linear equation in one variable class 8.pptx

linear equation in one variable class 8.pptx

• 1.
  CLASS VIII
• 2.
   Algebraic expressionis the expression having constants and variable. It can have multiple variable and multiple power of the variable.  We already know the below terms from previous class. Lets recall those terms. What is equation? An equation is a condition on a variable. What is variable? A variable takes on different numerical values; its value is not fixed. Variables are denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.
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   An algebraicequation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side.  Example(i)7x-9=16  (ii) 1.5 =y / 1.6
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   We willrestrict the above equation with two conditions a) algebraic equation in one variable b) variable will have power 1 only  Example(i)5x=25 (ii)2x−3=92x  Linear Equation is an equation which is in the form of ax+b=0,ax=b etc. where a and b is real number. x is variable of power one.
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   2x−3=5 3x−11=22 How tosolve Linear equation in one Variable Transpose (changing the side of the number) the numbers to the side where all number are present. We know the sign of the number changes when we transpose it to other side.  Now you will have an equation have variable on one side and number on other side. Add/subtract on both the side to get single term.  Now divide or multiply on both the side to get the value of the variable.
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   Solution: Transposing3 to other side 2x=5+3 or,2x=8 Dividing both the sides by 2 x=4  Ex2. y + 3 = 10 Solution:  Given y+ 3 = 10  By subtracting 3 from both sides, we get  y + 3 – 3 = 10 – 3  ⇒ y = 10 – 3  ⇒ y = 7
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   Linear equationcan be used to solve many word Problem. The procedure is simple First read the problem carefully. Write down the unknown and known  Assume one of the unknown to x and find the other unknown in term of that  Create the linear equation based on the condition given  Solve them by using the above method  Example: When five is added to three more than a certain number, the result is 19. What is the  number? Sol: Let certain number be x.According to the given condition we have 3x+5=19 now we can solve for x.
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   Example: Theperimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool? Solution: The unknown are length and breadth. Let the breadth be x m. Then as per question the length will be (2x + 2) m. Perimeter of swimming pool = 2(l + b) = 154 m or,2(2x+2+x)=154 or,2(3x+2)=154 Dividing both sides by 2 3x+2=77 Transposing 2 to R.H.S, we obtain 3x=77−2 or,3x=75 Dividing 3 on both the sides x=25 So, Breadth is 25 m Length =2x + 2 = 2 × 25 + 2 = 52m Hence, the breadth and length of the pool are 25 m and 52 m respectively.
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  Example: 2x−3=6−x Solution: Transposing3 to RHS and x to LHS 2x+x=6+3 or,3x=9 Dividing both the sides by 3 x=3.Ans
• 10.
   You willfind many situations where the linear equation may be having number in denominator. We can perform the below steps to simplify them and solve it.  Take the LCM of the denominator of both the LHS and RHS.  Multiple the LCM on both the sides, this will reduce the number without denominator and we can solve using the method described above.
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   8x +4 = 3(x – 1) + 7  Solution:  Given,8x + 4 = 3(x – 1) + 7 By removing bracket from RHS, we get 8x + 4 = 3x – 3 + 7 By transposing 3x to LHS, we get 8x-3x + 4 = -3 + 7 ⇒ 8x – 3x + 4 = 4 By transposing 4 to RHS, we get 8x – 3x = 4 – 4 ⇒ 5x = 0 After dividing both sides by 5, we get or, 5x/5=0/5 or, X=0
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   Ques:A positivenumber is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?  Solution: Let the given positive number = a  Therefore, another number which is 5 times of it = 5a  Now, after adding 21 to both of the number .First number = a + 21  Second number = 5a + 21  According to question, one new number becomes twice of the other new number.Therefore,Second number = 2 x first number  i.e. 5a + 21 = 2 (a + 21)  ⇒ 5a + 21 = 2a + 42  By transposing ‘2a’ to LHS, we get  ⇒ 5a + 21 – 2a = 42  Now, after transposing 21 to RHS, we get  ⇒ 5a – 2a = 42 – 21  ⇒ 3a = 21  After dividing both sides by 3, we get  Therefore, another number 5a = 5 x 7 = 35  Thus, required numbers are 7 and 35
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  Solve the equation: (5x+3)/4 - (2x -4)/3=5 Or,[(5x +3)/4] x 12 – [(2x -4)/3] x 12= 5 x12 Or,3(5x +3) - 4(2x -4) = 60 Or, 15x + 9 -8x +16 =60 Or, 7x +25 =60 Or, 7x = 60-25 Or, 7x = 35 Or, x= 35/7 Or, x= 5
• 14.
  Solve the followingquestion: 1. Find the solution of 3x-4 = 12 2. Solve: 5x-9 = 8 3. What should be subtracted from thrice the rational number -8/3 to get 5/2? 4. The sum of three consecutive multiples of 7 is 63. Find these multiples. 5. Solve 3x/4 – 7/4 = 5x + 12 6. Perimeter of a rectangle is 13cm. if its width is 11/4 cm, find its length. 7. The present age of Seta’s father is three times the present age of Sita. After six years sum of their ages will be 69 years. Find their present ages. 8. The digits of a two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number. 9. (x-2)/(x+1) = ½. Find x 10. Sanjay will be 3 times as old as he was 4 years ago after 18 years. Find his present age. 11. If the sum of two numbers is 30 and their ratio is 2/3 then find the numbers. 12. The numerator of a fraction is 2 less than the denominator. If one is added to its denominator, it becomes 1/2 find the fraction.
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  Thank you