RS Aggarwal Solutions for Class 7 Maths Exercise 6A Chapter 6 Algebraic Expressions in simple PDF are available here. This exercise of RS Aggarwal Solutions for Class 7 Chapter 6 has the topics related to constants and variables. A symbol having a fixed numerical value is constant, whereas a symbol which takes various numerical values is variables. The exercise also contains addition and subtraction of algebraic expressions. Students are suggested to try solving the questions from RS Aggarwal Solutions for Class 7 Maths Chapter 6 Algebraic Expressions.

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**1. Add the following expressions:**

**(i) 5x, 7x, -6x**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= 5x + 7x + (-6x) â€¦ [âˆµ + Ã— – = -]

Add terms having same sign first,

= 5x + 7x â€“ 6x

= 12x â€“ 6x

= 6x

**(ii) (3/5)x, (2/3)x, (-4/5)x**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

= (3/5)x + (2/3)x + (-4/5)x

LCM of 5, 3, and 5 is 15

= (9x + 10x â€“ 12x)/ 15

= (19x â€“ 12x)/ 15

= (7/15)x

**(iii) 5a ^{2}b, -8a^{2}b, 7a^{2}b**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= 5a^{2}b – 8a^{2}b + 7a^{2}b

= 12 a^{2}b – 8a^{2}b

= 4a^{2}b

**(iv) (3/4)x ^{2}, 5 x^{2}, -3x^{2}, -(1/4) x^{2}**

**Solution:-**

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= (3/4)x^{2} + 5x^{2} – 3x^{2} – (1/4) x^{2}

= (3/4)x^{2}– (1/4) x^{2} + 5x^{2} – 3x^{2}

= ((3-1)/4)x^{2} + 2x^{2}

= (2/4)x^{2} + 2x^{2}

= (1/2)x^{2} + 2x^{2}

= ((1 + 4)/2)x^{2}

= (5/2)x^{2}

**(v) x â€“ 3y + 4z, y â€“ 2x â€“ 8z, 5x â€“ 2y â€“ 3z**

**Solution:-**

Required sum,

= (x â€“ 3y + 4z) + (y â€“ 2x â€“ 8z) + (5x â€“ 2y â€“ 3z)

Collecting like terms,

= x â€“ 2x + 5x â€“ 3y + y â€“ 2y + 4z â€“ 8z â€“ 3z

Adding like terms,

= (1 â€“ 2 + 5)x + (- 3 + 1 â€“ 2)y + (4 â€“ 8 â€“ 3)z

= 4x â€“ 4y â€“ 7z

**(vi) 2x ^{2} â€“ 3y^{2}, 5x^{2} + 6y^{2}, -3x^{2} â€“ 4y^{2}**

**Solution:-**

Required sum,

= (2x^{2} â€“ 3y^{2}) + (5x^{2} + 6y^{2}) + (-3x^{2} â€“ 4y^{2})

Collecting like terms,

= 2x^{2} + 5x^{2} â€“ 3x^{2} â€“ 3y^{2} + 6y^{2} â€“ 4y^{2}

Adding like terms,

= (2 + 5 â€“ 3)x^{2} + (â€“ 3 + 6 â€“ 4)y^{2}

= 4x^{2} â€“ y^{2}

**(vii) 5x â€“ 2x ^{2} â€“ 8, 8x^{2 }â€“ 7x â€“ 9, 3 + 7x^{2} â€“ 2x**

**Solution:-**

Required sum,

= (5x â€“ 2x^{2} â€“ 8) + (8x^{2 }â€“ 7x â€“ 9) + (3 + 7x^{2} â€“ 2x)

Collecting like terms,

= 2x^{2} + 8x^{2} + 7x^{2} + 5x â€“ 7x â€“ 2x â€“ 8 â€“ 9 + 3

= (-2 + 8 + 7)x^{2} + (5 â€“ 7 â€“ 2)x + ( â€“ 8 â€“ 9 + 3)

= 13x^{2} â€“ 4x â€“ 14

**(viii) (2/3)a â€“ (4/5)v + (3/5)c, -(3/4)a â€“ (5/2)b + (2/3)c, (5/2)a + (7/4)b â€“ (5/6)c**

**Solution:-**

Required sum,

= [(2/3)a â€“ (4/5)b + (3/5)c] + [-(3/4)a â€“ (5/2)b + (2/3)c] + [(5/2)a + (7/4)b â€“ (5/6)c]

Collecting like terms,

= (2/3)a -(3/4)a + (5/2)a â€“ (4/5)b â€“ (5/2)b + (7/4)b + (3/5)c + (2/3)c â€“ (5/6)c

= [(2/3) – (3/4) + (5/2)]a + [(-4/5) â€“ (5/2) + (7/4)]b + [(3/5) + (2/3) â€“ (5/6)]c

= [(8 – 9 + 30)/12)a] + [(-16 -15 +30)/20)b] + [(18 + 20 â€“ 25)/ 30)c]

= (29/12)a +- (31/20)b + (13/30)c

**(ix) (8/5)x + (11/7)y + (9/4)xy, (-3/2)x â€“ (5/3)y â€“ (9/5)xy**

**Solution:-**

Required sum,

[(8/5)x + (11/7)y + (9/4)xy] + [(-3/2)x â€“ (5/3)y â€“ (9/5)xy]Collecting like terms,

= [(8/5)x – (3/2)x] + [(11/7)y â€“ (5/3)y] + [(9/4)xy â€“ (9/5)xy]

= (1/10)x â€“ (2/21)y + (9/20)xy

**(x) (3/2)x ^{3} – (1/4)x^{2} + (5/3), (-5/4)x^{3} + (3/5)x^{2} â€“ x + (1/5), -x^{2} + (3/8)x â€“ (8/15)**

**Solution:-**

Required sum,

= (3/2)x^{3} – (1/4)x^{2} + (5/3), (-5/4)x^{3} + (3/5)x^{2} â€“ x + (1/5), -x^{2} + (3/8)x â€“ (8/15)

Collecting like terms,

= [(3/2)x^{3} – (5/4)x^{3}] + [- (1/4)x^{2} + (3/5)x^{2} -x^{2}] + [â€“ x + (3/8)x] + [(5/3) + (1/5) â€“ (8/15)]

= (1/4)x^{3} â€“ (13/20)x^{2} â€“ (5/8)x + (4/3)

**2. Subtract :**

**(i) -8xy from 7xy**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

Then,

= (7 â€“ (-8))xy

= 15xy

**(ii) X ^{2} from -3x^{2}**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

Then,

= (-3 â€“ 1)x^{2}

= -4x^{2}

**(iii) (x â€“ y) from (4y â€“ 5x)**

**Solution:-**

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (4y â€“ 5x) â€“ (x â€“ y)

Change the sign of each term of the expression to be subtracted and then add.

= 4y â€“ 5x â€“ x + y

= (-5x â€“ x) + (4y – y)

= -5x + 3y)

= 3y â€“ 5x

**(iv) (a ^{2} + b^{2} â€“ 2ab) from (a^{2} + b^{2} + 2ab)**

**Solution:-**

We have,

= (a^{2} + b^{2} + 2ab) – (a^{2} + b^{2} â€“ 2ab)

Change the sign of each term of the expression to be subtracted and then add.

= a^{2} + b^{2} + 2ab â€“ a^{2} â€“ b^{2} + 2ab

= (1 -1)a^{2} + (1 â€“ 1)b^{2} + (2 + 2)ab

= (0)a^{2} + (0)b^{2} + (4)ab

= 4ab

**(v) (x ^{2} â€“ y^{2}) from (2x^{2} â€“ 3y^{2} + 6xy)**

**Solution:-**

We have,

= (2x^{2} â€“ 3y^{2} + 6xy) – (x^{2} â€“ y^{2})

Change the sign of each term of the expression to be subtracted and then add.

= 2x^{2} â€“ 3y^{2} + 6xy â€“ x^{2} + y^{2}

= (2x^{2} – x^{2}) + (â€“ 3y^{2} + y^{2}) + 6xy

= (2 â€“ 1)x^{2} + (â€“ 3 + 1)y^{2} + 6xy

= 1x^{2} + (â€“ 2y^{2}) + 6xy

= 1x^{2} â€“ 2y^{2} + 6xy

**(vi) (x â€“ y + 3z) from (2z â€“ x – 3y)**

**Solution:-**

We have,

= (2z â€“ x – 3y) – (x â€“ y + 3z)

Change the sign of each term of the expression to be subtracted and then add.

= 2z â€“ x – 3y â€“ x + y â€“ 3z

= (2z â€“ 3z) + (â€“ x â€“ x) + (- 3y + y)

= (2 â€“ 3)z + (â€“ 1 â€“ 1)x + (- 3 +1)y

= -1z â€“ 2x – 2y

**3. Subtract (2a â€“ 3b + 4c) from the sum of (a + 3b â€“ 4c), (4a â€“ b + 9c) and (-2b + 3c â€“ a)**

**Solution:-**

First we find the sum of (a + 3b â€“ 4c), (4a â€“ b + 9c) and (-2b + 3c â€“ a)

= (a + 3b â€“ 4c) + (4a â€“ b + 9c) + (-2b + 3c â€“ a)

= (a + 3b â€“ 4c + 4a â€“ b + 9c – 2b + 3c â€“ a)

= (a + 4a â€“ a) + (3b â€“ b â€“ 2b) + (-4c + 9c + 3c)

= (1 + 4 â€“ 1)a + (3 â€“ 1 â€“ 2)b + (-4 + 9 + 3)c

= 4a + (0)b + 8c

=4a + 8c

Then,

Subtract (2a â€“ 3b + 4c) from (4a + 8c)

= (4a + 8c) – (2a â€“ 3b + 4c)

= 4a + 8c â€“ 2a + 3b â€“ 4c

= (4a â€“ 2a) + (3b) + (8c â€“ 4c)

= 2a + 3b + 4c