# ML Aggarwal CBSE Solutions Class 6 Math Twelveth Chapter Algebra Exercise 12.4

### 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 ….

Updated: December 4, 2019 — 2:32 pm

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