Equations are statements of equality.

x + 1 = 2 is a statement of equality between x + 1 and 2.

The statement is:

For what value of x, will x + 1 become equal to 2 ?

Answer: for x = 1.

Replacing x by 1 makes x + 1 equal to 2,

i.e. 1 + 1 = 2.

We say, x = 1 is solution of the equation:

x + 1 = 2.

Replacing x by 1 is called substitution

In the equation x + 1 = 2:

x + 1 is said to be in the Left Hand Side, in short LHS, and 2, in the Right Hand Side, abbreviated as RHS

x is called variable, i,e what changes or varies.

The values of x for which the LHS and the RHS become equal are called Solution Set of the Equation.

We also say solution of x satisfies the equation.

To satisfy an equation means the values of x which make LHS and RHS equal.

“To solve for x” is to find the solution of an equation.

An equation in which the highest exponent of the variable is 1 is called a linear equation.

In x + 1 = 2, the highest exponent of the variable x is 1.

So, we say x + 1 = 2 is a linear equation in variable x.

Other examples of linear equations are:

x + 3 = 5, in which the highest exponent of the variable x is 1

p + 9 = 2, in which the highest exponent of the variable p is 1

a + b = 1, in which the highest exponents of the two variables a and b are 1

Equations in which the highest exponent of the variable is 2 are called quadratic equations .

Examples of quadratic equations are:

x^{2} + 2x + 1 = 0, in which the highest exponent of the variable x is 2

x^{2} – 4 = 0, in which the highest exponent of the variable x is 2

y^{2} – 4ax = 0, in which the highest exponent of the variable y is 2

6p^{2} – p - 1 =0, in which the highest exponent of the variable p is 2.

If you wish to set off with your lesson on **Equations**, then click on any of the links below:

**Linear Equation in One Variable:****Solution of an Equation:****How to Find the Solution of Linear Equations in One Variable:****By the Method of Transposition:****Simultaneous Linear Equations****Elimination of one of the variables through substitution****Elimination of one of the variables by making coefficients of any one variable equal****Consistent Equations having unique solutions****Consistent Equations having infinitely many solutions****Inconsistent Equations having No solution****How to check if equations do not have any solution:****Homogenous Equations:**

- Homogenous equations have infinitely many solutions, if ratios of x and y coefficients are equal.
- Homogenous equations have only one solution, namely x = 0 and y = 0, if the ratios of the x and y coefficients are not equal.
- Equations Reducible to Simultaneous Liner Equations:

Or, if you wish to capture a terse overview of each Equations Formula, then go through each of the following header-links.

You can also click the header-links to take you to the page discussing in detail the specific Equations concept:

Example 1:

Solve for x:

x + 2 = 3

Solution:

Transpose 2 to RHS, and reverse its sign.

x = 3 – 2 = 1, so x = 1 is the solution.

Solving two equations x + y = 3 and x – 2y = 1 for the two different variables x and y so that the solution satisfies both the given equations are called simultaneous equations.

**Methods of solving simultaneous equations:**

Solve the equations:

x – y = 2 and x + y = 1.

Solution:

From the first equation, transpose y to the right by inversing its sign:

x = y + 2,

substitute this value of x, i.e. y + 2 in x in the second equation:

y + 2 + y = 1,

so that, 2y = 1 – 2, 2y = -1, y = -1/2

now, substitute y = -1/2 in any of the two equations to solve for x:

(let us substitute in the first equation)

x – (-1/2) = 2, i.e. x + ½ = 2, so that x = 2 – ½ = 3/2

so, the solution, i.e. values of the two variables x and y, for the pair of equations is x = 3/2 and y = -1/2

Solve the system of equations:

2x – 3y = 4 and 3x – 4y = 8.

Solution:

Let us eliminate y by making coefficients of y in the two equations equal.

Multiply first equation with 4, the coefficient of y (disregard the minus sign) in the 2^{nd} equation, and the second with 3, the coefficient of y (here too disregard the minus sign) in the 1^{st} equation as below:

4 × (2x – 3y = 4), i.e. 8x – 12y = 16, and

3 × (3x – 4y = 8), i.e. 9x – 12y = 24.

Now subtract the new 2^{nd} equation from the new 1st, and you will see the y-term gone.

[8x – 12y – 16] – [9x – 12y – 24] = 0

i.e. –x + 8 = 0, i.e. –x = - 8 , i.e. x = 8.

Now, substitute x = 8 in any of the two equations, say in the 1st (the original) to solve for y:

2 × 8 – 3y = 4,

i.e. -3y = 4 – 16,

i.e. -3y = -12,

3y = 12, i.e. y = 12/3 = 4.