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Simple Equations
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in mathematics and it is going to be simple equations. Appu, Sarita and Ameena have revised what they learnt in algebra chapter in Class VI. Have you? Appu, Sarita and Ameena are excited because they have constructed a game which they call mind reader and they want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena begins; she asks Sara to think of a number, multiply it by 4 and add 5 to the product. Then, she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now. He asks Balu to think of a number, multiply it by 10 and subtract 20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and Ameena works. Can you see how it works? After studying this chapter and chapter 12, you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number. Ameena does not know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us denote this unknown number by a letter, say x. You may use y or t or some other letter in 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 | (4.1) |
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 | (4.2) |
The solution of this equation will give us the number Balu had thought of.
4.3 REVIEW OF WHAT WE KNOW
Note, (4.1) and (4.2) are equations. Let us recall what we learnt about equations in Class VI. An equation is a condition on a variable. In equation (4.1), the variable is x; in equation (4.2), 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). For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we formed the expression (10y – 20). For this, we multiplied y by 10 and then subtracted 20 from the product. All these are examples of expressions.
The value of an expression thus formed depends upon the chosen value of the variable.
As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly,
when | x = 15, 4 x + 5 = 4×15 + 5 = 65; | |
when | x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on. |
Equation (4.1) 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 equation 4x + 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 than 15 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 condition 10y – 20 = 50 is met.
4.4 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 (4.1), the L.H.S. is (4x + 5) and the R.H.S. is 65. In equation (4.2), 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 (4.1), 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
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.
Write the following statements in the form of equations:
- The sum of three times x and 11 is 32.
- If you subtract 5 from 6 times a number, you get 7.
- One fourth of m is 3 more than 7.
- One third of a number plus 5 is 8.
- Three times x is 3x.
Sum of 3x and 11 is 3x + 11. The sum is 32.
The equation is 3x + 11 = 32. - Let us say the number is z; z multiplied by 6 is 6z.
Subtracting 5 from 6z, one gets 6z – 5. The result is 7.
The equation is 6z – 5 = 7 - One fourth of m is
It is greater than 7 by 3. This means the difference (-7) is 3.
The equation is -7= 3 - Take the number to be n. One third of n is .
The number plus 5 is + 5. It is 8.
The equation is + 5 = 8.
Convert the following equations in statement form:
- x – 5 = 9
- 5p = 20
- 3n + 7 = 1
- – 2 = 6
- Taking away 5 from x gives 9.
- Five times a number p is 20.
- Add 7 to three times n to get 1.
- You get 6, when you subtract 2 from one fifth of a number m.
What is important to note is that for a given equation, not just one, but many statements forms can be given. For example, for Equation (i) above, you can say:
Subtract 5 from x, you get 9.
or | The number x is 5 more than 9. |
or | The number x is greater by 5 than 9. |
or | The difference between x and 5 is 9, and so on. |
Write at least one other form for each Equation (ii)= ,
(iii)=
and (iv)= .
Consider the following situation:
Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age.
We do not know Raju’s age. Let us take it to be y years. Three times Raju’s age is 3y years. Raju’s father’s age is 5 years more than 3y; that is, Raju’s father is (3y + 5) years old. It is also given that Raju’s father is 44 years old.
Therefore, | 3y + 5 = 44 | (4.3) |
This is an equation in y. It will give Raju’s age when solved.
A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100.
Let a small box contain m mangoes. A large box contains 4 more than 8 times m, that is, 8m + 4 mangoes. But this is given to be 100. Thus
8m + 4 = 100 | (4.4) |
You can get the number of mangoes in a small box by solving this equation.
4.4.1 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).
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.
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.
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 =
R.H.S. = (4+1) ÷ 2 = 5 ÷ 2 = = 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. Specificially, 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 (4.5),
add 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.
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, ofcourse, 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.
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 |
Why should we subtract 3, and not some other number? Try adding 3. Will it help? Why not? It is because subtracting 3 reduces the L.H.S. to x. |
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 (4.6).
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.
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.
5y = 35 | (4.8) |
(4.9) |
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. = | , | New R.H.S. = |
Therefore, | y = 7 |
This is the required solution. We can substitute y = 7 in Eq. (4.8) 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. = 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 depends on 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.
Solve:
(a) | 3n + 7 = 25 | (4.10) |
(b) | 2p – 1 = 23 | (4.11) |
- We go stepwise to separate the variable n on the L.H.S. of the equation. The L.H.S. is 3n + 7. We shall first subtract 7 from it so that we get 3n. From this, in the next step we shall divide by 3 to get n. Remember we must do the same operation on both sides of the equation. Therefore, subtracting 7 from both sides,
- What should we do here? First we shall add 1 to both the sides:
3n + 7 – 7 = 25 – 7 | (Step 1) | |
or | 3n = 18 |
Now divide both sides by 3,
(Step 2) | |
or | n = 6, which is the solution. |
2p – 1 + 1 = 23 + 1 | (Step 1) |
or | 2p = 24 |
Now divide both sides by 2, we get | (Step 2) |
or | p = 12, which is the solution. |
One good practice you should develop is to check the solution you have obtained.
Although we have not done this for (a) above, let us do it for this example.
Let us put the solution p = 12 back into the equation.
= 23 = R.H.S.
The solution is thus checked for its correctness.
Why do you not check the solution of (a) also?
We are now in a position to go back to the mind-reading game presented by Appu, Sarita, and Ameena and understand how they got their answers. For this purpose, let us look at the equations (4.1) and (4.2) which correspond respectively to Ameena’s and Appu’s examples.
Subtract 5 from both sides, 4x + 5 – 5 = 65 – 5.
i.e. | 4x = 60 |
Divide both sides by 4; this will separate x. We get
or . | x = 15, which is the solution. (Check, if it is correct.) |
Adding 20 to both sides, we get 10y – 20 + 20 = 50 + 20 or 10y = 70
Divide both sides by 10, we get
or . | y = 7, which is the solution. (Check if it is correct.) |
You will realise that exactly these were the answers given by Appu, Sarita and Ameena. They had learnt to set-up equations and solve them. That is why they could construct their mind reader game and impress the whole class. We shall come back to this in Section 4.7.
4.5 MORE EQUATIONS
Let us practise solving some more equations. While solving these equations, we shall learn about transposing a number, i.e., moving it from one side to the other. We can transpose a number instead of adding or subtracting it from both sides of the equation.
Solve: 12p – 5 = 25 | (4.12) |
12p – 5 + 5 = 25 + 5 | or | 12p = 30 |
or |
Check Putting p = in the L.H.S. of equation 4.12,
L.H.S. = 12× = 6 × 5 – 5
= 30 – 5 = 25 = R.H.S.
Note adding 5 to both sides is the same as changing side of (– 5). 12p – 5 = 25
12p = 25 + 5
Changing side is called transposing. While transposing a number, we change its sign. |
As we have seen, while solving equations one commonly used operation is adding or subtracting the same number on both sides of the equation. Transposing a number (i.e. changing the side of the number) is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed. What applies to numbers also applies to expressions. Let us take two more examples of transposing.
Adding or Subtracting on both sides | Transposing |
---|---|
(i) 3p – 10 = 5 | (i)3p – 10 = 5 |
Add 10 to both sides | Transpose (–10) from L.H.S. to R.H.S. |
3p – 10 + 10 = 5 + 10 | (On transposing – 10 becomes + 10). |
or 3p = 15 | 3p = 5 + 10 or 3p = 15 |
(ii) 5x + 12 = 27 | (ii) 5x + 12 = 27 |
Subtract 12 from both sides | Transposing + 12 |
(On transposing + 12 becomes – 12) | |
5x + 12 – 12 = 27 – 12 | 5x = 27 – 12 |
or 5x = 15 | or 5x = 15 |
We shall now solve two more equations. As you can see they involve brackets, which have to be solved before proceeding.
Solve
- 4 (m + 3) = 18
- – 2(x + 3) = 5
- 4(m + 3) = 18
m+ 3 = or m+ 3 = or m =(transposing 3 to R.H.S.) or m =(required solution) Check L.H.S. = + 4×3 = 2×3 + 4×3[ put m = ]
= 6 + 12 = 18 = R.H.S.
- –2(x + 3) = 5
We divide both sides by (– 2), so as to remove the brackets in the L.H.S. We get,
x + 3 = or x =-3 (transposing 3 to R.H.S.) i.e. x = or x = (required solution) Check L.H.S. =
=5 = R.H.S. as required.
4.6 FROM SOLUTION TO EQUATION
Atul always thinks differently. He looks at successive steps that one takes to solve an equation. He wonders why not follow the reverse path:
Equation | Solution | (normal path) | |
Solution | Equation | (reverse path) |
He follows the path given below:
Start with | x = 5 | |||
Multiply both sides by 4, | 4x = 20 | Divide both sides by 4. | ||
Subtract 3 from both sides, | 4x – 3 = 17 | Add 3 to both sides. |
This has resulted in an equation. If we follow the reverse path with each step, as shown on the right, we get the solution of the equation. Hetal feels interested. She starts with the same first step and builds up another equation.
Multiply both sides by 3 | 3x = 15 |
Add 4 to both sides | 3x + 4 = 19 |
Start with y = 4 and make two different equations. Ask three of your friends to do the same. Are their equations different from yours?
Start with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5.
Is it not nice that not only can you solve an equation, but you can make equations? Further, did you notice that given an equation, you get one solution; but given a solution, you can make many equations? |
Now, Sara wants the class to know what she is thinking. She says, “I shall take Hetal’s
equation and put it into a statement form and that makes a puzzle. For example,
Think of a number; multiply it by 3 and add 4 to the product. Tell me the sum you get.
If the sum is 19, the equation Hetal got will give us the solution to the puzzle. In fact, we
know it is 5, because Hetal started with it.”
She turns to Appu, Ameena and Sarita to check whether they made their puzzle this way. All three say, “Yes!”
We now know how to create number puzzles and many other similar problems.
Try to make two number puzzles, one with the solution 11 and another with 100 =
4.7 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 (Example 1 (i) and (iii), Section 4.2)
The sum of three times a number and 11 is 32. Find the number.
3x = 32 – 11 | or | 3x = 21 |
Now, divide both sides by 3
So |
The required number is 7. (We may check it by taking 3 times 7 and adding 11 to it. It gives 32 as required.)
Find a number, such that one fourth of the number is 3 more than 7.
This number () is more than 7 by 3.
Hence we get the equation for y as – 7 = 3
= 3 + 7 = 10.
We then multiply both sides of the equation by 4, to get
× 4 = 10 × 4 | or | y = 40 | (the required number) |
Let us check the equation formed. Putting the value of y in the equation,
L.H.S. = – 7 = 10 – 7 = 3 = R.H.S., as required.
- When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is? =
- What is that number one third of which added to 5 gives 8? =
Raju’s father’s age is 5 years more than three times Raju’s age. Find Raju’s age, if his father is 44 years old.
Hence, the equation that gives Raju's age is | 3y + 5 = 44 |
Dividing both sides by 3, we get | y = 13 |
That is, Raju's age is 13 years. (You may check the answer.)
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box? =
What have we discussed?
- An equation is a condition on a variable such that two expressions in the variable should have equal value.
- The value of the variable for which the equation is satisfied is called the solution of the equation.
- An equation remains the same if the L.H.S. and the R.H.S. are interchanged.
- In case of the balanced equation, if we (i) add the same number to both the sides, or (ii) subtract the same number from both the sides, or (iii) multiply both sides by the same number, or (iv) divide both sides by the same number, the balance remains undisturbed, i.e., the value of the L.H.S. remains equal to the value of the R.H.S.
- The above property gives a systematic method of solving an equation. We carry out a series of identical mathematical operations on the two sides of the equation in such a way that on one of the sides we get just the variable. The last step is the solution of the equation.
- Transposing means moving to the other side. Transposition of a number has the same effect as adding same number to (or subtracting the same number from) both sides of the equation. When you transpose a number from one side of the equation to the other side, you change its sign. For example, transposing +3 from the L.H.S. to the R.H.S. in equation x + 3 = 8 gives x = 8 – 3 (= 5). We can carry out the transposition of an expression in the same way as the transposition of a number.
- We have learnt how to construct simple algebraic expressions corresponding to practical situations.
- We also learnt how, using the technique of doing the same mathematical operation (for example adding the same number) on both sides, we could build an equation starting from its solution. Further, we also learnt that we could relate a given equation to some appropriate practical situation and build a practical word problem/puzzle from the equation.