RD SHARMA SOLUTION CHAPTER –7 Algebraic Expressions | CLASS 7TH MATHEMATICS-EDUGROWN

Exercise 7.1

Question: 1

Identify the monomials, binomials, trinomials and quadrinomials from the following expressions:

(i) a²

(ii) a² − b²

(iii) x³ + y³ + z³

(iv) x³ + y³ + z³ + 3xyz

(v) 7 + 5

(vi) abc + 1

(vii) 3x – 2 + 5

(viii) 2x – 3y + 4

(ix) xy + yz + zx

(x) ax³ + bx² + cx + d

Solution:

The monomials, binomials, trinomials and quadrinomials are as follows.

(i) a² is a monomial expression as it contains one term only.

(ii) a² − b² is a binomial expression as it contains two terms.

(iii) x³ + y³ + z³ is a trinomial expression as it contains three terms.

(iv) x³ + y³ + z³ + 3xyz is a quadrinomial expression as it contains four terms.

(v) 7 + 5 = 12 is a monomial expression as it contains one term only.

(vi) abc + 1 is a binomial expression as it contains two terms.

(vii) 3x – 2 + 5 = 3x + 3 is a binomial expression as it contains two terms.

(viii) 2x – 3y + 4 is a trinomial expression as it contains three terms.

(ix) xy + yz + zx is a trinomial expression as it contains three terms.

(x) ax³ + bx² + cx + d is a quadrinomial expression as it contains four terms.

Question: 2

Write all the terms of each of the following algebraic expressions:

(i) 3x

(ii) 2x – 3

(iii) 2x² − 7

(iv) 2x² + y² − 3xy + 4

Solution:

The terms of each of the given algebraic expressions are as follows.

(i) 3x is the only term of the given algebraic expression.

(ii) 2x and -3 are the terms of the given algebraic expression.

(iii) 2x² and −7 are the terms of the given algebraic expression.

(iv) 2x², y², −3xy and 4 are the terms of the given algebraic expression.

Question: 3

Identify the terms and also mention the numerical coefficients of those terms:

(i) 4xy, -5x²y, -3yx, 2xy²

(ii) 7a²bc,-3ca²b,-(5/2) abc², 3/2abc²,-4/3cba²

Solution:

(i) Like terms – 4xy, -3yx and Numerical coefficients – 4, -3

(i) Like terms – {7a²bc, −3ca²b} and Numerical coefficients – 7, -3

{−5/2abc²}   {−5/2}

{3/2 abc²}     {3/2}

{−4/3cba²}    {−4/3}

Question: 4

Identify the like terms in the following algebraic expressions:

(i) a² + b² -2a² + c² + 4a

(ii) 3x + 4xy − 2yz + 52zy

(iii) abc + ab²c + 2acb² + 3c²ab + b²ac − 2a²bc + 3cab²

Solution:

The like terms in the given algebraic expressions are as follows.

(i) The like terms in the given algebraic expressions are a² and −2a².

(ii) The like terms in the given algebraic expressions are -2yz and 5/2zy.

(iii) The like terms in the given algebraic expressions are ab²c, 2acb², b²ac and 3cab².

Question: 5

Write the coefficient of x in the following:

(i) –12x

(ii) –7xy

(iii) xyz

(iv) –7ax

Solution:

The coefficients of x are as follows.

(i) The numerical coefficient of x is -12.

(ii) The numerical coefficient of x is -7y.

(iii) The numerical coefficient of x is yz.

(iv) The numerical coefficient of x is -7a.

Question: 6

Write the coefficient of 2 in the following:

(i) −3x²

(ii) 5x²yz

(iii) 5/7x²z

(iv) –(3/2) ax² + yx

Solution:

The coefficient of x² are as follows.

(i) The numerical coefficient of x² is -3.

(ii) The numerical coefficient of x² is 5yz.

(iii) The numerical coefficient of x² is 57z.

(iv) The numerical coefficient of x² is – (3/2) a.

Question: 7

Write the coefficient of:

(i) y in –3y

(ii) a in 2ab

(iii) z in –7xyz

(iv) p in –3pqr

(v) y² in 9xy²z

(vi) x³ in x³ +1

(vii) x² in − x²

Solution:

The coefficients are as follows.

(i) The coefficient of y is -3.

(ii) The coefficient of a is 2b.

(iii) The coefficient of z is -7xy.

(iv) The coefficient of p is -3qr.

(v) The coefficient of y² is 9xz.

(vi) The coefficient of x³ is 1.

(vii) The coefficient of −x² is -1.

Question: 8

Write the numerical coefficient of each in the following

(i) xy

(ii) -6yz

(iii) 7abc

(iv) -2x3y2z

Solution:

The numerical coefficient of each of the given terms is as follows.

(i) The numerical coefficient in the term xy is 1.

(ii) The numerical coefficient in the term – 6yz is – 6.

(iii) The numerical coefficient in the term 7abc is 7.

(iv) The numerical coefficient in the term −2x³y²z is -2.

Question: 9

Write the numerical coefficient of each term in the following algebraic expressions:

(i) 4x²y – (3/2)xy + 5/2 xy²

(ii) –(5/3)x²y + (7/4)xyz + 3

Solution:

The numerical coefficient of each term in the given algebraic expression is as follows.

Question: 10

Write the constant term of each of the following algebraic expressions:

(i) x²y − xy² + 7xy − 3

(ii) a³ − 3a² + 7a + 5

Solution:

The constant term of each of the given algebraic expressions is as follows.

(i) The constant term in the given algebraic expressions is -3.

(ii) The constant term in the given algebraic expressions is 5.

Question: 11

Evaluate each of the following expressions for x = -2, y = -1, z = 3:

Solution:

Question: 12

Evaluate each of the following algebraic expressions for x = 1, y = -1, z = 2, a = -2, b = 1, c = -2:

(i) ax + by + cz

(ii) ax² + by² – cz

(iii) axy + byz + cxy

Solution:

We have x = 1, y = -1, z = 2, a = -2, b = 1 and c = -2.

Thus,

(i) ax + by + cz

= (-2)(1) + (1)(-1) + (-2)(2)

= –2 – 1 – 4

= –7

(ii) ax² + by² – cz

= (-2) × 1² + 1 × (-1)² – (-2) × 2

= 4 + 1 – (-4)

= 5 + 4

= 9

(iii) axy + byz + cxy

= (-2) × 1 × -1 + 1 × -1 × 2 + (-2) × 1 × (-1)

= 2 + (-2) + 2

= 4 – 2

= 2

Exercise 7.2

Question: 1

Add the following:

(i) 3x and 7x

(ii) -5xy and 9xy

Solution:

We have

(i) 3x + 7x = (3 + 7) x = 10x

(ii) -5xy + 9xy = (-5 + 9)xy = 4xy

Question: 2

Simplify each of the following:

(i) 7x³y +9yx³

(ii) 12a²b + 3ba²

Solution:

Simplifying the given expressions, we have

(i) 7x³y + 9yx³ = (7 + 9)x³y = 16x³y

(ii) 12a²b + 3ba² = (12 + 3)a²b =15a²b

Question: 3

Add the following:

(i) 7abc, -5abc, 9abc, -8abc

(ii) 2x²y, – 4x²y, 6x²y, -5x²y

Solution:

Adding the given terms, we have

(i) 7abc + (-5abc) + (9abc) + (-8abc)

= 7abc – 5abc + 9abc – 8abc

= (7 – 5 + 9 – 8)abc

= (16 – 13)abc

= 3abc

(ii) 2x²y +(-4x²y) + (6x²y) + (-5x²y)

= 2x²y – 4x²y + 6x²y – 5x2y

= (2- 4 + 6 – 5) x 2y

= (8 – 9) x 2y

= -x²y

Question: 4

Add the following expressions:

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Solution:

Adding the given expressions, we have

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

Collecting positive and negative like terms together, we get

x³ +2x³ – 2x²y + 3x²y + 3xy² – 5xy² – y³– 4y³

= 3x³ + x²y – 2xy² – 5y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴ – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Collecting positive and negative like terms together, we get

a⁴ – 2a⁴– 2a³b + 7a³b + 3ab³ – 5ab³ + 4a²b² – 6a²b² + 3b⁴ + b⁴

= – a⁴ + 5a³b – 2ab³ – 2a²b² + 4b⁴

Question: 5

Add the following expressions:

(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b

(ii) 5x³ + 7 + 6x – 5x², 2x² – 8 – 9x, 4x – 2x² + 3 x 3, 3 x 3 – 9x – x² and x – x² – x³ – 4

Solution:

(i) Required expression = (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)

Collecting positive and negative like terms together, we get

8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b

= 8a – 10a – 7ab + 2ab + 8b – 8b

= –2a – 5ab

(ii) Required expression = (5 x 3 + 7+ 6x – 5x²) + (2 x 2 – 8 – 9x) + (4x – 2x² + 3 x 3) + (3 x 3 – 9x-x²) + (x – x² – x³ – 4)

Collecting positive and negative like terms together, we get

5x³ + 3x³ + 3x³ – x³ – 5x² + 2x² – 2x²– x² – x² + 6x – 9x + 4x – 9x + x + 7 – 8 – 4

= 10x³ – 7x² – 7x – 5

Question: 6

Add the following:

(i) x – 3y – 2z

5x + 7y – 8z

3x – 2y + 5z

(ii) 4ab – 5bc + 7ca

–3ab + 2bc – 3ca

5ab – 3bc + 4ca

Solution:

(i) Required expression = (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)

Collecting positive and negative like terms together, we get

x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z

= 9x – 5y + 7y – 10z + 5z

= 9x + 2y – 5z

(ii) Required expression = (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)

Collecting positive and negative like terms together, we get

4ab – 3ab + 5ab – 5bc + 2bc – 3bc + 7ca – 3ca + 4ca

= 9ab – 3ab – 8bc + 2bc + 11ca – 3ca

= 6ab – 6bc + 8ca

Question: 7

Add 2x² – 3x + 1 to the sum of 3x² – 2x and 3x + 7.

Solution:

Sum of 3x² – 2x and 3x + 7

= (3x² – 2x) + (3x +7)

=3x² – 2x + 3x + 7

= (3x² + x + 7)

Now, required expression = 2x² – 3x + 1+ (3x² + x + 7)

= 2x² + 3x² – 3x + x + 1 + 7

= 5x² – 2x + 8

Question: 8

Add x² + 2xy + y² to the sum of x² – 3y²and 2x² – y² + 9.

Solution:

Question: 9

Add a³+ b³ – 3 to the sum of 2a³ – 3b³ – 3ab + 7 and -a³ + b³ + 3ab – 9.

Solution:

Question: 10

Subtract:

(i) 7a²b from 3a²b

(ii) 4xy from -3xy

Solution:

(i) Required expression = 3a²b -7a²b

= (3 -7)a²b

= – 4a²b

(ii) Required expression = –3xy – 4xy

= –7xy

Question: 11

Subtract:

(i) – 4x from 3y

(ii) – 2x from – 5y

Solution:

(i) Required expression = (3y) – (–4x)

= 3y + 4x

(ii) Required expression = (-5y) – (–2x)

= –5y + 2x

Question: 12

Subtract:

(i) 6x³ −7x² + 5x − 3 from 4 − 5x + 6x² − 8x³

(ii) − x² −3z from 5x² – y + z + 7

(iii) x³ + 2x²y + 6xy² − y³ from y³−3xy²−4x²y

Solution:

Question: 13

From

(i) p3 – 4 + 3p², take away 5p² − 3p³ + p − 6

(ii) 7 + x − x², take away 9 + x + 3x² + 7x³

(iii) 1− 5y², take away y³ + 7y² + y + 1

(iv) x³ − 5x² + 3x + 1, take away 6x² − 4x³ + 5 + 3x

Solution:

Question: 14

From the sum of 3x² − 5x + 2 and − 5x² − 8x + 9 subtract 4x² − 7x + 9.

Solution:

Question: 15

Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and   2x + 4y – 7.

Solution:

Sum of (13x – 4y + 7z) and (–6z + 6x + 3y)

= (13x – 4y + 7z) + (–6z + 6x + 3y)

= (13x – 4y + 7z – 6z + 6x + 3y)

= (13x + 6x – 4y + 3y + 7z – 6z)

= (19x – y + z)

Sum of (6x – 4y – 4z) and (2x + 4y – 7)

= (6x – 4y – 4z) + (2x + 4y – 7)

= (6x – 4y – 4z + 2x + 4y – 7)

= (6x + 2x – 4z – 7)

= (8x – 4z – 7)

Now, required expression = (8x – 4z – 7) – (19x – y + z)

= 8x – 4z – 7 – 19x + y – z

= 8x – 19x + y – 4z – z – 7

= –11x + y – 5z – 7

Question: 16

From the sum of x² + 3y² − 6xy, 2x² − y² + 8xy, y² + 8 and x² − 3xy subtract −3x² + 4y² – xy + x – y + 3.

Solution:

Question: 17

What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?

Solution:

The required expression can be got by subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7.

Therefore, required expression = (4xy – 3zx + 4yz + 7) – (xy – 3yz + 4zx)

= 4xy – 3zx + 4yz + 7 – xy + 3yz – 4zx

= 4xy – xy – 3zx – 4zx + 4yz + 3yz + 7

= 3xy – 7zx + 7yz + 7

Question: 18

What should be subtracted from x² – xy + y² – x + y + 3 to obtain −x² + 3y² − 4xy + 1?

Solution:

Question: 19

How much is x – 2y + 3z greater than 3x + 5y – 7?

Solution:

Required expression = (x – 2y + 3z) – (3x + 5y – 7)

= x – 2y + 3z – 3x – 5y + 7

Collecting positive and negative like terms together, we get

x – 3x – 2y + 5y + 3z + 7

= –2x – 7y + 3z + 7

Question: 20

How much is x² − 2xy + 3y² less than 2x² − 3y² + xy?

Solution:

Question: 21

How much does a² − 3ab + 2b² exceed 2a² − 7ab + 9b²?

Solution:

Question: 22

What must be added to 12x³ − 4x² + 3x − 7 to make the sum x³ + 2x² − 3x + 2?

Solution:

Question: 23

If P = 7x² + 5xy − 9y², Q = 4y² − 3x² − 6xy and R = −4x² + xy + 5y², show that P + Q + R = 0.

Solution:

Question: 24

If P = a² − b² + 2ab, Q = a² + 4b² − 6ab, R = b² + b, S = a² − 4ab and T = −2a² + b² – ab + a. Find P + Q + R + S – T.

Solution:

Exercise 7.3

Question: 1

Place the last two terms of the following expressions in parentheses preceded by a minus sign:

(i) x + y – 3z + y    

(ii) 3x – 2y – 5z – 4

(iii) 3a – 2b + 4c – 5

(iv) 7a + 3b + 2c + 4

(v) 2a² – b² – 3ab + 6

(vi) a² + b² – c² + ab – 3ac

Solution:

We have

(i) x + y – 3z + y = x + y – (3z – y)

(ii) 3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)

(iii) 3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)

(iv) 7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)

(v) 2a² – b² – 3ab + 6 = 2a² – b² – (3ab – 6)

(vi) a² + b² – c² + ab – 3ac = a² + b² – c² – (- ab + 3ac)

Question: 2

Write each of the following statements by using appropriate grouping symbols:

(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 2b – 7.

(ii) Three times the sum of 2x + y – [5 – (x – 3y)] and 7x – 4y + 3 is subtracted from 3x – 4y + 7

(iii) The subtraction of x² – y² + 4xy from 2x² + y² – 3xy is added to 9x² – 3y²– xy.

Solution:

(i) The sum of a – b and 3a – 2b + 5 = [(a – b) + (3a – 2b + 5)].

This is subtracted from 4a + 2b – 7.

Thus, the required expression is (4a + 2b – 7) – [(a – b) + (3a – 2b + 5)]

(ii) Three times the sum of 2x + y – {5 – (x – 3y)} and 7x – 4y + 3 = 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

This is subtracted from 3x – 4y + 7.

Thus, the required expression is (3x – 4y + 7) – 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

(iii) The product of subtraction of x²– y² + 4xy from 2x² + y² – 3xy is given by {(2x² + y² – 3xy) – (x²-y² + 4xy)}

When the above equation is added to 9x² – 3y² – xy, we get

{(2x² + y² – 3xy) – (x² – y² + 4xy)} + (9x² – 3y²– xy))

Exercise 7.4

Question: 1

Simplify, the algebraic expressions by removing grouping symbols.

2x + (5x – 3y)

Solution:

We have

2x + (5x – 3y)

Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.

= 2x + 5x – 3y

= 7x – 3y

Question: 2

Simplify, the algebraic expressions by removing grouping symbols.

3x – (y – 2x)

Solution:

We have

3x – (y – 2x)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

3x – y + 2x

= 5x – y

Question: 3

Simplify, the algebraic expressions by removing grouping symbols.

5a – (3b – 2a + 4c)

Solution:

We have

5a – (3b – 2a + 4c)

Since the ‘-‘ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them.

= 5a – 3b + 2a – 4c

= 7a – 3b – 4c

Question: 4

Simplify, the algebraic expressions by removing grouping symbols.

-2(x² – y² + xy) – 3(x² +y² – xy)

Solution:

We have

– 2(x² – y² + xy) – 3(x² +y² – xy)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= -2x² + 2y² – 2xy – 3x² – 3y² + 3xy

= -2x² – 3x² + 2y² – 3y² – 2xy + 3xy

= -5x² – y² + xy

Question: 5

Simplify, the algebraic expressions by removing grouping symbols.

3x + 2y – {x – (2y – 3)}

Solution:

We have

3x + 2y – {x – (2y – 3)}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 3x + 2y – {x – 2y + 3}

= 3x + 2y – x + 2y – 3

= 2x + 4y – 3

Question: 6

Simplify, the algebraic expressions by removing grouping symbols.

5a – {3a – (2 – a) + 4}

Solution:

We have

5a – {3a – (2 – a) + 4}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 5a – {3a – 2 + a + 4}

= 5a – 3a + 2 – a – 4

= 5a – 4a – 2

= a – 2

Question: 7

Simplify, the algebraic expressions by removing grouping symbols.

a – [b – {a – (b – 1) + 3a}]

Solution:

First we have to remove the parentheses, or small brackets, ( ), then the curly brackets, { }, and then the square brackets [ ].

Therefore, we have

a – [b – {a – (b – 1) + 3a}]

= a – [b – {a – b + 1 + 3a}]

= a – [b – {4a – b + 1}]

= a – [b – 4a + b – 1]

= a – [2b – 4a – 1]

= a – 2b + 4a + 1

= 5a – 2b + 1

Question: 8

Simplify, the algebraic expressions by removing grouping symbols.

 a – [2b – {3a – (2b – 3c)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

a – [2b – {3a – (2b – 3c)}]

= a – [2b – {3a – 2b + 3c}]

= a – [2b – 3a + 2b – 3c]

= a – [4b – 3a – 3c]

= a – 4b + 3a + 3c

= 4a – 4b + 3c

Question: 9

Simplify, the algebraic expressions by removing grouping symbols.

 -x + [5y – {2x – (3y – 5x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets { }, and then the square brackets, [ ].

Therefore, we have

– x + [5y – {2x – (3y – 5x)}]

= – x + [5y – {2x – 3y + 5x)]

= – x + [5y – {7x – 3y}]

= – x + [5y – 7x + 3y]

= – x + [8y – 7x]

= – x + 8y – 7x

= – 8x + 8y

Question: 10

Simplify, the algebraic expressions by removing grouping symbols.

 2a – [4b – {4a – 3(2a – b)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2a – [4b – {4a – 3(2a – b)}]

= 2a – [4b – {4a – 6a + 3b}]

= 2a – [4b – {- 2a + 3b}]

= 2a – [4b + 2a – 3b]

= 2a – [b + 2a]

= 2a – b – 2a

= – b

Question: 11

Simplify, the algebraic expressions by removing grouping symbols.

-a – [a + {a + b – 2a – (a – 2b)} – b]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

– a – [a + {a + b – 2a – (a – 2b)} – b]

= – a – [a + {a + b – 2a – a + 2b} – b]

= – a – [a + {- 2a + 3b} – b]

= – a – [a – 2a + 3b – b]

= – a – [- a + 2b]

= – a + a – 2b

= – 2b

Question: 12

Simplify, the algebraic expressions by removing grouping symbols.

2x – 3y – [3x – 2y -{x – z – (x – 2y)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2x – 3y – [3x – 2y – {x – z – (x – 2y)})

= 2x – 3y – [3x – 2y – {x – z – x + 2y}]

= 2x – 3y – [3x – 2y – {- z + 2y}]

= 2x – 3y – [3x – 2y + z – 2y]

= 2x – 3y – [3x – 4y + z]

= 2x – 3y – 3x + 4y – z

= – x + y – z

Question: 13

Simplify, the algebraic expressions by removing grouping symbols.

 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]

= 5 + [x – {y – 4x + 4} – {x – y + 2}]

= 5 + [x – y + 4x – 4 – x + y – 2]

= 5 + [4x – 6]

= 5 + 4x – 6

= 4x – 1

Question: 14

Simplify, the algebraic expressions by removing grouping symbols.

x² – [3x + [2x – (x² – 1)] + 2]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

x² – [3x + [2x – (x² – 1)] + 2]

= x² – [3x + [2x – x² + 1] + 2]

= x² – [3x + 2x – x² + 1 + 2]

= x² – [5x – x² + 3]

= x² – 5x + x² – 3

= 2x² – 5x – 3

Question: 15

Simplify, the algebraic expressions by removing grouping symbols.

20 – [5xy + 3[x² – (xy – y) – (x – y)]]

Solution:

20 – [5xy + 3[x² – (xy – y) – (x – y)]]

= 20 – [5xy + 3[x² – xy + y – x + y]]

= 20 – [5xy + 3[x² – xy + 2y – x]]

= 20 – [5xy + 3x² – 3xy + 6y – 3x]

= 20 – [2xy + 3x² + 6y – 3x]

= 20 – 2xy – 3x² – 6y + 3x

= – 3x² – 2xy – 6y + 3x + 20

Question: 16

Simplify, the algebraic expressions by removing grouping symbols.

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

= 85 – [12x – 56x + 21 – 2{10x – 10 + 20x}]

= 85 – [12x – 56x + 21 – 2{30x – 10}]

= 85 – [12x – 56x + 21 – 60x + 20]

= 85 – [12x – 116x + 41]

= 85 – [- 104x + 41]

= 85 + 104x – 41

= 44 + 104x

Question: 17

Simplify, the algebraic expressions by removing grouping symbols.

 xy[yz – zx – {yx – (3y – xz) – (xy – zy)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]

= xy – [yz – zx – {yx – 3y + xz – xy + zy}]

= xy – [yz – zx – {- 3y + xz + zy}]

= xy – [yz – zx + 3y – xz – zy]

= xy – [- zx + 3y – xz]

= xy – [- 2zx + 3y]

= xy + 2xz – 3y