Algebra Equations

Instructions: Use this calculator to solve algebra equations, showing all the steps. Please type in the equation you want to solve (Type a one or two variable equation).

Enter the equation you want to solve (Ex: sin(2x)= 1/2, etc.)

Algebra Equations

No doubt equations are one of the main elements to pay attention in Algebra. This calculator will allow you to solve an Algebra equation that you provide, either linear or non-linear

All you need to do is to type or paste the equation you want to solve, and click on the "Solve" button in order to get all the steps of the solution shown.

One caveat right off the bat, not all algebra equations will be easily solved, and some of them will not be solved at all. Of course, some easy examples like linear equations or quadratic equations are quite straightforward, but that is about it.

Anything that does not fit those categories simply will not have a standard/straightforward method to be solved. It does not mean that you CANNOT solve them, it just mean that there is not a "roadmap" for it.

What is an Algebra Equation?

An Algebra equation, also known as Algebraic Equation are an umbrella term to refer to the different kind of math equations that you will find when working with algebra.

They will range from trivial linear equations such as

$\displaystyle \frac{1}{2} x + \frac{2}{3} = \frac{5}{6}$

to more complicated equations such as

$\displaystyle \sin \left(\frac{1}{2} x^2 + x + 1 \right)= \frac{\sqrt{2}}{2}$

to equations that cannot be solved by elementary methods, such as

$\displaystyle x e^x = \sin x$

What are the basic equation/formulas of Algebra?

There are many, perhaps too many to mention:

• Step 1: We have different types of equations, such as linear, quadratic and polynomial equations
• Step 2: Aside from equations (which are satisfied by only some values of x), we have different algebraic identities, which hold for all values
• Step 3: The basic identities in Algebra are the binomial expansion (a+b)2 = a 2 + 2ab + b2, the difference of squares: a 2 - b2 = (a+b)(a-b), just to mention a few

The big difference between algebra equations and identities is that identities are expressions that hold for all values that you plug in, whereas equations will hold only for selected few values. Usually, you will USE identities to SOLVE equations.

What is a basic Algebra equation?

There many types of most basic algebra equation is the linear equation. For example, for one variable, the linear equation is:

$\displaystyle a x + b = c$

Observe that the left side corresponds to $$ax + b$$, which is a linear function. This type of function has a strong geometric interpretation, as it is tightly related to a geometric line, where $$a$$ corresponds to the slope, and $$b$$ to the y-intercept.

What are some uses for Algebra Equations

• Step 1: Algebra Equations encapsule relationship between variables. Solving an equation usually lead to a very singular point in the interaction of elements
• Step 2: By using equations, we get to quantify things, and are able to talk specific about variables
• Step 3: Equations are usually the key to great things: points of equilibrium, points of maximum gain, points of least resistence, etc.

Therefore, we want to have equations. One little problem is that equations can be hard to solve. Using an equation solver with steps can prove crucial at the time of tackling the harder equations that we will inevitably will find.

What is the most popular equation in algebra?

Depends on who asks. For some, the most popular equation is that easiest one, which without a doubt is the linear equation. But if you ask a mathematician, they will tell you something different.

Some purists will tell you that this is the most popular formula in Algebra:

$\displaystyle e^{-i \pi} + 1 = 0$

because it uses ALL the most important math symbols. Points of view, huh?

Example: Linear Equations

Solve the following linear equation: $$2x + 3y = \frac{1}{6}$$

Solution: We need to solve the following given linear equation:

$2x+3y=\frac{1}{6}$

The linear equation has two variables, which are $$x$$ and $$x$$, so the objective is to solve for $$x$$.

Putting $$y$$ on the left hand side and $$x$$ and the constant on the right hand side we get

$\displaystyle 3y = -2x + \frac{1}{6}$

Now, solving for $$y$$, by dividing both sides of the equation by $$3$$, the following is obtained

$\displaystyle y=-\frac{2}{3}x+\frac{\frac{1}{6}}{3}$

and simplifying we finally get the following

$\displaystyle y=-\frac{2}{3}x+\frac{1}{18}$

Therefore, the solving for $$x$$ for given linear equation leads to $$y=-\frac{2}{3}x+\frac{1}{18}$$.

Example: Quadratic Equations

Resolve the following quadratic equation: $$2x^2 + \frac{5}{4}x - \frac{1}{6} = 0$$

Solution: We need to solve the following given polynomial equation:

$2x^2+\frac{5}{4}x-\frac{1}{6}=0$

The equation we need to solve has only one variable, which is $$x$$, so the objective is to solve for it.

Observe that the degree of the given polynomial is $$\displaystyle deg(p) = 2$$, its leading coefficient is $$\displaystyle a_{2} = 2$$ and its constant coefficient is $$\displaystyle a_0 = -\frac{1}{6}$$.

We need to solve the following given quadratic equation $$\displaystyle 2x^2+\frac{5}{4}x-\frac{1}{6}=0$$.

Using the Quadratic Formula

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 $$\displaystyle 2x^2+\frac{5}{4}x-\frac{1}{6} = 0$$, which implies that corresponding coefficients are:

$a = 2$ $b = \frac{5}{4}$ $c = -\frac{1}{6}$

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

$\Delta = b^2 - 4ac = \displaystyle \left( \frac{5}{4}\right)^2 - 4 \cdot \left(2\right)\cdot \left(-\frac{1}{6}\right) = \frac{139}{48}$

Since in this case we get the discriminant is $$\Delta = \displaystyle \frac{139}{48} > 0$$, 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:

$x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \displaystyle \frac{-\frac{5}{4} \pm \sqrt{\left(\frac{5}{4}\right)^2-4\left(2\right)\left(-\frac{1}{6}\right)}}{2\cdot 2} = \displaystyle \frac{-\frac{5}{4} \pm \sqrt{\frac{139}{48}}}{4}$

so then, we find that:

${x}_1 = -\frac{\frac{5}{4}}{4}-\frac{1}{4}\sqrt{\frac{139}{48}}=\frac{-5}{4\cdot 4}-\frac{1}{4}\sqrt{\frac{139}{48}}=-\frac{5}{16}-\frac{1}{4}\sqrt{\frac{139}{48}}=-\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$ ${x}_2 = -\frac{\frac{5}{4}}{4}+\frac{1}{4}\sqrt{\frac{139}{48}}=\frac{-5}{4\cdot 4}+\frac{1}{4}\sqrt{\frac{139}{48}}=-\frac{5}{16}+\frac{1}{4}\sqrt{\frac{139}{48}}=\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$

In this case, the quadratic equation $$\displaystyle 2x^2+\frac{5}{4}x-\frac{1}{6} = 0$$, has two real roots, so then:

$\displaystyle 2x^2+\frac{5}{4}x-\frac{1}{6} = 2 \left(x+\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)\left(x-\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)$

so then the original polynomial is factored as $$\displaystyle p(x) = 2x^2+\frac{5}{4}x-\frac{1}{6} = 2 \left(x+\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)\left(x-\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)$$, which completes the factorization.

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

$\displaystyle p(x) = 2x^2+\frac{5}{4}x-\frac{1}{6} = 2 \left(x+\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)\left(x-\frac{1}{48}\sqrt{139}\sqrt{3}+\frac{5}{16}\right)$

The roots found using the factorization process are $$-\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$$ and $$\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$$ .

Therefore, solving for $$x$$ for the given polynomial equation leads to the solutions $$x = \,$$$$-\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$$, $$\frac{1}{48}\sqrt{139}\sqrt{3}-\frac{5}{16}$$, using factorization mathods.

Other useful equation calculators calculators

Linear equations are by far the easiest ones. You will find a lot more difficulties solving trigonometric equations, or any non-linear equation that is not a polynomial equation, even though polynomial equations still can be very hard to solve.

You will learn that different types of equations follow different rules. You can use for example an exponential equation calculator so to exploit the properties of exponents to solve specific equations.

The same goes if you try to solve a logarithmic equation, where specific structures of the logarithmic function will make the equation solving process easier.

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