11.10 Solution of an Equation
NCERT Class 6 Mathematics Text book for Blind Students made Screen Readable by Professor T K Bansal.
We saw in the earlier section that the equation
2n = 10 .. .. (1)
was satisfied by n = 5. No other value of n satisfies the equation. The value of the variable in an equation which satisfies the equation is called a solution to the equation. Thus, n = 5 is a solution to the equation 2 n = 10.
Note, n = 6 is not a solution to the equation 2n = 10; because for n = 6, 2n = 2 × 6 = 12 and not 10.
Also, n = 4 is not a solution. Tell, why not?
Let us take the equation x − 3 = 11 .. .. (2)
This equation is satisfied by x = 14, because for, x = 14, LHS of the equation = 14 − 3 = 11 = RHS
It is not satisfied by x = 16, because for, x = 16, LHS of the equation = 16 − 3 = 13, which is not equal to RHS.
Thus, x = 14 is a solution to the equation x − 3 = 11 and x = 16 is not a solution to the equation. Also, x = 12 is not a solution to the equation. Explain, why not?
Now complete the entries in the following table and explain why your answer is Yes/No.
In finding the solution to the equation 2n = 10, we prepare a table for various values of n and from the table, we pick up the value of n which is the solution to the equation (i.e. which satisfies the equation). What we used is a trial and error method. It is not a direct and practical way of finding a solution of the equation.
| S.No. | Equation | Value of the variable | Solution (Yes/No) |
|---|---|---|---|
| 1. | x + 10 = 30 | x = 10 | No |
| 2. | x + 10 = 30 | x = 30 | No |
| 3. | x + 10 = 30 | x = 20 | Yes |
| 4. | p − 3 = 7 | p = 5 | No |
| 5. | p − 3 = 7 | p = 15 | - |
| 6. | p − 3 = 7 | p = 10 | - |
| 7. | 3n = 21 | n = 9 | - |
| 8. | 3n = 21 | n = 7 | - |
| 9. | t/5 = 4 | t = 25 | - |
| 10. | t/5 = 4 | t = 20 | - |
| 11. | 2l + 3 = 7 | l = 5 | - |
| 12. | 2l + 3 = 7 | l = 1 | - |
| 13. | 2l + 3 = 7 | l = 2 | - |
solution. We need a direct way of solving an equation, i.e. finding the solution of the equation. We shall learn a more systematic method of solving equations only next year.
Start of blue box
Beginning of Algebra
It is said that algebra as a branch of Mathematics began about 1550 BC, i.e. more than 3500 years ago, when people in Egypt started using symbols to denote unknown numbers.
Around 300 BC, use of letters to denote unknowns and forming expressions from them was quite common in India. Many great Indian mathematicians, Aryabhatt (born 476AD), Brahmagupta (born 598AD), Mahavira (who lived around 850AD) and Bhaskara II (born 1114AD) and others, contributed a lot to the study of algebra. They gave names such as Beeja, Varna etc. to unknowns and used first letters of colour names [e.g., ka from kala (black), nee from neela (blue)] to denote them. The Indian name for algebra, Beejaganit, dates to these ancient Indian mathematicians.
The word ‘algebra’ is derived from the title of the book, ‘Aljebar w’al almugabalah’, written about 825AD by an Arab mathematician, Mohammed Ibn Al Khowarizmi of Baghdad.
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EXERCISE 11.5
Q1. State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t−7) > 5
(c) 4/2 = 2
(d) (7×3) − 19 = 8
(e) 5 × 4 − 8 = 2x
(f) x − 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) − (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) 3q/2 < 5
(m) z + 12 > 24
(n) 20 − (10 − 5) = 3 × 5
(o) 7 − x = 5
Q2. Complete the entries in the third column of the table.
| S.No. | Equation | Value of variable | Equation satisfied Yes/No |
|---|---|---|---|
| (a) | 10y = 80 | y = 10 | |
| (b) | 10y = 80 | y = 8 | |
| (c) | 10y = 80 | y = 5 | |
| (d) | 4l = 20 | l = 20 | |
| (e) | 4l = 20 | l = 80 | |
| (f) | 4l = 20 | l = 5 | |
| (g) | b + 5 = 9 | b = 5 | |
| (h) | b + 5 = 9 | b = 9 | |
| (i) | b + 5 = 9 | b = 4 | |
| (j) | h − 8 = 5 | h = 13 | |
| (k) | h − 8 = 5 | h = 8 | |
| (l) | h − 8 = 5 | h = 0 | |
| (m) | p + 3 = 1 | p = 3 | |
| (n) | p + 3 = 1 | p = 1 | |
| (o) | p + 3 = 1 | p = 0 | |
| (p) | p + 3 = 1 | P = − 1 | |
| (q) | p + 3 = 1 | P = − 2 |
Q3. Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
| (a) | 5m = 60 | (10, 5, 12, 15) |
| (b) | n + 12 = 20 | (12, 8, 20, 0) |
| (c) | p − 5 = 5 | (0, 10, 5, − 5) |
| (d) | q/2 = 7 | (7, 2, 10, 14) |
| (e) | r− 4 = 0 | (4, − 4, 8, 0) |
| (f) | x + 4 = 2 | (− 2, 0, 2, 4) |
Q4.
(a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.
| M | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | - | - | - |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| m + 10 | - | - | - | - | - | - | - | - | - | - | - | - | - |
(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.
| t | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | - | - | - |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 5t | - | - | - | - | - | - | - | - | - | - | - | - |
(c) Complete the table and find the solution of the equation z/3 =4 using the table.
| z | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | - | - | - | - |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| z/3 | 8/3 | 3 | 10/3 | - | - | - | - | - | - | - | - | - | - |
(d) Complete the table and find the solution to the equation m − 7 = 3.
| m | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | - | - |
|---|---|---|---|---|---|---|---|---|---|---|---|
| m − 7 | - | - | - | - | - | - | - | - | - | - | - |
Q5. Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square Counting every corner Thrice and no more! Add the count to me To get exactly thirty four!
(ii) For each day of the week Make an upcount from me If you make no mistake You will get twenty three!
(iii) I am a special number Take away from me a six! A whole cricket team You will still be able to fix!
(iv) Tell me who I am I shall give a pretty clue! You will get me back If you take me out of twenty two!