Studies on the Internet usage patterns among students show that they spend an average of 34 hours a week on the Internet, which is more time than what they spend on any other activity including studying and socializing. While some experiences cannot be replicated online, academic work can be. Free step by step algebra solvers is a handy online tool for the mathematically challenged. Many are equipped with calculators into which you can input your problem and get the answer within seconds. Free algebra problem solver with steps also shows you the various steps from beginning to end.

Word problems in algebra present a challenge to many students. The terminology and relation between different variables takes a bit of practice to understand. Algebra problem solver with steps free also explains how to go about writing those statements as equations. Once you put the equation into words, things become a easier and you just need to solve the equation. Practice sheets available with most online algebra problem solvers present students with the opportunity to get familiar with a wide variety of problems and scenarios.

Solved Example

Let two parts be x and y, and x is greater than y.

**Step 1:**

Divide 16 into two equal parts

=> x + y = 16 ..............................(1)

and twice the square of the greater part = 164 + square of the smaller part

=> 2$x^2$ = 164 + $y^2$ .................................(2)

Step 2:

Solve (1) and (2)

(1) => y = 16 - x, put in (2)

=> 2$x^2$ = 164 + $(16 - x)^2$

=> 2$x^2$ = 164 + $16^2$ + $x^2$ - 2 * 16 * x

[(a - b)^{2} = a^{2} + b^{2 }- 2ab]

=> 2$x^2$ = 164 + 256 + $x^2$ - 32x

=> 2$x^2$ = 420 + $x^2$ - 32x

=> 2$x^2$ - $x^2$ + 32x - 420 = 0

=> $x^2$ + 32x - 420 = 0

Step 3:

Solve for x, $x^2$ + 32x - 420 = 0

=> $x^2$ + 42x - 10x - 420 = 0

=> x(x + 42) - 10(x + 42) = 0

=> (x - 10)(x + 42) = 0

Step 4:

either x - 10 = 0 or x + 42 = 0

=> x = 10 or x = - 42, neglected

So x = 10 is only the value of x.

(1) => 10 + y = 16

=> y = 16 - 10 = 6

Hence, the parts are: 10 and 6.**answer**

Divide 16 into two equal parts

=> x + y = 16 ..............................(1)

and twice the square of the greater part = 164 + square of the smaller part

=> 2$x^2$ = 164 + $y^2$ .................................(2)

Step 2:

Solve (1) and (2)

(1) => y = 16 - x, put in (2)

=> 2$x^2$ = 164 + $(16 - x)^2$

=> 2$x^2$ = 164 + $16^2$ + $x^2$ - 2 * 16 * x

[(a - b)

=> 2$x^2$ = 164 + 256 + $x^2$ - 32x

=> 2$x^2$ = 420 + $x^2$ - 32x

=> 2$x^2$ - $x^2$ + 32x - 420 = 0

=> $x^2$ + 32x - 420 = 0

Step 3:

Solve for x, $x^2$ + 32x - 420 = 0

=> $x^2$ + 42x - 10x - 420 = 0

=> x(x + 42) - 10(x + 42) = 0

=> (x - 10)(x + 42) = 0

Step 4:

either x - 10 = 0 or x + 42 = 0

=> x = 10 or x = - 42, neglected

So x = 10 is only the value of x.

(1) => 10 + y = 16

=> y = 16 - 10 = 6

Hence, the parts are: 10 and 6.

Also algebra problem solver free with steps shows users how to solve problems while explaining how they are done. All in all, algebra problem solvers are an easy and quick way for students of all grades and levels to get assistance with their algebra homework. Free algebra problem solvers use screen shots or in some cases interactive white boards which allows you to work on the problems and the algebra problem solver to monitor your work.

Since x : y = 3 : 2

Let x = 3k and y = 2k

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{4 * 3k + 6 * 2k}{2 * 3k + 7 * 2k}$

= $\frac{12k + 12k}{6k + 14k}$

= $\frac{24k}{20k}$

= $\frac{6}{5}$

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{6}{5}$

Let x = 3k and y = 2k

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{4 * 3k + 6 * 2k}{2 * 3k + 7 * 2k}$

= $\frac{12k + 12k}{6k + 14k}$

= $\frac{24k}{20k}$

= $\frac{6}{5}$

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{6}{5}$

Given (x - 5)(40 - x) = 800

Step 1:

(x - 5)(40 - x) = 800

=> x(40 - x) - 5(40 - x) = 800

=> 40x - $x^2$ - 200 + 5x = 800

=> 45x - $x^2$ - 1000 = 0

or $x^2$ - 45x + 1000 = 0

Which is quadratic equation

**Step 2:**

Solve for x, $x^2$ - 45x + 1000 = 0

=> $x^2$ - 45x + 1000 = 0

Compare with ax^{2} + bx + c = 0

a = 1, b = - 45 and c = 1000

b^{2} - 4ac = (- 45)^{2} - 4 * 1 * 1000

= 2025 - 4000

= - 1975

< 0

=> b^{2} - 4ac < 0 ( no real roots)

Hence this quadratic equation doesn't have any real root.

Step 1:

(x - 5)(40 - x) = 800

=> x(40 - x) - 5(40 - x) = 800

=> 40x - $x^2$ - 200 + 5x = 800

=> 45x - $x^2$ - 1000 = 0

or $x^2$ - 45x + 1000 = 0

Which is quadratic equation

Solve for x, $x^2$ - 45x + 1000 = 0

=> $x^2$ - 45x + 1000 = 0

Compare with ax

a = 1, b = - 45 and c = 1000

b

= 2025 - 4000

= - 1975

< 0

=> b

Hence this quadratic equation doesn't have any real root.

Given, $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$

Step 1:

Solve numerator and denominator

15a^{2} - 10ab = 5a(3a - 2b)

and

9$a^2$ - 4$b^2$ = $(3a)^2$ - $(2b)^2$

= (3a - 2b)(3a + 2b)

[ $a^2$ - $b^2$ = (a - b)(a + b)]

**Step 2**:

$\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a(3a - 2b)}{(3a - 2b)(3a + 2b)}$

= $\frac{5a}{3a + 2b}$

=> $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a}{3a + 2b}$.

Step 1:

Solve numerator and denominator

15a

and

9$a^2$ - 4$b^2$ = $(3a)^2$ - $(2b)^2$

= (3a - 2b)(3a + 2b)

[ $a^2$ - $b^2$ = (a - b)(a + b)]

$\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a(3a - 2b)}{(3a - 2b)(3a + 2b)}$

= $\frac{5a}{3a + 2b}$

=> $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a}{3a + 2b}$.