** Instructions: ** Use this equation simplifier to simplify the terms and solve an equation you provide, showing all the steps. Please type an equation in the box below.

## More about this Equation Simplifier

This equation simplifier will aid you in the task of taking an algebraic equation and to first simplify and then solve it. If the equation you provide comes already simplified, the calculator will tell you so and will proceed to compute the solution if possible.

Then, the process is completed once you have the equation provided in the box above, and you clicking on "Solve". Then, the steps of the process are shown, and all the steps in the calculation of the solution are shown, if indeed a solution exists.

Simplifying the expressions included in the equation is usually the simple part, as there are many rules we can follow, like PEMDAS and such. The task starts once you have simplified as much as possible, and are left with the task of finding the right strategy to actually solve the equation, if possible.

## How to simplify an equation?

The answer is: it really depends. The broad answer is, "simplify by collecting like terms", and that is great advice, but like terms will be collected, grouped and reduced in a way that is highly dependent on the type of terms we are dealing with.

For example, radicals and roots behave differently to exponents and logarithms. Or, when collecting radicals we prefer to have multiplications in order to group the radicals, same as with exponents. But by the log rules, you would prefer to have sums and subtractions to simplify logarithms.

## Steps to simplify an equation

**Step 1:**It is usually a good first step to pass everything to one side of the equation, though you have to tread carefully as you may want to cross multiply first if needed**Step 2:**Collect like terms based on their structure: polynomials with polynomials, radicals with radicals, etc**Step 3:**Reduce each type as much as possible. Ideally many terms will happen to cancel**Step 4:**If the equation allows and there is not a too difficult mix of equation types, you can try a substitution, in case the resulting equation is not a simple kind to solve (linear or quadratic)

Naturally, these rules are way too broad, but reality is, there is no way to give more precise advice in the general case.

## Why do I simplify before solving

You need to simplify for good measure, as you don't want to deal with terms that are not necessary in the equation, and they add unnecessary complexity to it.

For example, if you have

\[\displaystyle x^3 + x^2 = x^3 + 1 \]

you will definitely want to simplify, as if you don't, you will say that you have a cubic equation, when actually, after simplifying you get

\[\displaystyle x^2 = 1 \]

which is a very simple quadratic equation.

## Why use this Equation Simplifier Calculator

This equation simplifier online is a tool that aids you with the simplification of complex expression. But the main thing is does not it only give you the final simplification, but it also shows you the steps of the process

This is very important because it will guide you to better understand what are best practices and how to get started, and what are the common tricks of the trade.

### Example: Simplifying equations

Simplify and find the solution for: \(x = x^2 + 2x - 1\)

Solution:

We need to solve the following given polynomial equation:\[x=x^2+2x-1\]

Simplifying directly, we observe that we need to solve the following given quadratic equation \(\displaystyle -x^2-x+1=0\).

For a quadratic equation of the form \(a x^2 + bx + c = 0\), the roots are computed using the following formula:

\[x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

In this case, we have that the equation we need to solve is \(\displaystyle -x^2-x+1 = 0\), which implies that corresponding coefficients are:

\[a = -1\]\[b = -1\]\[c = 1\]

First, we will compute the discriminant to assess the nature of the roots. The discriminating is computed as:

\[\Delta = b^2 - 4ac = \displaystyle \left( -1\right)^2 - 4 \cdot \left(-1\right)\cdot \left(1\right) = 5\]

Since in this case we get the discriminant is \(\Delta = \displaystyle 5 > 0\), which is positive, we know that the equation has two different real roots.

Now, plugging these values into the formula for the roots we get:

\[x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \displaystyle \frac{1 \pm \sqrt{\left(-1\right)^2-4\left(-1\right)\left(1\right)}}{2\cdot -1} = \displaystyle \frac{1 \pm \sqrt{5}}{-2}\]

so then, we find that:

\[ {x}_1 = \frac{1}{-2}-\frac{1}{-2}\sqrt{5}=\frac{1}{2}\sqrt{5}-\frac{1}{2} \]\[{x}_2 = \frac{1}{-2}+\frac{1}{-2}\sqrt{5}=-\frac{1}{2}\sqrt{5}-\frac{1}{2}\]

In this case, the quadratic equation \( \displaystyle -x^2-x+1 = 0 \), has two real roots, so then:

\[\displaystyle -x^2-x+1 = - \left(x-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(x+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\]

so then the original polynomial is factored as \(\displaystyle p(x) = -x^2-x+1 = - \left(x-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(x+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right) \), which completes the factorization.

**Conclusion**: Therefore, the final factorization that we obtain is:

\[\displaystyle p(x) = -x^2-x+1 = - \left(x-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(x+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\]

The roots found using the factorization process are \(\frac{1}{2}\sqrt{5}-\frac{1}{2}\) and \(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\) .

Therefore, solving for \(x\) for the given polynomial equation leads to the solutions \(x = \, \)\(\frac{1}{2}\sqrt{5}-\frac{1}{2}\), \(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\), using factorization mathods.

## Other useful Algebra calculators

The main goal of solving equations is to actually reduce a hard to solve equation to an easier-to-solve one. Usually, via a good substitution, we are able to transform a complicated equation into something simpler, like a polynomial equation.

Sometimes, the structure involves solving a trigonometric equation, in which case it all revolves about our ability to write all trig expressions in term of just one, and use a suitable substitution.