Solving
equations would be much easier if someone would just explain them in
plain English right? And of course explain the nuances of positive
and negative signs traversing either side of the equal to sign. This
is the kind of help that an **online algebra equation solver**
offers. Online
algebra equation solvers are the answer to many students who are
discovering that algebra becomes much easier with clear simple
explanations and regular practice. Accessed from the comfort of your
home, everything you need to learn algebra is brought to you by **free
algebra equation solvers**.

## Algebra Equation Solver Free

Online
algebra equation solvers cater to students in all grades,
including college students. Algebra 1 equation solver covers
the introductory topics in algebra like using rational numbers,
algebra properties and rules. **Algebra 2 equation solvers** go a
step further and introduce you to linear equations, quadratic
equations, exponents and logarithms. Online tutorials which
offer **algebra equation solvers **are like having your personal
algebra helpers available whenever you need them with umpteen
examples and step by step demonstrations.** College
algebra equation solver**s cover advanced topics and are a real
help when you are looking for more examples or varied practice
worksheets.

### Solved Examples

**Question 1:**Solve for x, 13x + 45 = - 20

**Solution:**

**Step 1:**

Given, 13x + 45 = - 20

13x + 45 = - 20

Subtract 45 from each side

=> 13x + 45 - 45 = - 20 - 45

=> 13x = - 65

Step 2:

Step 2:

Divide each side by 13

=> $\frac{13x}{13} = \frac{- 65}{13}$

=> x = - 5.

**Question 2:**Solve log $\frac{xy^3}{z^2}$

**Solution:**

Step 1:

Step 1:

log $\frac{xy^3}{z^2}$

=> log xy

^{3}- log z

^{2}

[log$\frac{m}{n}$ = log m - log n]

Step 2:

Step 2:

Again

log xy

^{3}- log z

^{2}= log x + log y

^{3}- log z

^{2}

[log mn = log m + log n]

Step 3:

Step 3:

log x + log y

^{3}- log z

^{2}= log x + 3 log y - 2 log z

[log m

^{n}= n log m]

=> $\frac{xy^3}{z^2}$ = log x + 3 log y - 2 log z.

**answer**

## Step by Step Algebra Equation Solver

### Solved Examples

**Question 1:**Solve the quadratic equation by using formula

2x

^{2}+ 7x - 4 = 0

**Solution:**

^{2}+ 7x - 4 = 0

Step 1:

Step 1:

2x

^{2}+ 7x - 4 = 0

Comparing with general quadratic equation

ax

^{2}+ bx + c = 0

a = 2, b = 7, c = - 4

**Step 2:**

b

^{2}- 4ac = (7)

^{2 }- 4 * 2 * - 4

= 49 + 32

= 81

and

$\sqrt{b^2 - 4ac}$ = $\sqrt{81}$

= 9

=> $\sqrt{b^2 - 4ac}$ = 9

Step 3:

x = $\frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$

=> x = $\frac{- 7 \pm9}{2*2}$

=> x = $\frac{- 7 + 9}{4}$ and x = $\frac{- 7 - 9}{4}$

=> x = $\frac{2}{4}$ and x = $\frac{- 16}{4}$

=> x = $\frac{1}{2}$ and x = - 4

Hence the values of x are: $\frac{1}{2}$ , - 4.

**answer**

**Question 2:**Find the value of p, 2x

^{2}+ p(2 - x) - 1 = 3, when x = 1.

**Solution:**

**Step 1:**

Given 2x

^{2}+ p(2 - x) - 1 = 3

2x

^{2}+ p(2 - x) - 1 = 3

=> 2x

^{2}+ 2p - px - 1 = 3

=> 2x

^{2}+ 2p - px = 3 + 1

=> 2x

^{2}+ 2p - px = 4

Step 2:

Step 2:

To find the value of p, put x = 1

=> 2 * 1

^{2}+ 2p - p * 1 = 4

=> 2 * 1 + 2p - p = 4

=> 2 + p = 4

Subtract 2 from both sides

=> 2 + p - 2 = 4 - 2

=> p = 2

**Answer:**The value of p is 2.

**Question 3:**Solve $\frac{1}{m + 1} + \frac{2}{2m + 3}$

**Solution:**

**:**

Step 1

Step 1

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

LCM of the fractions = (m + 1)(2m + 3)

Step 2:

Step 2:

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

= $\frac{2m + 3 + 2(m + 1)}{(m + 1)(2m + 3)}$

= $\frac{2m + 3 + 2m + 2}{(m + 1)(2m + 3)}$

= $\frac{4m + 5}{(m + 1)(2m + 3)}$

=> $\frac{1}{m + 1} + \frac{2}{2m + 3}$ = $\frac{4m + 5}{(m + 1)(2m + 3)}$