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IT WAS A GREAT EXPERIENCE, WHILE MAKING THIS PROJECT.I LEARNT LOT ABOUT THE GIVEN TOPIC. I WOULD ALSO LIKE TO THANK MY PARENTS WHO HELPED AND ENCOURAGED ME TO COLLECT THE EQUIRED INFORMATION AND MATERIALS. I AM REALLY THANKFUL TO THEM AS I GOT A CHANCE TO SHOW MY INTEREST IN THE SUBJECT. THANK YOU Slide 3: Simple Equations INTRODUCTION :- We have already come across simple equations like x + 3, y – 5, 4x + 5, 10y – 5 and so on. In Class VI, we have seen how these equations are useful in formulating puzzles and problems. Equations are a central concept in algebra. This Chapter is devoted to simple equations. When you have studied this Chapter, you will know how simple equations are formed, how they can be combined, how we can find their values and how they can be used. Slide 4: SETTING UP OF AN EQUATION:- Let us take Ameena’s example. Ameena asks Sara to think of a number. Ameena does not denote this unknown number by a letter, say x. You may use y or t or some other letter in know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us place of x. It does not matter which letter we use to denote the unknown number Sara has thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus if Sara had chosen 5, the result would have been 25. To find the number thought by Sara let us work backward from her answer 65. We have to find x such that 4x + 5 = 65 Solution to the equation will give us the number which Sara held in her mind. Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y, Balu first gets 10y and from there (10y – 20). The result is known to be 50. Therefore, 10y – 20 = 50 The solution of this equation will give us the number Balu had thought of. . Slide 5: REVIEW OF WHAT WE KNOW:- Note, and are equations. Let us recall what we learnt about equations in Class VII. An equation is a condition on a variable. In equation , the variable is x; in equation the variable is y. The word variable means something that can vary, i.e. change. A variable takes on different numerical values; its value is not fixed. Variables are denoted usually by letters of the alphabet, such as x, y, z, l, m, n, p etc. From variables, we form expressions. The expressions are formed by performing operations like addition, subtraction, multiplication and division on the variables. From x, we formed the expression (4x + 5). from the product. All these are examples of expressions. The value of an expression thus formed depends upon the chosen value of the variable. when x = 15, 4 x + 5 = 4×15 + 5 = 65; when x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on. Equation is a condition on the variable x. It states that the value of the expression(4x + 5) is 65. The condition is satisfied when x = 15. It is the solution to the equation4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the equation. Similarly, x = 0 is not a solution to the equation. No other value of x other than15 satisfies the condition 4x + 5 = 65. The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10 y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition10y-20=50 is met. Slide 6: WHAT EQUATION IS? In an equation there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign (the left hand side or L.H.S.) is equal to the value of the expression to the right of the sign (the right hand side or R.H.S.). In Equation , the L.H.S. is (4x + 5) and the R.H.S. is 65. In equation , the L.H.S. is (10y – 20) and the R.H.S. is 50. If there is some sign other than the equality sign between the L.H.S. and the R.H.S., it is not an equation. Thus, 4x + 5 > 65 is not an equation. It says that, the value of (4x + 5) is greater than 65. Similarly, 4x + 5 < 65 is not an equation. It says that the value of (4x + 5) is smaller than 65. In equations, we often find that the R.H.S. is just a number. In Equation , it is 65 and in Equation (4.2), it is 50. But this need not be always so. The R.H.S. of an equation may be an expression containing the variable. For example, the equation 4x + 5 = 6x – 25 has the expression (4x + 5) on the left and (6x – 25) on the right of the equality sign. In short, an equation is a condition on a variable. The condition is that two expressions should have equal value. Note that at least one of the two expressions must contain the variable. We also note a simple and useful property of equations. The equation 4x +5 = 65 is the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x +5 is the same as 4x + 5 = 6x – 25. An equation remains the same, when the expression on the left and on the right are interchanged. This property is often useful in solving equations. Slide 7: Solving an Equation:- Consider 8 – 3 = 4 + 1 (4.5) Since there is an equality sign between the two sides, so, at present we call it a numerical equation. You will study about its formal terminology in the later classes. The equation (4.5) is true. Let us call it balanced, since both sides of the equation are equal. (Each is equal to 5). Let us now add 2 to both sides; as a result L.H.S. = 8 – 3 + 2 = 5 + 2 = 7, R.H.S. = 4 + 1 + 2 = 5 + 2 = 7. Again we have an equation that is balanced. We say that the balance is retained or undisturbed. Thus if we add the same number to both sides of a balance equation, the balance is undisturbed. Let us now subtract 2 from both the sides; as a result, L.H.S. = 8 – 3 – 2 = 5 – 2 = 3, R.H.S. = 4 + 1 – 2 = 5 – 2 = 3. Again, we get a balanced equation. Thus if we subtract the same number from both sides of a balance equation, the balance is undisturbed. Similarly, if we multiply or divide both sides of the equation by the same number, the balance is undisturbed. For example let us multiply both the sides of the equation by 3, we get L.H.S. = 3 × (8 – 3) = 3 × 5 = 15, R.H.S. = 3 × (4 + 1) = 3 × 5 = 15. The balance is undisturbed. Let us now divide both sides of the equation by 2. L.H.S. = (8 – 3) ÷ 2 = 5 ÷ 2 =5 R.H.S. = (4+1) ÷ 2 = 5 ÷ 2 =5 = L.H.S. Again, the balance is undisturbed. If we take any other numerical equation, we shall find the same conclusions. Suppose, we do not observe these rules. Specifically, suppose we add different numbers, to the two sides of a balanced equation. We shall find in this case that the balance is disturbed. For example, let us take again Equation , 8 – 3 = 4 + 1add 2 to the L.H.S. and 3 to the R.H.S. The new L.H.S. is 8 – 3 + 2 = 5 + 2 = 7 and the new R.H.S. is 4 + 1 + 3 = 5 + 3 = 8. The balance is disturbed, because the new L.H.S. and R.H.S. are not equal. Thus if we fail to do the same mathematical operation on both sides of a balanced equation, the balance is disturbed. Slide 8: These conclusions are also valid for equations with variables as, in each equation variable represents a number only. Often an equation is said to be like a weighing balance. Doing a mathematical operation on an equation is like adding weights to or removing weights from the pans of a weighing balance. A balanced equation is like a weighing balance with equal weights on both its pans, in which case the arm of the balance is exactly horizontal. If we add the same weights to both the pans, the arm remains horizontal. Similarly, if we remove the same weights from both the pans, the arm remains horizontal. On the other hand if we add different weights to the pans or remove different weights from them, the balance is tilted; that is, the arm of the balance does not remain horizontal. We use this principle for solving an equation. Here, of course, the balance is imaginary and numbers can be used as weights that can be physically balanced against each other. This is the real purpose in presenting the principle. Let us take some examples. Consider the equation: x + 3 = 8 (4.6) We shall subtract 3 from both sides of this equation. The new L.H.S. is x + 3 – 3 = x and the new R.H.S. is 8 – 3 = 5 Since this does not disturb the balance, we have New L.H.S. = New R.H.S. or x = 5 which is exactly what we want, the solution of the equation To confirm whether we are right, we shall put x = 5 in the original equation. We get L.H.S. = x + 3 = 5 + 3 = 8, which is equal to the R.H.S. as required. By doing the right mathematical operation (i.e., subtracting 3) on both the sides of the equation, we arrived at the solution of the equation. Let us look at another equation x – 3 = 10 What should we do here? We should add 3 to both the sides, By doing so, we shall retain the balance and also the L.H.S. will reduce to just x. New L.H.S. = x – 3 + 3 = x , New R.H.S. = 10 + 3 = 13 Therefore, x = 13, which is the required solution. By putting x = 13 in the original equation (4.7) we confirm that the solution is correct: L.H.S. of original equation = x – 3 = 13 – 3 = 10 This is equal to the R.H.S. as required. Similarly, let us look at the equations5y = 35m2 = 5 In the first case, we shall divide both the sides by 5. This will give us just y on L.H.S. New L.H.S. = 5y = × y = y , New R.H.S. =355= × = 7Therefore, y = 7. Slide 9: This is the required solution. We can substitute y = 7 in Eq. and check that it is satisfied. In the second case, we shall multiply both sides by 2. This will give us just m on the L.H.S. The new L.H.S. =m2× 2 = m. The new R.H.S. = 5 × 2 = 10. Hence, m = 10 (It is the required solution. You can check whether the solution is correct). One can see that in the above examples, the operation we need to perform depend son the equation. Our attempt should be to get the variable in the equation separated. Sometimes, for doing so we may have to carry out more than one mathematical operation. Let us solve some more equations with this in mind. Slide 10: APPLICATIONS OF SIMPLE EQUATIONS TO PRACTICAL SITUATIONS :- We have already seen examples in which we have taken statements in everyday language and converted them into simple equations. We also have learnt how to solve simple equations. Thus we are ready to solve puzzles/problems from practical situations. The method is first to form equations corresponding to such situations and then to solve those equations to give the solution to the puzzles/problems. We begin with what we have already seen. Slide 11: THANK YOU You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.