ML Aggarwal CBSE Solutions Class 6 Math 12th Chapter Algebra Exercise 12.4
Ex 12.4
(1) State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.
(i) 17 + x = 5
Solution: It is an equation with variable x.
(ii) 2b – 3 = 7
Solution: It is an equation with variable b.
(iii) (y -7) > 5
Solution: it is not an equation.
(iv) 9/3 = 3
Solution: It is a numerical equation. It has no variable.
(v) 7 x 3 – 19 = 2
Solution: It is a numerical equation. It has no variable.
(vi) 5 x 4 – 8 = 3t
Solution: It is an equation with variable t.
(vii) 2p < 15
Solution: It is not an equation.
(viii) 7 = 11 x 5 – 12 x 4
Solution: It is a numerical equation. It has no variable.
(ix) 3/2 q = 5
Solution: It is an equation with variable q.
(2) (i) x + 12 = 20 (12, 8, 20, 0)
Solution: The equation is x + 12 = 20
For x = 12, LHS = 12 + 12 = 24 ≠ RHS Therefore, x = 12 is not a solution
For x = 8, LHS 8 + 12 = 20 = RHS Therefore, x = 8 is a solution
For x = 20, LHS 20 + 12 = 32 ≠ RHS Therefore x = 20 is not a solution
For x = 0, LHS 0 + 12 = 12 ≠ RHS Therefore x = 0 is not a solution.
Hence, x = 8 is the solution of the given equation.
(ii) y/2 = 7 (7, 2, 10, 14)
Solution: The given equation y/2 = 7
For y = 7, LHS = 7/2 = 3.5 ≠ RHS Therefore y = 7 is not a solution.
For y = 2, LHS = 2/2 = 1 ≠ RHS Therefore, y = 2 is not a solution
For y = 10, LHS = 10/2 = 5 ≠ RHS Therefore y = 10 is not a solution
For y = 14, LHS = 14 / 2 = 7 = RHS Therefore y = 14 is a solution
Hence, y = 14 is the solution of the given equation.
(iii) p – 4 = 0 (4. -4, 8, 0)
Solution: The given equation p – 4 = 0
For p = 4, LHS = 4 – 4 = 0 = RHS Therefore p = 4 is the solution
For p = -4, LHS = -4 – 4 = -8 ≠ RHS Therefore, p = -4 is not a solution
For p = 8, LHS = 8 – 4 = 4 ≠ RHS Therefore p = 8 is not a solution
For p = 0, LHS = 0 – 4 = -4 ≠ RHS Therefore p = o is not a solution
Hence, p = 4 is the solution of the given equation
(iv) n + 4 = 2 (-2, 0, 2, 4)
Solution: The given equation n + 4 = 2
For n = -2, LHS = -2 + 4 = 2 = RHS Therefore n = -2 is the solution
For n = 0, LHS = 0 + 4 = 4 ≠ RHS Therefore n = 0 is not a solution
For n = 2, LHS = 2 + 4 = 6 ≠ RHS Therefore n = 2 is not a solution
For n = 4, LHS = 4 + 4 = 8 ≠ RHS Therefore n = 4 is not a solution
Hence, n = -2 is the solution of the given equation.
(v) 2y – 3 = 7 (0,3,5,7)
Solution: For y = 0, LHS = 2×0 – 3 = 0 – 3 = -3 ≠ RHS Therefore y = 0 is not a solution
For y = 3, LHS = 2×3 – 3 = 6 – 3 = 3 ≠ RHS Therefore, y = 3 is not a solution.
For y = 5, LHS = 2 x 5 – 3 = 10 – 3 = 7 = RHS Therefore, y = 5 is a solution.
For y = 7, LHS = 7 x 5 – 3 = 35 – 3 = 32 ≠ RHS Therefore, y = 7 is not a solution.
(3) Complete the entries in the fourth of the following table:
Solution:
S. No. | Equation | Value of Variable | Equation Satisfied Yes / No |
(i) | 5n = 40 | 10 | No |
(ii) | 5n = 40 | 8 | Yes |
(iii) | 5n = 40 | 5 | No |
(iv) | X + 10 = 30 | 10 | No |
(v) | X + 10 | 20 | Yes |
(vi) | X + 20 = 30 | 30 | No |
(vii) | 2p – 3 = 7 | 5 | Yes |
(viii) | 2p – 3 = 7 | 10 | No |
(ix) | 2p – 3 = 7 | 15 | No |
(x) | Y + 3 = 1 | 1 | No |
(xi) | Y + 3 = 1 | 0 | No |
(xii) | Y + 3 = 1 | -1 | No |
(xiii) | Y + 3 = 1 | -2 | yes |
(5) Complete the following table and by inspection of the table, find the solution of the equation 2x + 3 = 17
(i) Solution: The given equation is 2x + 3 = 17. We complete the table as under:
x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | …. |
2x + 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 | 21 | …. |
(ii) Solution:
x | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | …. |
25 – 3x | 16 | 13 | 10 | 7 | 4 | 1 | -2 | -5 | -8 | …. |
It’s is right solution but teachers are get wrong it is wrong solution
Where it is wrong?