Equations are found everywhere in mathematics. Students of middle school are introduced by the equations in algebra. An **algebraic equation** is a combination of one or more terms separated with "**equal**" symbol **"="**. The terms are the expressions or monomials made up of constants and variables. The terms can be numerical, alpha numerical, expression etc. The terms are connected with one another with the help of addition (+) or subtraction (-) symbols.**For example -(1)** x + 2 = -3

There are different types of equations, such as - linear equations in one and two variables, logarithmic equations, exponential equations, fractional equations, polynomial equations etc. Equations represent the relationship between variables. There may be one or more variables in an equation.

By solving equations, we mean to find all the possible values of one or more variables contained in it. Equations can be solved either algebraically or graphically. There are various algebraic methods that can be utilized in order to get the solution of an equation. The choice of these methods may depend upon the types of equation. In order to find the values of all the variables in the equation, we need as many number of same types of equations as the total number of variables.

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Equation is a mathematical statement that states that the values of two mathematical expressions are equal (denoted by the sign **=**). An equation says that two things are the same.

For Example: 2 + 5 = 1 + 6

=> 2 + 5 = 7

and 1 + 6 = 7

Both the values are same.

### Linear Equations:

### Quadratic Equations:

### Radical Equations:

### Rational Equations

As we being solving equations, whatever we do to one side, we**must **do to the other side of the equation.

Let us solve for x, x + 4 = 10

To keep the scale balanced, add or remove anything on one side, and must do the same thing on the other side.

Given equation is x + 4 = 10

To isolate x, subtract 4 from both side.

=> x + 4 - 4 = 10 - 4

=> x = 6

The value of x is 6.

Two step equations are the equations which are solved in two steps. Solving a two-step equation requires the same procedure as a one-step equation.

Let us solve for x, 3x + 4 = 12

To isolate x, subtract 4 from both side, then divide each side by 3.

**Step 1:**

Subtract 4 from both sides

=> 3x + 4 - 4 = 12 - 4

=> 3x = 8

**Step 2:**

Divide each side by 3

=> $\frac{3x}{3}$ = $\frac{8}{3}$

=> x = $\frac{8}{3}$.

Multi step equations are just as solving one-step or two-step eqautions. Solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. These equations can make use of any letter as a variable.

### Solving Multistep Equations

Let us solve 3x = 2x + 5

Given equation is 3x = x + 5

The equation 3x = x + 5 shows multiplication and addition. To find the value of x, we will need to use their opposites, division and subtraction.

Step 1:

Subtract 5 from both sides

=> 3x - 5 = x + 5 - 5

=> 3x - 5 = x

Step 2:

Subtract x from both sides

=> 3x - 5 - x = x - x

=> 2x - 5 = 0

or 2x = 5

Step 3:

Divide each side by 2

=> $\frac{2x}{2} = \frac{5}{2}$

=> x = $\frac{5}{2}$.

→ Read More The process of writing a equation as a product is called factoring. Let us see with the help of an example how to find the factors of equation.

Let us find the factors of x^{2} - x - 2 = 0

Given equation is a quadratic, x^{2} - x - 2 = 0

x^{2} - x - 2 = 0

Factors of 2 are $\pm$1 and $\pm$2. Choose factors whose sum is equal to middle term of the equation.

=> -2 + 1 = -1 = middle term

=> x^{2} - 2x + x - 2 = 0

=> x(x - 2) + (x - 2) = 0

=> (x + 1)(x - 2) = 0

Therefore, factors of the given equation are (x + 1) and (x - 2).

To solve equation with fraction, firstly transform it into an equation without fractions. Find the lowest common multiple of the denominators and remove the fractions by multiplying both sides of the equation by the LCD. Solve the equation for the unknown variable by performing the same operations to both sides of the equation.

Let us simplify $\frac{2}{x}$ = $\frac{1}{2}$

Given equation is $\frac{2}{x}$ = $\frac{1}{2}$

Multiply both side by 2

=> 2 * $\frac{2}{x}$ = 2 * $\frac{1}{2}$

=> $\frac{4}{x}$ = 1

Multiply each side by x

=> 4 = x

x = 4.

Cubic equation is a equation with degree 3. The general form of cubic equation is f(x) = ax^{3} + bx^{2} + cx + d, a $\neq$ 0. Let us see with the help of example how to solve cubic equations.

Let us find the values of x, 2x^{3} - 2x^{2} - 12x.

Given cubic equation is 2x^{3} - 2x^{2} - 12x.

Step 1:

Factor the GCF from the given equation

GCF of 2x^{3} , 2x^{2} , and - 12x is 2x

2x^{3} - 2x^{2} - 12x = 2x(x^{2} - x - 6)

Step 2:

Solve the quadratic equation:

x^{2} - x - 6 = x^{2} - 3x + 2x - 6

= x(x - 3) + 2(x - 3)

= (x + 2)(x - 3)

Step 3:

Factors of cubic equation is the product of GCF and factors of quadratic equation.

=> 2x^{3} - 2x^{2} - 12x = 2x(x + 2)(x - 3) = 0

2x = 0, x + 2 = 0 or x - 3 = 0

=> x = 0, x = -2 or x = 3

Thus, the values of cubic equation are -2, 0 and 3.

Pre algebra Help Online is the best way to learn the subject. It is convenient, effective, personalized and easy and it can be done from the comfort of home. Pre algebra equation solvers are a huge help for exams and tests as well. Some sites also have test prep sections with mock tests which are scored to give you regular feedback about your performance. Pre algebra practice problems are the most effective way to learn and understand these concepts.

Let us solve x^{2} - 4 = 0

Given x^{2} - 4 = 0

x^{2} - 4 = 0 = x^{2} - 2^{2} = 0

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

=> (x - 2)(x + 2) = 0

x - 2 = 0 or x + 2 = 0

=> x = 2 or x = -2

The values of x are -2 and 2.

→ Read More A system of equations will have two or more variables with the same variables. A solution of the system is the ordered values for the variables which satisfy all the equations forming the system.

The systems of equations are categorized into two types on the basis of their solutions.

** Inconsistent system**

Let us see with the help of example how to solve a system of linear equations in two variables.

Solve linear system: 4x - 2y = 9 and 6x + y = 70

Given system:

4x - 2y = 9 .............................(i)

6x + y = 70 ............................(ii)

Apply Addition Method:

Step 1:

Multiply equation (ii) by 2, the coefficient of y in the resulting equation and equation (i) will be opposites.

ii) => 12x + 2y = 140 ...........................(iii)

Adding (i) and (iii), we have

4x - 2y = 9

12x + 2y = 140

----------------------------

16x + 0y = 149

----------------------------

=> 16x = 149

=> x = $\frac{149}{16}$

Step 2:

Plug the value of x in equation (1), we get

4 * $\frac{149}{16}$ - 2y = 9

=> $\frac{149}{4}$ - 2y = 9

2y = $\frac{149}{4}$ - 9 = $\frac{149 - 36}{4}$ = $\frac{113}{4}$

=> y = $\frac{113}{8}$

The solution for the linear system is ($\frac{149}{16}$, $\frac{113}{8}$).

→ Read More Graphs are very important for giving a visual representation of the relationship between two variables in an equation. There are many types of equations. A linear equation is a equation whose graph is straight line. Quadratic equation, whose graph is "U" shaped. From the form of the equation we can determine the shape of function.

Let us draw the graph for y = x^{2} - 1.

Given equation is y = x^{2} - 1

Draw the function table for the values of x and y

**Graph:**

Given below are some of the examples in solving equations:### Solved Examples

**Question 1: **Solve 10x + 2 = 8 - 2x

** Solution: **
**Question 2: **Factorize x^{2} + 6x - 40 = 0

** Solution: **
Given below are some of the word problems in equations. ### Solved Examples

**Question 1: **The perimeter of an isosceles triangle is 20 cm and length of one of the sides is 4 cm. Find the lengths of the other two equal sides.

** Solution: **
**Question 2: **Find two numbers whose sum is 20 and whose difference is 9.

** Solution: **

For Example:

=> 2 + 5 = 7

and 1 + 6 = 7

Both the values are same.

Equation is one of the most important calculations in algebra and the most essential concept of math. It deals mainly with algebraic expressions. **There are different types of equations:** Linear, radical, rational, exponential and quadratic equations. Learn about all these concepts with our tutor, understand the depth of the topics and get a hold over the whole concept of equations. Below stated are the different types of algebra equations frequently asked by students and how our tutors help you make them easier.

Linear equation is any equation that when graphed produces a straight line. There are various types of equations; none except linear equations will produce a straight line when graphed. Get help in solving linear equations and find easy steps to follow. Students can learn about graphing of various equations like the graph of linear equations also.

The equations that are having the highest degree term x^{2} or second degree are called as Quadratic Equations. Learn how to solve quadratic equations with an online tutor, and make the topic easy and interesting.

Radical are also called as roots of an expression. Get help in solving radical expressions and solve with easy steps and methods.

Rational equations are the equations having a variable in the denominator. Get help from tutors in solving linear equations and find easy steps to follow.

Type of Equation | General Form | Description | Example |

Linear |
y = ax + b | Equation with variables but no exponents | y = x - 1 |

Quadratic |
y = ax^{2} + bx + c |
Equation with a squared term | y = 2x^{2} - x + 5 |

Radical | y = $\sqrt{x}$ | Equation with exponent $\frac{1}{2}$ | y= $\sqrt{50}$ |

Rational | y = $\frac{a}{x}$ + b | Equation with a variable in the denominator | y = $\frac{2}{x + 1}$ |

Exponential | y = $a^x$ | Equation with a variable as an exponent | y = $2^x$ |

A number which satisfies the given equation is called a solution or root of that. **"Satisfying the equation"** means that if the variable involved in this is replaced by the number, then both sides of that become equal. The process of finding the particular value of the variable which makes both sides of the equation equal is called solving the equation. Below are rules that need to be followed in solving equations.

**Rules for Solving Equations**

**Rule 1:** If equals are added to equals, the sums are equals. In other words, you can add the same number to both sides of an equation.

**Rule 2: **If equals are subtracted from equals, the remainders are equal. In other words, you can subtract the same number from both sides of an equation.

**Rule 3:** The two sides of an equation may be multiplied by same non-zero number For example, if $\frac{x}{5}$ = 7, then $\frac{x}{5}$ × 5 = 7 × 5.

**Rule 4:** The two sides of an equation may be divided by same non-zero number.

As we being solving equations, whatever we do to one side, we

Let us solve for x, x + 4 = 10

Given equation is x + 4 = 10

To isolate x, subtract 4 from both side.

=> x + 4 - 4 = 10 - 4

=> x = 6

The value of x is 6.

Let us solve for x, 3x + 4 = 12

Subtract 4 from both sides

=> 3x + 4 - 4 = 12 - 4

=> 3x = 8

=> $\frac{3x}{3}$ = $\frac{8}{3}$

=> x = $\frac{8}{3}$.

Multi step equations are just as solving one-step or two-step eqautions. Solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. These equations can make use of any letter as a variable.

Let us solve 3x = 2x + 5

The equation 3x = x + 5 shows multiplication and addition. To find the value of x, we will need to use their opposites, division and subtraction.

Step 1:

Subtract 5 from both sides

=> 3x - 5 = x + 5 - 5

=> 3x - 5 = x

Step 2:

Subtract x from both sides

=> 3x - 5 - x = x - x

=> 2x - 5 = 0

or 2x = 5

Step 3:

Divide each side by 2

=> $\frac{2x}{2} = \frac{5}{2}$

=> x = $\frac{5}{2}$.

→ Read More The process of writing a equation as a product is called factoring. Let us see with the help of an example how to find the factors of equation.

x

Factors of 2 are $\pm$1 and $\pm$2. Choose factors whose sum is equal to middle term of the equation.

=> -2 + 1 = -1 = middle term

=> x

=> x(x - 2) + (x - 2) = 0

=> (x + 1)(x - 2) = 0

Therefore, factors of the given equation are (x + 1) and (x - 2).

To solve equation with fraction, firstly transform it into an equation without fractions. Find the lowest common multiple of the denominators and remove the fractions by multiplying both sides of the equation by the LCD. Solve the equation for the unknown variable by performing the same operations to both sides of the equation.

Multiply both side by 2

=> 2 * $\frac{2}{x}$ = 2 * $\frac{1}{2}$

=> $\frac{4}{x}$ = 1

Multiply each side by x

=> 4 = x

x = 4.

Cubic equation is a equation with degree 3. The general form of cubic equation is f(x) = ax

Step 1:

Factor the GCF from the given equation

GCF of 2x

2x

Step 2:

Solve the quadratic equation:

x

= x(x - 3) + 2(x - 3)

= (x + 2)(x - 3)

Step 3:

Factors of cubic equation is the product of GCF and factors of quadratic equation.

=> 2x

2x = 0, x + 2 = 0 or x - 3 = 0

=> x = 0, x = -2 or x = 3

Thus, the values of cubic equation are -2, 0 and 3.

Pre algebra Help Online is the best way to learn the subject. It is convenient, effective, personalized and easy and it can be done from the comfort of home. Pre algebra equation solvers are a huge help for exams and tests as well. Some sites also have test prep sections with mock tests which are scored to give you regular feedback about your performance. Pre algebra practice problems are the most effective way to learn and understand these concepts.

x

[Using Identity a

=> (x - 2)(x + 2) = 0

x - 2 = 0 or x + 2 = 0

=> x = 2 or x = -2

The values of x are -2 and 2.

→ Read More A system of equations will have two or more variables with the same variables. A solution of the system is the ordered values for the variables which satisfy all the equations forming the system.

The systems of equations are categorized into two types on the basis of their solutions.

**Consistent system**

A consistent system is subdivided into two types

- System with unique solution
- System with many solutions

- Inconsistent systems have no solution

Let us see with the help of example how to solve a system of linear equations in two variables.

Solve linear system: 4x - 2y = 9 and 6x + y = 70

4x - 2y = 9 .............................(i)

6x + y = 70 ............................(ii)

Apply Addition Method:

Step 1:

Multiply equation (ii) by 2, the coefficient of y in the resulting equation and equation (i) will be opposites.

ii) => 12x + 2y = 140 ...........................(iii)

Adding (i) and (iii), we have

4x - 2y = 9

12x + 2y = 140

----------------------------

16x + 0y = 149

----------------------------

=> 16x = 149

=> x = $\frac{149}{16}$

Step 2:

Plug the value of x in equation (1), we get

4 * $\frac{149}{16}$ - 2y = 9

=> $\frac{149}{4}$ - 2y = 9

2y = $\frac{149}{4}$ - 9 = $\frac{149 - 36}{4}$ = $\frac{113}{4}$

=> y = $\frac{113}{8}$

The solution for the linear system is ($\frac{149}{16}$, $\frac{113}{8}$).

→ Read More Graphs are very important for giving a visual representation of the relationship between two variables in an equation. There are many types of equations. A linear equation is a equation whose graph is straight line. Quadratic equation, whose graph is "U" shaped. From the form of the equation we can determine the shape of function.

Let us draw the graph for y = x

Given equation is y = x

Draw the function table for the values of x and y

x | y = x^{2} - 1 |
y |

0 | = 0^{2} - 1 |
-1 |

1 | = 1^{2} - 1 |
0 |

2 | = 2^{2} - 1 |
3 |

-1 | = (-1)^2 - 1 |
0 |

-2 | = (-2)^2 - 1 |
3 |

Given below are some of the examples in solving equations:

Given 10x + 2 = 8 - 2x

Subtract 2 from both the sides

=> 10x + 2 - 2 = 8 - 2x - 2

=> 10x = 6 - 2x

Add 2x to both the sides

=> 10x + 2x = 6 - 2x + 2x

=> 12x = 6

To isolate x, divide each side by 12

=> x = $\frac{6}{12}$ = $\frac{1}{2}$.

Subtract 2 from both the sides

=> 10x + 2 - 2 = 8 - 2x - 2

=> 10x = 6 - 2x

Add 2x to both the sides

=> 10x + 2x = 6 - 2x + 2x

=> 12x = 6

To isolate x, divide each side by 12

=> x = $\frac{6}{12}$ = $\frac{1}{2}$.

Given equation is quadratic, x^{2} + 6x - 40 = 0

**Step 1:**

=> x^{2} + 6x - 40 = 0

=> x^{2} + 10x - 4x - 40 = 0

=> x(x + 10) - 4(x + 10) = 0

=> (x - 4)(x + 10) = 0

Therefore, factors of given equation are (x - 4) and (x - 10).

Step 2:

Solve for x,

x - 4 or x + 10 = 0

=> x = 4 or x = -10

The values of quadratic equation are -10 and 4.

=> x

=> x

=> x(x + 10) - 4(x + 10) = 0

=> (x - 4)(x + 10) = 0

Therefore, factors of given equation are (x - 4) and (x - 10).

Step 2:

Solve for x,

x - 4 or x + 10 = 0

=> x = 4 or x = -10

The values of quadratic equation are -10 and 4.

Let the lengths of equal sides of the triangle = x

and length of other side of the triangle = y = 4

Step 1:

Given, perimeter of isosceles triangle = 20

Perimeter of isosceles triangle = 2x + y

=> 2x + y = 20

=> 2x + 4 = 20

Step 2:

Solve for x,

Subtract 4 from both the sides

=> 2x + 4 - 4 = 20 - 4

=> 2x = 16

Divide each side by 2

=> x = 8

Hence, the lengths of the other two equal sides = 8 cm.

and length of other side of the triangle = y = 4

Step 1:

Given, perimeter of isosceles triangle = 20

Perimeter of isosceles triangle = 2x + y

=> 2x + y = 20

=> 2x + 4 = 20

Step 2:

Solve for x,

Subtract 4 from both the sides

=> 2x + 4 - 4 = 20 - 4

=> 2x = 16

Divide each side by 2

=> x = 8

Hence, the lengths of the other two equal sides = 8 cm.

Let the larger number = x

Then, the smaller number = 20 - x

(Sum of two numbers is 20)

Now, x - (20 - x) = 9

=> x - 20 + x = 9

=> 2x = 9 + 29

=> 2x = 38

=> x = 19

and 20 - 19 = 1

Hence, the numbers are 1 and 19.

Then, the smaller number = 20 - x

(Sum of two numbers is 20)

Now, x - (20 - x) = 9

=> x - 20 + x = 9

=> 2x = 9 + 29

=> 2x = 38

=> x = 19

and 20 - 19 = 1

Hence, the numbers are 1 and 19.

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Line of Best Fit | Equation Word Problems |

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