Algebra equations can be useful to solve everyday problems. You've probably used the principles of algebraic equations without being aware of it. For example, consider the following problem:
What's the number that should go into the space? Of course, you figured out that it was 2. However, it was likely that you were drawing from your own experience or filling in the blank by trial and error. How do you use equations to solve this problem quickly and reliably?
Step 1: Use an unknown/variable to represent the blank space.
A unknown can be written as an alphabet. We'll use "x" for this example.
Step 2: Subtract 1 from both sides of the equation.
$$\begin{array}{ccc}x+11& =& 31\\ x& =& 2\end{array}$$So we know that x (or blank space) is 2.
What happened?
In an algebraic equation, both sides of the equal sign (=) must be balanced in value. So if you add something to one side, you need to add the same thing to the other. If you subtract from, divide or multiply to one side, you need to do the identical step to the other. The idea is that if you do an operation on the terms on one side, you need to do it on the other. To solve an equation, use this rule to make sure the unknown is on one side of the equation by itself, and everything else is on the right.
Example 1

Solution$$\begin{array}{ccc}x5& =& 1\\ x5+5& =& 1+5\\ x& =& 6\end{array}$$ 
Example 2

Solution$$\begin{array}{ccc}6+x& =& 2\\ 6+x6& =& 26\\ x& =& 4\end{array}$$ 
Example 3

Solution$$\begin{array}{ccc}\frac{x}{9}& =& 5\\ \frac{x}{9}\times 9& =& 5\times 9\\ x& =& 45\end{array}$$ 
Example 4

Solution$$\begin{array}{ccc}5x& =& 10\\ \frac{5x}{5}& =& \frac{10}{5}\\ x& =& 2\end{array}$$ 
Example 5

Solution$$\begin{array}{ccc}{x}^{\frac{1}{3}}& =& 3\\ {({x}^{\frac{1}{3}})}^{3}& =& {3}^{3}\\ x& =& 27\end{array}$$ 
Example 6

Solution$$\begin{array}{ccc}{x}^{2}& =& 4\\ \sqrt{{x}^{2}}& =& \sqrt{4}\\ x& =& \pm 2\end{array}$$ 
Example 7
Solve, $$2x\frac{x}{3}=x7$$ 
Solution$$\begin{array}{ccc}2x\frac{x}{3}& =& x7\\ 2x\frac{x}{3}x& =& x7x\\ 2x\frac{x}{3}x& =& 7\\ x(2\frac{1}{3}1)& =& 7\\ x(\frac{6}{3}\frac{1}{3}\frac{3}{3})& =& 7\\ \frac{2}{3}x& =& 7\\ \frac{2}{3}x\cdot \frac{3}{2}& =& 7\cdot \frac{3}{2}\\ x& =& \frac{21}{2}\\ & & \end{array}$$ 
Example 8
Solve, 
Solution$$\begin{array}{ccc}\sqrt{\frac{3y7}{5}}& =& 2\\ \frac{3y7}{5}& =& {2}^{2}\\ \frac{3y7}{5}\cdot 5& =& 4\cdot 5\\ 3y7& =& 20\\ 3y7+7& =& 20+7\\ 3y& =& 27\\ \frac{\mathit{3y}}{3}& =& \frac{27}{3}\end{array}$$Hence, 
The following method also solves algebra equations, with less writing.
Basically, each mathematical operator has what I call an "equation pair". That is, if you bring a an algebraic term from one side of the equation to the other, you need to swap the operator to its pair.
+ is paired with –
× is paired with ÷
Let's try an example. Follow each step and explanation carefully, and you'll get to see how these pairs can be used to solve the algebraic equation.
ExampleSolve for the following 
Solution$$\begin{array}{ccc}\frac{9x4}{x+1}& =& 3\\ 9x4& =& 3(x+1)\\ 9x4& =& 3x\text{}+\text{}3\\ 9x3x& =& 3+4\\ 6x& =& 7\\ x& =& \frac{7}{6}\end{array}$$ 
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