**Selina Concise Mathematics Class 6 ICSE Solutions – Substitution(Including Use of Brackets as Grouping Symbols)**

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**IMPORTANT POINTS**

**Substitution :**The value of an expression depends on the value of its variable (s).**Use of Brackets :**

The Symbols —, ( ), { }, [ ] are called brackets.

If an expression is enclosed within a bracket, it is considered a single quantity, even if it is made up of many terms.

**Keep in Mind :**- While simplifying an expression containing a bracket, first of all, the terms inside the bracket are operated (combined).
- ( ) is called a small bracket or Parenthesis.
- { } is called a middle bracket or Curly bracket.
- [ ] is called big or square bracket.
- If one more bracket is needed, then we use the bar bracket.

i.e. a line ———— is drawn over a group of terms.

Thus, in , the line over 4y – 5z serves as the bar bracket and is called Vinculum.

**EXERCISE 20(A)**

**Question 1.**

Fill in the following blanks, when :

x = 3,y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.

**Solution:**

**Question 2.**

Find the value of :

**Solution:**

**Question 3.**

Find the value of :

**Solution:**

**Question 4.**

If a = 3, b = 0, c = 2 and d = 1, find the value of :

**Solution:**

**Question 5.**

Find the value of 5x^{2} – 3x + 2, when x = 2.

**Solution:**

**Question 6.**

Find the value of 3x^{3} – 4x^{2} + 5x – 6, when x = -1.

**Solution:**

**Question 7.**

Show that the value of x^{3} – 8x^{2} + 12x – 5 is zero, when x = 1.

**Solution:**

**Question 8.**

State true and false :

(i) The value of x + 5 = 6, when x = 1

(ii) The value of 2x – 3 = 1, when x = 0

(iii) = -1,when x = 1

**Solution:**

**Question 9.**

If x = 2, y = 5 and z = 4, find the value of each of the following :

**Solution:**

**Question 10.**

If a = 3, find the values of a^{2} and 2^{a}.

**Solution:**

a^{2} = (3)^{2} = 3 x 3 = 9

2^{a} = (2)^{3} = 2 x 2 x 2 = 8

**Question 11.**

If m = 2, find the difference between the values of 4m^{3} and 3m^{4}.

**Solution:**

4m^{3} = 4 (2)^{3} = 4 x 2 x 2 x 2 = 32

3m^{4} = 3 (2)^{4} = 3 x 2 x 2 x 2 x 2 = 48

Now, a difference 3m^{4} – 4m^{3} = 48 – 32 = 16

**EXERCISE 20(B)**

**Question 1.**

Evaluate :

(i) (23 – 15) + 4

(ii) 5x + (3x + 7x)

(iii) 6m – (4m – m)

(iv) (9a – 3a) + 4a

(v) 35b – (16b + 9b)

(vi) (3y + 8y) – 5y

**Solution:**

(i) (23 – 15) + 4 = 8 + 4 = 12

(ii) 5x + (3x + 7x) = 5x + 10x = 15x

(iii) 6m – (4m – m) = 6m – 3m = 3m

(iv) (9a – 3a) + 4a = 6a + 4a = 10a

(v) 35b – (16b + 9b)= 35b – 25b = 10b

(vi) (3y + 8y) – 5y = 11y – 5y = 6y

**Question 2.**

Simplify :

**Solution:**

**Question 3.**

Simplify :

**Solution:**

**EXERCISE 20(C)**

**Question 1.**

**Fill in the blanks :**

**Solution:**

(viii) 2t + r – p – q + s = 2t + r – (p + q – s)

**Question 2.**

Insert the bracket as indicated :

**Solution:**

**REVISION EXERCISE**

**Question 1.**

Find the value of 3ab + 10bc – 2abc when a = 2, b = 5 and c = 8.

**Solution:**

**Question 2.**

If x = 2, = 3 and z = 4, find the value of 3x^{2} – 4y^{2} + 2z^{2}.

**Solution:**

**Question 3.**

If x = 3, y = 2 and z = 1; find the value of:

(i) x^{y}

(ii) y^{x}

(iii) 3x^{2} – 5y^{2}

(iv) 2x – 3y + 4z + 5

(v) y^{2} – x^{2} + 6z^{2}

(vi) xy + y^{2}z – 4zx

**Solution:**

**Question 4.**

If P = -12x^{2} – 10xy + 5y^{2}, Q = 7x^{2} + 6xy + 2y^{2}, and R = 5x^{2} + 2xy + 4y^{2} ; find :

(i) P – Q

(ii) Q + P

(iii) P – Q + R

(iv) P + Q + R

**Solution:**

**Question 5.**

If x = a^{2} – bc, y = b^{2} – ca and z = c^{2} – ab ; find the value of :

(i) ax + by + cz

(ii) ay – bx + cz

**Solution:**

**Question 6.**

Multiply and then evaluate :

(i) (4x + y) and (x – 2y); when x = 2 and y = 1.

(ii) (x^{2} – y) and (xy – y^{2}); when x = 1 and y = 2.

(iii) (x – 2y + z) and (x – 3z); when x = -2, y = -1 and z = 1.

**Solution:**

**Question 7.**

Simplify :

(i) 5 (x + 3y) – 2 (3x – 4y)

(ii) 3x – 8 (5x – 10)

(iii) 6 {3x – 8 (5x – 10)}

(iv) 3x – 6 {3x – 8 (5x – 10)}

(v) 2 (3x^{2} – 4x – 8) – (3 – 5x – 2x^{2})

(vi) 8x – (3x – )

(vii) 12x^{2} – (7x – )

**Solution:**

**Question 8.**

If x = -3, find the value of : 2x^{3} + 8x^{2} – 15.

**Solution:**

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