Simple Equations

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.

Simple Equations

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 alphabets, 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 value of x other than 15 satisfies the condition 4x + 5 = 65.

Simple Equations

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.

Simple Equations

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 LHS) is equal to the value of the expression to the right of the sign (the right hand side or RHS). In equation (4.1), the LHS is (4x + 5) and the RHS is 65. In equation (4.2), the LHS is (10y – 20) and the RHS is 50

If there is some sign other than the equality sign between the LHS and the RHS, 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 RHS 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 RHS 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 expressions on the left and on the right are interchanged. This property is often useful in solving equations.

EXAMPLE  Write the following statements in the form of equations:

 (i) The sum of three times x and 11 is 32.

 (ii) If you subtract 5 from 6 times a number, you get 7.

 (iii) One fourth of m is 3 more than 7. (iv) One third of a number plus 5 is 8.

(iii) One fourth of m is \(m \over 4\) . It is greater than 7 by 3. This means the difference ( \(m \over 4\) – 7) is 3. The equation is \(m \over 4\) – 7 = 3.

(iv) Take the number to be n. One third of n is \(n \over 3\) . This one-third plus 5 is  \(n \over 3\)+ 5. It is 8. The equation is  \(n \over 3\)+ 5 = 8.

EXAMPLE : Convert the following equations in statement form: (i) x – 5 = 9 (ii) 5p = 20 (iii) 3n + 7 = 1 (iv) m 5 – 2 = 6

SOLUTION :(i) Taking away 5 from x gives 9.

(ii) Five times a number p is 20.

 (iii) Add 7 to three times n to get 1.

 (iv) 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 statement 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

EXAMPLE :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.

SOLUTION: 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.

EXAMPLE :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.

 SOLUTION: 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

Simple Equations

Solving an Equation

 Consider an equality 8 – 3 = 4 + 1…….. (4.5)

 The equality (4.5) holds, since both its sides are equal (each is equal to 5)

* Let us now add 2 to both sides; as a result

LHS = 8 – 3 + 2 = 5 + 2 = 7 RHS = 4 + 1 + 2 = 5 + 2 = 7.

Again the equality holds (i.e., its LHS and RHS are equal).

Thus if we add the same number to both sides of an equality, it still holds.

* Let us now subtract 2 from both the sides; as a result,

LHS = 8 – 3 – 2 = 5 – 2 = 3  RHS = 4 + 1 – 2 = 5 – 2 = 3.

Again, the equality holds. Thus if we subtract the same number from both sides of an equality, it still holds.

 *Similarly, if we multiply or divide both sides of the equality by the same non-zero number, it still holds. For example, let us multiply both the sides of the equality by 3, we get

 LHS = 3 × (8 – 3) = 3 × 5 = 15, RHS = 3 × (4 + 1) = 3 × 5 = 15.

The equality holds. Let us now divide both sides of the equality by 2.

LHS = (8 – 3) ÷ 2 = 5 ÷ 2 = \(5 \over 2\)  RHS = (4+1) ÷ 2 = 5 ÷ 2 = \(5 \over 2\) = LHS

Again, the equality holds

If we take any other equality, we shall find the same conclusions. Suppose, we do not observe these rules. Specificially, suppose we add different numbers, to the two sides of an equality. We shall find in this case that the equality does not hold (i.e., its both sides are not equal). For example, let us take again equality (4.5),

8 – 3 = 4 + 1

add 2 to the LHS and 3 to the RHS. The new LHS is 8 – 3 + 2 = 5 + 2 = 7 and the new RHS is 4 + 1 + 3 = 5 + 3 = 8. The equality does not hold, because the new LHS and RHS are not equal.

Thus if we fail to do the same mathematical operation with same number on both sides of an equality, the equality may not hold.

The equality that involves variables is an equation.

These conclusions are also valid for equations, 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. An 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.

*Consider the equation: x + 3 = 8……….(4.6)

 We shall subtract 3 from both sides of this equation. The new

LHS is x + 3 – 3 = x and the new RHS is8 – 3 = 5

Since this does not disturb the balance, we have

New LHS = New RHS 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 LHS = x + 3 = 5 + 3 = 8, which is equal to the RHS 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…(4.7) What should we do here? We should add 3 to both the sides, By doing so, we shall retain the balance and also the LHS will reduce to just x.

New LHS = x – 3 + 3 = x , New RHS = 10 + 3 = 13

Therefore, x = 13, which is the required solution.

* Similarly, let us look at the equations

5y = 35…... (4.8)

\(m \over 2\) = 5…. (4.9)

In the first case, we shall divide both the sides by 5. This will give us just y on LHS

\(\text{New LHS }=\text{ }\frac{\text{5 y}}{5}\text{=}\frac{\text{5 }\times y\text{ }}{5}=\text{ y },\text{New RHS }=\text{ }\frac{\text{35 }}{5}\text{=}\frac{\text{5 }\times \text{7}}{5}\text{ }=\text{ 7}\)

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 LHS The new LHS = m 2 × 2 = m.

 The new RHS = 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.

EXAMPLE Solve: (a) 3n + 7 = 25          (4.10)

(b) 2p – 1 = 23                          (4.11)

SOLUTION

(a)We go stepwise to separate the variable n on the LHS of the equation. The LHS 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,

3n + 7 – 7 = 25 – 7                    (Step 1)

 Or    3n = 18

Now divide both sides by 3,

\(\frac{3n}{3}=\frac{18}{3}\)                                     (Step 2)

or n = 6, which is the solution.

(b) What should we do here? First we shall add 1 to both the sides:

2p – 1 + 1 = 23 + 1                    (Step 1)

or 2p = 24

Now divide both sides by 2, we get

\(\frac{2p}{2}=\frac{24}{2}\)                                        (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.

LHS = 2p – 1 = 2 × 12 – 1 = 24 – 1 = 23 = RHS

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.

* First consider the equation

4x + 5 = 65…………………..(4.1)

 Subtracting 5 from both sides, 4x + 5 – 5 = 65 – 5. i.e. 4x = 60

Divide both sides by 4; this will separate x.

We get \(\frac{4x}{4}=\frac{60}{4}\) or x = 15, which is the solution. (Check, if it is correct.)

*Now consider,10y – 20 = 50 ………………(4.2)

Adding 20 to both sides, we get 10y – 20 + 20 = 50 + 20 or 10y = 70

Dividing both sides by 10, we get \(\frac{10y}{10}=\frac{70}{10}\)

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

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

EXAMPLE  Solve: 12p – 5 = 25 …..(4.12)

SOLUTION

* Adding 5 on both sides of the equation,

12p – 5 + 5 = 25 + 5 or 12p = 30

* Dividing both sides by 12,

\(\frac{12p}{12}=\frac{30}{12}(or)p=\frac{5}{2}\)

Check Putting p = \(\frac{5}{2}\) in the LHS of equation 4.12,

LHS = \(12\times \frac{5}{2}-5\)= 6 × 5 – 5

= 30 – 5 = 25 = RHS

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.

We shall now solve two more equations. As you can see they involve brackets, which have to be solved before proceeding

EXAMPLE :

Solve (a) 4(m + 3) = 18 (b) – 2(x + 3) = 8

SOLUTION (a) 4(m + 3) = 18

Let us divide both the sides by 4. This will remove the brackets in the LHS We get,

\(\text{m}+\text{3 =}\frac{\text{18}}{4}\text{  or m }+3=\text{ }\frac{\text{9 }}{2}\)
\(\begin{align}   & \text{m}=\text{ }\frac{\text{9 }}{2}-3\left( \text{transposing 3 to RHS} \right)\text{ } \\  & \text{or m }=\text{ }\frac{\text{3}}{2}\text{  }\left( \text{required solution} \right)\text{ }\left( \text{as }\frac{\text{9 }}{2}\text{-3 =}\frac{\text{9}}{2}\text{ -}\frac{6}{2}\text{=}\frac{3}{2}\text{ } \right) \\ \end{align}\)

\(\begin{align}   & \text{Check LHS }=\text{ 4}\left[ \frac{3}{2}+3 \right]\text{  }=\text{ 4}\times \frac{3}{2}\text{ }+4\text{ }\times 3\text{  }\left[ \text{put m }=\text{ }\frac{\text{3}}{2}\text{  } \right]\text{ } \\  & =\text{ 6 }+\text{ 12 }=\text{ 18 }=\text{ RHS} \\ \end{align}\)

\(\begin{align}   & \left( \text{b} \right)\text{ }\text{2}\left( \text{x }+\text{ 3} \right)\text{ }=\text{ 8 } \\  & \text{We divide both sides by }\left( \text{ 2} \right),\text{ so as to remove the brackets in the LHS},\text{ we get}, \\  & \text{ x }+\text{ 3 }=\text{ }\text{ 8 2 or x }+\text{ 3 }=\text{ }\text{ 4 i}.\text{e}.,\text{ x }=\text{ }\text{ 4 }\text{ 3 }\left( \text{transposing 3 to RHS} \right)\text{ } \\  & \text{or x }=\text{ }\text{7 }\left( \text{required solution} \right) \\ \end{align}\)

Check LHS = -2(-7+3)=-2(-4)

=8= RHS as required.

Simple Equations

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]

EXAMPLE :The sum of three times a number and 11 is 32. Find the number.

SOLUTION :

 If the unknown number is taken to be x, then three times the number is 3x and the sum of 3x and 11 is 32

That is, 3x + 11 = 32

To solve this equation, we transpose 11 to RHS, so that 3x = 32 – 11 or 3x = 21 Now, divide both sides by 3

So x =\(21 \over 3\) = 7 The required number is 7. (We may check it by taking 3 times 7 and adding 11 to it. It gives 32 as required.)

EXAMPLE :Find a number, such that one-fourth of the number is 3 more than 7. This equation was obtained earlier in Section 4.2, Example 1.

SOLUTION

* Let us take the unknown number to be y; one-fourth of y is \(y \over 4\) .

This number \(y \over 4\) is more than 7 by 3.

Hence we get the equation for y as \(y \over 4\)– 7 = 3

* To solve this equation, first transpose 7 to RHS We get, \(y \over 4\) = 3 + 7 = 10.

 We then multiply both sides of the equation by 4, to get

\(y \over 4\) × 4 = 10 × 4 or y = 40 (the required number)

Let us check the equation formed. Putting the value of y in the equation,

 LHS = \(40 \over 4\) – 7 = 10 – 7 = 3 = RHS, as required.

EXAMPLE :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.

SOLUTION

* As given in Example 3 earlier, the equation that gives Raju's age is 3y + 5 = 44

*To solve it, we first transpose 5, to get 3y = 44 – 5 = 39

Dividing both sides by 3, we get y = 13

That is, Raju’s age is 13 years. (You may check the answer.)