Linear equations in one variable
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This document provides examples and explanations for solving linear equations in one variable. It begins by defining a linear equation as an equation that can be written in the form Ax + B = C, where A, B, and C are real numbers. It then gives 4 examples of linear equations. The document explains that a linear equation has a left-hand side (LHS) and right-hand side (RHS) that are always equal. It provides steps for solving linear equations, including using inverse operations and the order of operations. Finally, it works through 5 example problems of solving linear equations.
Linear equations in one variable
- 1. By Anuja Gupta VIII C Roll.No.04 KV, OFD Raipur, Dehradun
- 2. Any equation that can be written in the form Ax + B = C, where A, B, C are real numbers, is referred to as a linear equation in one variable. For example:- 5x + 15 = 35 4x + 12 = 3x + 2 (x – ½) * ½ = 1/8 3x – 6x = -12 are linear equations in one variable.
- 3. The linear equation may be divided in two parts the RHS or the Right Hand Side and the LHS or the Left Hand Side. RHS LHS5x + 15 = 35 Both the sides of a Linear equation are always equal. 5x + 15 35
- 4. To solve a linear equation our aim is to make LHS equal to the variable than the RHS is our solution…. As simple as that !!! To do this remember that Addition and subtraction are inverse operations, and so are Division and multiplication. To open a bracket use the distributive property. Follow the BODMAS rule. Now let us take some examples ….
- 5. Problem 1: 5x + 9 = 29 find x To make LHS equal to the variable that is ‘x’ we have to eliminate 5 and 9 To eliminate 9 subtract 9 from both sides 5x + 9 – 9 = 29 – 9 => 5x = 20 To eliminate 5 divide both sides by 5 5x/5 = 20/5 => x = 4 Now when LHS = the variable ‘x’ RHS is the answer i.e. 4 To check the validity of the answer substitute x = 4 in the original equation 5 * 4 + 9 = 20 + 9 = 29 Hence x = 4 is the correct solution.
- 6. Problem 2: (2 + a) * 6 = (14 + a) 2 find ‘a’ Firstly we will open the brackets 12 +6a = 28 + 2a To remove 2a from RHS Subtract 2a from both the sides 12 + 6a -2a = 28 + 2a – 2a => 12 + 4a = 28 To eliminate 12 Subtract 12 from both the sides 12 - 12 +4a = 28 – 12 => 4a = 16 To eliminate 4 divide both sides by 4 4a/4 = 16/4 => a = 4 Check the answer by substituting a = 4 in the original equation.
- 7. Problem 3: I have some 2 rupee coins and 5 rupee coins in my pocket. The number of 2 rupee coins is double than that of 5 rupee coins. If I have total 45 Rs. In my pocket find the number of each type of coins. Let I have ‘x’ coins of Rs. 5 then the number of Rs. 2 coins will be double that ‘x’ i.e. 2x. Also I have Rs. 45 in total… => 5 * x + 2 * 2x = 45 => 5x + 4x = 45 => 9x = 45 => x = 45/9 or x = 5 That I have 5 coins of Rs. 5 and 10 coins of Rs. 2 Now check the answer 5 * 5 +2 * 10 = 25 + 20 = 45 that was what I had in my pocket.
- 8. Problem 4: The length of a swimming pool is 9/4 times its width. If its perimeter is 26 meters calculate its area. x Solution: Let the length of the pool be ‘x’ meter. Then its width will be x*9/4 or 9x/4 meter. Perimeter of the pool is 26 meter, and the Relation between perimeter and area is 2(l+b) So now we have a linear equation => 2(x+9x/4) = 26 => 2x + 9x/2 = 26 => 4x + 9x = 52 => 13x = 52 => 13x/13 = 52/13 => x = 4 Hence width of the pool is 4 meter while length is 9 meter.
- 9. Problem 5: The age of father is 6 times than the son, after 12 years father will be thrice as old as his son. Calculate their present ages. Solution: Present age Age after 12 years Son x x + 12 Father 6x 6x + 12 According to the question (x + 12) *3 = 6x + 12 => 3x + 36 = 6x + 12 => 3x = 24 => x = 24/3 or x = 8 So the present age of son is 8 years and that of father is 48 years.
- 10. Solve: 5x + 6x = 9 7x – 2x = -25 y/12 = 5 1/3 z = 19 6/7 t = 9/5 5x – 6 = 17 -4x + 2 = 9
- 11. To solve x + 3 = 7, you subtract 3 from both sides. To solve x - 3 = 7, what do you do? Addition and subtraction are inverse operations. We always use inverse operations to solve equations. Back
- 12. A Project Made by Anuja Gupta VIII C Roll No. 04 KV, OFD Raipur Dehradun I am thankful to Wikipedia Microsoft Art Anuja Gupta Narrator Ananya Gupta Direction A.K. Gupta The End