Exercise 7.4
Simplify, the algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
We get,
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
2. 3x – (y – 2x)
Solution:
Given 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
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given: 5a – (3b – 2a + 4c) = 5a – 3b + 2a – 4c
= 7a – 3b – 4c
4. -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
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x- (2y – 3)
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x-2y + 3}
= 3x + 2y – x+ 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
We get,
5a – {3a – (2 – a) + 4}
First we remove to parentheses, than the braces and then the square brackets.
Therefore,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
First we remove to parentheses, than the braces 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
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [ 2b – { 32 – ( 2b -3c)}]
First we remove to parentheses, than the braces 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
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
First we remove to parentheses, than the braces 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
10. Simplify, the algebraic expressions by removing grouping symbols.
2a – [4b – {4a – 3(2a – b)}]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then we have,
= 2a – [ 4b – {4a – 3(2a – b)}]
= 2a – [4b – { 4a – 6a – 6a + 3b})]
= 2a – [4b – { -2a + 3b}[
= 2a – [ 4b + 2a – 3b]
= 2a – [ b + 2a]
= 2a – b – 2a
= – b
11. Simplify, the algebraic expressions by removing grouping symbols.
-a – [a + {a + b – 2a – (a – 2b)} – b]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then 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
12. Simplify, the algebraic expressions by removing grouping symbols.
2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then 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
13. Simplify, the algebraic expressions by removing grouping symbols.
5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then 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
14. x2 – [3x + [2x – (x2 – 1)] + 2]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then, 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
15. 20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then, we have
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
= – 3×2 – 2xy – 6y + 3x + 20
16. 85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then, 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
17. xy[yz – zx – {yx – (3y – xz) – (xy – zy)}]
Solution:
First we remove to parentheses, than the braces and then the square brackets.
Then, 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