An easy to use, free and quick solution finder is every math student’s dream. You get the correct answers. However this doesn’t help much in a test where these online tools are not at your disposal. So what if there was an online math algebra solver who would help you find the right answers and figure out how you got there, in your way and at your pace? This is exactly the type of service that online tutors provide with online algebra solvers who are knowledgeable and provide customized service based on your abilities and aptitude.

Algebra solvers from online tutoring sites like www.tutorvista.com, www.smarthinking.com, www.tutornext.com, to name a few, provide easily accessible algebra help while helping you get a hang of those concepts which seem to baffle.

Algebra solvers from online tutoring sites like www.tutorvista.com, www.smarthinking.com, www.tutornext.com, to name a few, provide easily accessible algebra help while helping you get a hang of those concepts which seem to baffle.

Solved Example

4x + 10 = 4 * 2 + 10

= 8 + 10

= 18

= 8 + 10

= 18

Solved Example

Firstly, translate verbal language into algebraic language.

Let first number = x

Then the second number = x + 1.

The problem states:

Sum of two consecutive numbers = 25

=> x + (x + 1) = 25

=> x + x + 1 = 25

=> 2x + 1 = 25

Subtract 1 from both sides of the equation,

=> 2x + 1 - 1 = 25 - 1

=> 2x = 24

Divide from both sides by 2,

=> $\frac{2x}{2} = \frac{24}{2}$

=> x = 12

=> First number = $12$,

Second number = x + 1 = 12 + 1 = $13$

Let first number = x

Then the second number = x + 1.

The problem states:

Sum of two consecutive numbers = 25

=> x + (x + 1) = 25

=> x + x + 1 = 25

=> 2x + 1 = 25

Subtract 1 from both sides of the equation,

=> 2x + 1 - 1 = 25 - 1

=> 2x = 24

Divide from both sides by 2,

=> $\frac{2x}{2} = \frac{24}{2}$

=> x = 12

=> First number = $12$,

Second number = x + 1 = 12 + 1 = $13$

Solved Example

Let width of the rectangle = x

then length of the rectangle = x + 3

Since the area of a rectangle is, Area = length * width

=> Area of given rectangle = (x + 3) * x

The problem states:

=> (x + 3)x = 10

=> x * x + 3x = 10

=> x^{2} + 3x - 10 = 0

=> x^{2} + 5x - 2x - 10 = 0

=> x(x + 5) - 2(x + 5) = 0

=> (x - 2)(x + 5) = 0

=> x - 2 = 0 or x + 5 = 0

=> x = 2 or x = -5

x = -5 is discarded, because rectangle can not have a negative width.

so x = 2 is the answer.

=> Width of rectangle = $2$

and length = x + 3 = 2 + 3 = $5$.

then length of the rectangle = x + 3

Since the area of a rectangle is, Area = length * width

=> Area of given rectangle = (x + 3) * x

The problem states:

=> (x + 3)x = 10

=> x * x + 3x = 10

=> x

=> x

=> x(x + 5) - 2(x + 5) = 0

=> (x - 2)(x + 5) = 0

=> x - 2 = 0 or x + 5 = 0

=> x = 2 or x = -5

x = -5 is discarded, because rectangle can not have a negative width.

so x = 2 is the answer.

=> Width of rectangle = $2$

and length = x + 3 = 2 + 3 = $5$.

Solved Example

Given, 3x + 6 = x - 2

Subtract x from both sides of the equation,

=> 3x - x + 6 = x - x - 2

=> 2x + 6 = - 2

Subtract 6 from both sides of the equation,

=> 2x + 6 - 6 = -2 - 6

=> 2x = - 8

Divide both side by 2,

=> $\frac{2x}{2} = \frac{-8}{2}$

=> $x = - 4$, is the solution.

Check the solution by substituting x = - 4 in the original equation.

**Left Side **= 3x + 6 = 3(- 4) + 6 = -12 + 6 = $- 6$

**Right Side **= x - 2 = - 4 - 2 = $- 6$

Subtract x from both sides of the equation,

=> 3x - x + 6 = x - x - 2

=> 2x + 6 = - 2

Subtract 6 from both sides of the equation,

=> 2x + 6 - 6 = -2 - 6

=> 2x = - 8

Divide both side by 2,

=> $\frac{2x}{2} = \frac{-8}{2}$

=> $x = - 4$, is the solution.

Check the solution by substituting x = - 4 in the original equation.

Solved Example

Let x be the smaller number

Then the larger number is 5 more = x + 5

The problem states,

x + 3(x + 5) = 39

=> x + 3x + 15 = 39

=> 4x + 15 = 39

Subtract 15 from both sides of the equation,

=> 4x + 15 - 15 = 39 - 15

=> 4x = 24

Divide both side by 4,

=> $\frac{4x}{4} = \frac{24}{4}$

=> x = 6

=> x = $6$ is the smaller number.

So the larger number is x + 5 = 6 + 5 = $11$.

Then the larger number is 5 more = x + 5

The problem states,

x + 3(x + 5) = 39

=> x + 3x + 15 = 39

=> 4x + 15 = 39

Subtract 15 from both sides of the equation,

=> 4x + 15 - 15 = 39 - 15

=> 4x = 24

Divide both side by 4,

=> $\frac{4x}{4} = \frac{24}{4}$

=> x = 6

=> x = $6$ is the smaller number.

So the larger number is x + 5 = 6 + 5 = $11$.