RD Sharma Solutions Class 7 Algebraic Expressions Exercise 7.4

RD Sharma Solutions Class 7 Chapter 7 Exercise 7.4

RD Sharma Class 7 Solutions Chapter 7 Ex 7.4 PDF Download

Exercise 7.4

Simplify each of the following algebraic expressions by removing grouping symbols.

 Q1) 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

Q2) 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

Q3) 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

Q4) \(-2(x^{2}-y^{2}+xy)-3(x^{2}+y^{2}-xy)\)

Solution:

We have

\(-2(x^{2}-y^{2}+xy)-3(x^{2}+y^{2}-xy)\)

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

= \(-2x^{2}+2y^{2}-2xy-3x^{2}-3y^{2}+3xy\)

= \(-2x^{2}-3x^{2}+2y^{2}-3y^{2}-2xy+3xy\)

= \(-5x^{2}-y^{2}+xy\)

Q5) 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

Q6) 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

Q7) 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

Q8) 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

 Q9) -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

Q10) 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

Q11)  -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

Q12)  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

Q13) 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

Q14) \(x^{2}-[3x+[2x-(x^{2}-1)]+2]\)

Solution:

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

Therefore, we have

\(x^{2}-[3x+[2x-(x^{2}-1)]+2]\)

= \(x^{2}-[3x+[2x-x^{2}+1]+2]\)

= \(x^{2}-[3x+2x-x^{2}+1+2]\)

= \(x^{2}-[5x-x^{2}+3]\)

= \(x^{2}-5x+x^{2}-3\)

= \(2x^{2}-5x-3\)

Q15) \(20-[5xy+3[x^{2}-(xy-y)-(x-y)]]\)

Solution:

\(20-[5xy+3[x^{2}-(xy-y)-(x-y)]]\)

= \(20-[5xy+3[x^{2}-xy+y-x+y]]\)

= \(20-[5xy+3[x^{2}-xy+2y-x]]\)

= \(20-[5xy+3x^{2}-3xy+6y-3x]\)

= \(20-[2xy+3x^{2}+6y-3x]\)

= \(20-2xy-3x^{2}-6y+3x\)

= \(-3x^{2}-2xy-6y+3x+20\)

  

Q16) 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

Q17)  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