Algebra equation solvers make learning algebra easy. A lot of online
sites and resources exist to make algebra simple. With step-by-step
explanations and a lot of solved and unsolved practice problems, they
provide students with plenty of opportunities to hone their algebra
solving skills. As any student knows, practice is the key to scoring
perfect grades for algebra. This is where an Algebra equation solver
makes things easy. An algebra equation solver can be an online tutor who
works with a student to make the subject easy. Online tutors are
available round-the-clock for a one-on-one personalized session any time
a student needs help. Algebra equation solver online is convenient and quick. Students find
it especially helpful because these solvers are available 24x7 letting
you learn whenever they feel like.

## Equation Solver Online

### Solved Examples

**Question 1:**Factor 6x

^{2}- 7x + 2

**Solution:**

Given quadratic equation

6x

6x

= 3x(2x - 1) - 2(2x - 1)

= (3x - 2)(2x - 1)

=> 6x

6x

^{2}- 7x + 26x

^{2}- 7x + 2 = 6x^{2}- 3x - 4x + 2= 3x(2x - 1) - 2(2x - 1)

= (3x - 2)(2x - 1)

=> 6x

^{2}- 7x + 2 = (3x - 2)(2x - 1)**Question 2:**Find the sum of four consecutive integers such that the sum of first, second and third integer is equal to two times the fourth integer.

**Solution:**

First integer = x

Second integer = x + 1

Third integer = x + 2

Forth integer = x + 3

The statement states:

Sum of first, second and third integer is equal to two times the fourth integer

=> x + (x + 1) + (x + 2) = 2(x + 3)

=> 3x + 3 = 2x + 6

=> 3x - 2x = 6 - 3

=> x = 3

First integer = 3

Second integer = 3 + 1 = 4

Third integer = 3 + 2 = 5

Forth integer = 3 + 3 = 6

=> Sum of four consecutive integers is 18 (ie 3 + 4 + 5 + 6 = 18).

Second integer = x + 1

Third integer = x + 2

Forth integer = x + 3

The statement states:

Sum of first, second and third integer is equal to two times the fourth integer

=> x + (x + 1) + (x + 2) = 2(x + 3)

=> 3x + 3 = 2x + 6

=> 3x - 2x = 6 - 3

=> x = 3

First integer = 3

Second integer = 3 + 1 = 4

Third integer = 3 + 2 = 5

Forth integer = 3 + 3 = 6

=> Sum of four consecutive integers is 18 (ie 3 + 4 + 5 + 6 = 18).

**Question 3:**If Riffela travel 30 miles in two hours, how far he travel in four hours.

**Solution:**

Given, Riffela travel 30 miles in two hours

Let he traveled 'd' miles in 4 hours.

=> $\frac{30}{2} = \frac{d}{4}$

=> 15 = $\frac{d}{4}$

=> 15 * 4 = d

=> 60 = d

=> Hence, Riffela traveled 60 miles in 4 hours.

Let he traveled 'd' miles in 4 hours.

=> $\frac{30}{2} = \frac{d}{4}$

=> 15 = $\frac{d}{4}$

=> 15 * 4 = d

=> 60 = d

=> Hence, Riffela traveled 60 miles in 4 hours.

**Question 4:**Solve for p, p + (p + 4) + 6(p - 4) + 3 = 68

**Solution:**

Given, p + (p + 4) + 6(p - 4) + 3 = 68

=> p + (p + 4) + 6(p - 4) + 3 = 68

=> p + p + 4 + 6p - 24 + 3 = 68

=> 8p - 17 = 68

Add 17 both sides

=> 8p - 17 + 17 = 68 + 17

=> 8p = 85

Divide each side by 8

=> p = $\frac{85}{8}$

=> p + (p + 4) + 6(p - 4) + 3 = 68

=> p + p + 4 + 6p - 24 + 3 = 68

=> 8p - 17 = 68

Add 17 both sides

=> 8p - 17 + 17 = 68 + 17

=> 8p = 85

Divide each side by 8

=> p = $\frac{85}{8}$