 3x + 6 + 6

=> $x^2$ + 4x - 3x = 12

=> $x^2$ + x - 12 = 0

Step 2:

Solve for x, $x^2$ + x - 12 = 0

=> $x^2$ + x - 12 = 0

=> $x^2$ + 4x - 3x - 12 = 0

=> x(x + 4) - 3(x + 4) = 0

=> (x - 3)(x + 4) = 0

Step 3:

either x - 3 = 0    or   x + 4 = 0

Case 1:
When x - 3 = 0


=> x = 3

Then
First integer = x = 3

Second integer = x + 2 = 3 + 2 = 5

Third integer = x + 4 = 3 + 4 = 7

Case 2:
When x + 4 = 0


=> x = - 4, first integer

Since integers are odd, so neglect this case.

Hence three consecutive integers are: 3, 5, 7. answer