Class 6 Linear Equations
Introduction to Linear Equation
Solving a Linear Equation by Trial and Error Method
Rules to Follow for Linear Equation
Introduction to Linear Equation
An equation in which the highest power of the variables involved is 1, is known as linear equation. In other words, an equation of the form of ax + b = c, where a, b, c are constants, a ≠ 0 and 'x' is the variable, is called a linear equation in the variable 'x'.
The equal sign in an equation divides it into two sides. Both the sides are known as left hand side (LHS) and right hand side (RHS). A number which when substituted for the variable in an equation, makes LHS = RHS and is known as solution of the equation.
Solving a Linear Equation by Trial and Error Method
In this method, generally we make a guess of the solution of the equation. We try several values of the variable and find LHS and RHS. When LHS = RHS for a particular value of the variable, we say that it is a solution of the equation. Let’s see some examples.
Example 1. Find by trial and error method the solution of the equation 5p = 15.
Solution. We try all the values of 'p' and find the LHS and RHS. We stop when for a particular value of 'p', LHS = RHS.
p | LHS | RHS |
---|---|---|
1 | 5 x 1 | 5 |
2 | 5 x 2 | 10 |
3 | 5 x 3 | 15 |
Example 2. Find by trial and error method, the equation 2n + 4 = 6 + n.
Solution.
n | LHS | RHS |
---|---|---|
1 | 2 x 1 + 4 = 6 | 6 + 1 = 7 |
2 | 2 x 2 + 4 = 8 | 6 + 2 = 8 |
Rules to Follow for Linear Equation
- We can add the same number to both the sides of the equation.
- We can subtract same number from both sides of the equation.
- We can multiply non-zero number to both the sides of an equation.
- We can divide non-zero number to both the sides of an equation.
Let's solve some linear equations by using above rules.
Example 1. x + 1 = 10. Find the value of 'x'.
Solution x + 1 = 10
Add −1 to both side of the equation.
x + 1 − 1 = 10 − 1
x = 9
Example 2. y − 2 = 22, find the value of 'y'.
Solution. y − 2 = 22
Add 2 to both side of the equation
y − 2 + 2 = 22 + 2
y = 24
Example 3. 4 × a = 12, find the value of 'a'.
Solution. 4 × a = 12
Divide 4 to both side of the equation
4 x a⁄4 = 12⁄4
a = 3
Example 4. n ÷ 5 = 3, find the value of 'n'.
Solution. n ÷ 5 = 3
Multiply 5 to the both side of the equation
n⁄5 × 5 = 3 × 5
n = 15
Rule of Transposition
If term transposed from one side of the equation to the other side, then it's sign will change. Let's see some examples.
Example 1. p + 4 = 12. Find the value of p.
Solution. p + 4 = 12
Transposing +4 from LHS to RHS by changing it's sign
p = 12 - 4
p = 8
Example 2. 2n = 9 − n. Find the value of n.
Solution. 2n = 9 − n
Transposing −n from RHS to LHS by changing it's sign
2n + n = 9
3n = 9
Dividing 3 both sides of LHS and RHS, we shall get the value of n.
3n⁄3 = 9⁄3
n = 3
Example 3. m⁄5 − 4 = 7. Find the value of m.
Solution. m⁄5 − 4 = 7
Transposing -4 from LHS to RHS by changing it's sign
m⁄5 = 7 + 4
m⁄5 = 11
Multiplying 5 to both LHS and RHS
m x 5⁄5 = 11 x 5
m = 55
Example 4. 3(q − 2) = 2(q + 1) − 3. Find the value of q.
Solution.3(q − 2) = 2(q + 1) − 3
First we should remove the brackets.
3q − 6 = 2q + 2 − 3
3q − 2q = 2 − 3 + 6
q = 5
Class-6 Linear Equation Test
Class-6 Linear Equation Worksheet
Answer Sheet
Linear-Equation-AnswerDownload the pdf
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