Algebraic Expressions:
- Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
- Some examples of algebraic expressions are 5x, 2x – 3, x2 + 1, etc.
Equation:
- Any mathematical expression equating one algebraic expression to another is called as an equation.
- Some examples of equations are 5x = 25, 2x – 3 = 9, x2 + 1 = 0, etc.
Linear Equation:
- An equation of the form ax + b = 0 where a, b are real numbers such that ‘a’ should not be equal to zero is called a linear equation.
- Remember, the highest power of the variable in these expressions is 1.
- Examples of some linear equations are 2x, 2x+7, 16 – 7y, etc.
- Examples of some non-linear equations are x2 + 1, y + y2, etc. In these examples, the power of variable is greater than 1, thus they are non-linear equations.
olving Linear Equations:
There are two methods by which the linear equations can be solved.
(1) Balancing Method
(2) Transposing Method:
(1) Balancing Method
In this method, both the sides of equation are balanced. Let us understand it by an example:
Example: Solve 2x - 10 = 2.
Solution: To balance both the sides, firstly we will add 10 on both the sides of the equation.
- 2x – 10 + 10 = 2 +10,
On solving, we get
- 2x = 12
Further, to balance the equation we will divide both the sides by 2
- 2x / 2 = 12/2
On solving, we get
- x = 6.
- Thus, x = 6 is the required solution.
Verification: Substitute the value of solution obtained in the original equation. And if after substitution, the LHS becomes equal to RHS then the solution obtained is correct.
As for above example, we obtained x = 6.
Now, let us substitute this value into the original equation 2x – 10 = 2. Then,
LHS = 2(6) – 10 = 12 – 10 = 2
RHS = 2
Here, it is observed that LHS = RHS, thus the solution obtained is correct.
(2) Transposing Method:
In this method, constants or variables are transposed from one side of the equation to other until the solution is obtained. Let us understand it by an example:
Example: Solve 2x – 5 = 5
Solution: Firstly, we will transpose the integer -5 from LHS to RHS, thus we will get
2x = 5 + 5
On solving, we get
2x = 10
Now, we will transpose the 2 from LHS to RHS, thus we will get
x = 10 / 2
On solving, we get
x = 5.
Thus, x = 5 is the required solution.
Verification: Again, we will the substitute the obtained solution into the original equation.
Thus, LHS = 2 (5) – 5 = 10 - 5 = 5
RHS = 5
Since, it is observed that LHS = RHS, the solution obtained is correct.
Some Examples:
Example 1: Solve a + 5 = 15.
Solution:
Transposing 5 to RHS, thus we will get
a = 15 – 5
a = 10
Example 2: Solve x/ 2 + 2 = 9/4.
Solution: Try These
Answer :x = 2/4 i.e. x = 1/2
Example 1: Sum of two numbers is 70. One of the numbers is 10 more than the other number. What are the numbers?
Solution: Let us assume the two numbers to be x and y.
Now, it’s given that sum of these two numbers is 70, thus, we can write x + y = 70
Further, it is given that one of the numbers is 10 more than the other. So, let y be 10 more than x, thus, we can write
y = x +10
Substituting value of y, we get,
x + x + 10 = 70
2x + 10 = 70
Transposing 10 on RHS,
2x = 70 – 10
2x = 60
Transposing 2 on RHS,
2x = 60
x = 30.
Now, the other number y will be x + 10 = 30 + 10 = 40 i.e. y = 40
Hence, the two desired numbers are 30 and 40.
Example 2: The sum of three consecutive multiples is 60. What are these integers?
Solution: Let the three consecutive integers be a, a+1 and a+2.
Given, a + a + 1 + a + 2 = 60
3a + 3 = 60
Transposing 3 on LHS, we get,
3a = 60 – 3
3a = 57
On solving, a = 57/3, we get
a = 19
a + 1 = 20 and a + 2 = 21
Thus, the three consecutive integers are 19, 20 and 21
Example 3: Rahul’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution: Let the age of Rahul’s son be x. Hence, his age will be 3x.
Given, 10 years ago Rahul’s age was five times his son’s age. So, we can write
(3x – 10) = 5 (x – 10)
3x – 10 = 5x – 50
Transposing 5x to LHS and 10 to RHS, we get
3x – 5x = -50 + 10
-2x = -40
Dividing both the sides by 2, we get
So, x = 20
3x = 60.
Thus, present age of Rahul is 60 and his son’s age is 20.
Practice problems
1: x + 7 – 8x/3 = 17/6 – 5x/2 Ans :x= -5
2: Solve (7x + 4)/ (x + 2) = -4/3. Ans x = - 4/5.
3: The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2 . Find the rational number.
Ans: the rational number will be 13/21.
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