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A statement involving the ‘equals’ symbol (‘=’) is called a statement of equality or simply an equality.
Equation:
A statement of equality which involves one or more literals (variables) is an equation.
Clearly, each of the equalities x + 3 = 8, y – 5 = 12, x – 7 = 4, 5p = 32, 3m – 4 = 14 is an equation.
Every equation has two sides, namely, the left hand side (written as L.H.S.) and the right hand side (written as R.H.S.)
In the equation x + 3 = 8, x + 3 is L.H.S. and 8 is R.H.S, whereas in the equation 3m – 4 = 14, 3m – 4 is L.H.S. and 14 is R.H.S. The literals involved in an equation are called variables or unknowns. Usually the variables are denoted by letters from the later part of the English alphabet, e.g. x, y, z, u, u, w etc.
An equation may contain any number of variables. The equation x + 3 = 8 has only one variable whereas in equation 2x — 3y = 5 there are two variables x and y.
Linear Equation:
An equation in which the highest power of the variables involved is 1, is called a linear equation.
For example, equations like 4x – 9 = 7, \(\frac{y}{3} + 6 = 2\), \(\frac{x}{2} + \frac{y}{5} = 8\) are linear equations. The equations \(2x^2 + x = 1\), \(y + 5 = y^2\) and \(x^3 = 8\) are not linear equations, because the highest power of the variable in each equation is greater than one.
Solution of an equation:
In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation.
Consider the linear equation 2x – 8 = -14.
x |
L.H.S |
R.H.S |
-2 |
2(-2) – 8 = -12 |
-14 |
-3 |
2(-3) – 8 = -14 |
-14 |
-4 |
2(-4) – 8 = -16 |
-14 |
From the above table, we observe that the LHS equals the RHS only when we substitute -3 for x. For all other values of x, the two sides are not equal. In other words, the equation is satisfied by x = -3. Such a value of the variable is called the solution of the equation.
Linear Equations in One Variable Example1:
Verify that x = 5, is the solution of the equation 3x + 7 = 22.
Solution:
Putting x = 5 on L.H.S., we have,
L.H.S. = 3(5) + 7 = 15 + 7 = 22
R.H.S. = 22
Thus, for x = 5, we have L.H.S. = R.H.S.
Hence, x = 5 is the solution of the equation 3x + 7 = 22.
Linear Equations in One Variable Example2:
Verify that y = 8, is the solution of the equation \(\frac{y}{4} + 1 = 3\).
Solution:
Putting y = 8 on L.H.S., we have,
L.H.S. = \(\frac{y}{4} + 1\) = \(\frac{8}{4} + 1\)
= 2 + 1 = 3
R.H.S. = 3
Thus, for y = 8, we have L.H.S. = R.H.S.
Hence, y = 8 is the solution of the equation \(\frac{y}{4} + 1 = 3\).