# Linear equations in one variable

Abhaya GuptaCNI Boys Inter College, Dehra Dun
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### 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
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