Definition of an Equation
In an equation, there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign (the left hand side or LHS) is
In the equation: (4x + 5) = 65, the LHS is (4x + 5) and the RHS is 65. In the equation: (10y – 20) = 50, the LHS is (10y – 20) and the RHS is 50.
If there is some sign other than the equality sign between the LHS and the RHS, it is not an equation.
Thus, 4x + 5 > 65 is not an equation. It says that, the value of (4x + 5) is greater than 65.
Similarly, 4x + 5 < 65 is not an equation. It says that the value of (4x + 5) is smaller than 65.
In equations, we often find that the RHS is just a number. But this need not be always so. The RHS of an equation may be an expression containing the variable.
For example, the equation:
4x + 5 = 6x – 25
has the expression (4x + 5) on the left and (6x – 25) on the right of the equality sign.
Thus,
(1) An equation is a condition on a variable. The condition is that two expressions should have equal value.
(2) Note that at least one of the two expressions must contain the variable.
We also note a simple and useful property of equations. The equation 4x + 5 = 65 is the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x +5 is the same as 4x + 5 = 6x – 25. An equation remains the same, when the expressions on the left and on the right are interchanged. This property is often useful in solving equations.
Example 1: Write equations for
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv)One third of a number plus 5 is 8.
(iv)One third of a number plus 5 is 8.
Example 2: Convert the following equations into statement form:
Write alteast one other form for the above equations of: 5p = 20, 3n + 7 = 1 and
Example 3: Consider the following situation:
Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age.
- Let's denote Raju's age by r .Therefore, three times Raju's age is
- And five years more than three times Raju's age is
- So, the equation becomes:
= - We have found the equation.
Example 4: A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100.
Solution:
Let a small box contain 'm' mangoes.
A large box contains 4 more than 8 times m, that is:
Thus,
You can get the number of mangoes in a small box by solving this equation.
Solving An Equation
Consider the equation: 8 − 3 = 4 + 1.
What is the value of both the LHS and RHS ? Both have the value of
Let's try addition: Add 2 to both sides, which gives us
LHS: 8 − 3 + 2 =
RHS: 4 + 1 + 2 =
We see that the equality holds i.e. 7 = 7.
Thus, if we add the same number to both sides of an equality, the equality still holds.
Let's try subtraction: Subtract 2 from both sides, which gives us:
LHS : 8 − 3 − 2 =
RHS : 4 + 1 − 2 =
We see that the equality holds i.e. 3 = 3.
Thus, if we subtract the same number from both sides of an equality, the equality still holds.
Let's try multiplication: Multiply both the sides of the equality by 3, we get
LHS = 3 × (8 – 3) = 3 × 5 =
RHS = 3 × (4 + 1) = 3 × 5 =
We see that the equality holds i.e. 15 = 15.
Let's try division: Divide both sides of the equality by 2, we get
LHS = (8 – 3) ÷ 2 = 5 ÷ 2 =
RHS = (4+1) ÷ 2 = 5 ÷ 2 =
We see that the equality holds i.e.
Similarly, if we multiply or divide both sides of the equality by the same non-zero number, the equality still holds.
If we take any other equality, we shall find the same conclusions.
Suppose, we do not observe these rules. Specificially, suppose we add different numbers, to the two sides of an equality.
We shall find in this case that the equality
For example, let us take again equality 8 – 3 = 4 + 1
Adding 2 to the LHS and 3 to the RHS, we get:
LHS: 8 – 3 + 2 = 5 + 2 =
RHS: 4 + 1 + 3 = 5 + 3 =
The equality does not hold, because the new LHS and RHS are not equal.
Thus if we fail to do the same mathematical operation with same number on both sides of an equality, the equality may not hold.
The equality that involves variables is an equation.
Important Note: These conclusions are also valid for equations, as in each equation variable represents a number only.
As seen earlier at the beginning of the section, often an equation is said to be like a weighing balance.
Doing a mathematical operation on an equation is like adding weights to or removing weights from the pans of a weighing balance.
An equation is like a weighing balance with equal weights on both its pans, in which case the arm of the balance is exactly horizontal.
If we add the same weights to both the pans, the arm remains horizontal. Similarly, if we remove the same weights from both the pans, the arm remains horizontal. On the other hand if we add different weights to the pans or remove different weights from them, the balance is tilted; that is, the arm of the balance does not remain horizontal.
1. Consider the equation: x + 3 = 8
We shall subtract 3 from both sides of this equation.
LHS: x + 3 – 3 =
Thus, we earlier had: LHS =
Now, we have: LHS = x, RHS =
Since this does not disturb the balance, we have New LHS = New RHS (or) x = 5 which is exactly what we want, the solution of the equation.
Validation: To confirm whether we are right, we shall put x = 5 in the original equation. We get LHS = x + 3 = 5 + 3 =
2. Let us look at another equation: 5y = 35
We shall
New LHS =
Original LHS:
New LHS: y, New RHS:
Therefore, y = 7, which is the required solution.
By putting y = 7 in the original equation, we confirm that the solution is correct:
LHS of original equation =
This is equal to the RHS as required.
One can see that in the above examples, the operation we need to perform depends on the equation.
Our attempt should be to get the variable in the equation separated.
Sometimes, for doing so we may have to carry out more than one mathematical operation.
Example 5: Solve (a) 3n + 7 = 25
- Subtracting 7 from both sides, we get 3n =
- We divide 3 on both sides which gives us
= - Thus, we have found the value of n.
(b) 2p – 1 = 23
- Adding 1 to both sides, we get 2p =
- We divide 2 on both sides which gives us
= - Thus, we have found the value of p.
What about the Mind Game?
We are now in a position to go back to the mind-reading game and understand how it works.
We already know that the equation for the mind-game becomes:
Say, we have thought of the number:
Thus, the equation becomes:
Subtracting 8 from both sides:
Which gives us:
Divide both sides by 2: this will separate x.
We get