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) a b c + 1
(vii) 3x – 2 + 5
(viii) 2x – 3y + 4
(ix) x y + y z + z x
(x) ax3 + bx2 + cx + d
Solution 1:
(i) a2
a2 contains only one term. That is why it is a monomial expression.
(ii) a2 − b2
a2 − b2 contains two terms. That is why it is a binomial expression.
(iii) x3 + y3 + z3
x3 + y3 + z3 contains three terms. That is why it is a trinomial expression.
(iv) x3 + y3 + z3 + 3xyz
x3 + y3 + z3 + 3xyz contains four terms. That is why it is a quadrinomial expression.
(v) 7 + 5
7 + 5 = 12 contains one terms. That is why it is a monomial expression.
(vi) a b c + 1
a b c + 1 contains two terms That is why it is a binomial expression.
(vii) 3x – 2 + 5
3x – 2 + 5 contains two terms That is why it is a binomial expression.
(viii) 2x – 3y + 4
2x – 3y + 4 contains three terms That is why it is a trinomial expression.
(ix) x y + y z + z x
x y + y z + z x contains three terms is a trinomial expression.
(x) ax3 + bx2 + cx + d
ax3 + bx2 + cx + d contains four terms That is why it is a quadrinomial expression.
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 2:
(i) 3x = is the only term of the given algebraic expression.
(ii) 2x – 3 = 2x and -3 are the terms of the given algebraic expression.
(iii) 2x2 – 7 = 2x2 and −7 are the terms of the given algebraic expression.
(iv) 2x2 + y2 − 3xy + 4 = 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, (3/2)abc2, (-4/3)cba2
Solution 3:
(i) Like terms 4xy, -3yx and Numerical coefficients of these are 4, -3
(ii) Like terms (7a2bc, −3ca2b) and (-4/3)cba2 and their Numerical coefficients are 7, -3,(-4/3)
Like terms are (-5/2)abc2 and (3/2)abc2 and their numerical coefficients are (-5/2) and (3/2)
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 4:
(i) a2 + b2 -2a2 + c2 + 4a
The like terms in the a2 + b2 -2a2 + c2 + 4a algebraic expressions are a2 and −2a2.
(ii) 3x + 4xy − 2yz + 52zy
The like terms in the 3x + 4xy − 2yz + 52zy algebraic expressions are -2yz and 52zy.
(iii) abc + ab2c + 2acb2 + 3c2ab + b2ac − 2a2bc + 3cab2
The like terms in the abc + ab2c + 2acb2 + 3c2ab + b2ac − 2a2bc + 3cab2 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 5:
(i) -12x = numerical coefficient of x is -12.
(ii) -7xy = numerical coefficient of x is -7y.
(iii) xyz = numerical coefficient of x is yz.
(iv) -7ax = numerical coefficient of x is -7a.
Question 6: Write the coefficient of x2 in the following:
(i) −3x2
(ii) 5x2yz
(iii) 5/7 x2z
(iv) (-3/2)ax2 + yx
Solution 6:
(i) −3x2 = numerical coefficient of x2 is -3.
(ii) 5x2yz = numerical coefficient of x2 is 5yz.
(iii) 5/7 x2z = numerical coefficient of x2 is 5/7 z.
(iv) (-3/2)ax2 + yx = 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 7:
(i) –3y = coefficient of y is -3.
(ii) 2ab = coefficient of a is 2b.
(iii) -7xyz = coefficient of z is -7xy.
(iv) -3pqr = The coefficient of p is -3qr.
(v) 9xy2z = coefficient of y2 is 9xz.
(vi) x3 +1 = coefficient of x3 is 1.
(vii) − x2 = 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 8:
(i) xy = numerical coefficient in the term xy is 1.
(ii) -6yz = numerical coefficient in the term – 6yz is – 6.
(iii) 7abc = numerical coefficient in the term 7abc is 7.
(iv) -2x3y2z = numerical coefficient in the term −2x3y2z is -2.
Question 10: Write the constant term of each of the following algebraic expressions:
(i) x2y − xy2 + 7xy − 3
(ii) a3 − 3a2 + 7a + 5
Solution 10:
(i) x2y − xy2 + 7xy − 3
-3 is the constant term.
(ii) Given a3 − 3a2 + 7a + 5
5 is the constant term.
Question 11: Evaluate each of the following expressions for x = -2, y = -1, z = 3:
(i) (x/y) + (y/z) + (z/x)
(ii) x2 + y2 + z2 – xy – yz – zx
Solution 11:
(i) x = -2, y = -1, z = 3
We take it as: (x/y) + (y/z) + (z/x)
On putting the given values we get,
= ((-2)/(-1)) + ((-1)/3) + (3/(-2))
The LCM of 3 and 2 is 6
= (((12-2-9))/6) = (1/6)
(ii) x = -2, y = -1, z = 3
We take it as: x2 + y2 + z2 – xy – yz – zx
On putting the given values we get,
= (-2)2 + (-1)2 + 32 – (-2) (-1) – (-1) (3) – (3) (-2)
= 4 + 1 + 9 – 2 + 3 + 6
= 23 – 2 = 21
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 12:
(i) x = 1, y = -1, z = 2, a = -2, b = 1, c = -2
We take it as: ax + by + cz
On putting the given values
= (-2) (1) + (1) (-1) + (-2) (2)
= –2 – 1 – 4 = –7
(ii) x = 1, y = -1, z = 2, a = -2, b = 1, c = -2
We take it as: ax2 + by2 – cz
On putting the given values
= (-2) × 12 + 1 × (-1)2 – (-2) × 2
= -2 + 1 – (-4)
= -1 + 4
= 3
(iii) x = 1, y = -1, z = 2, a = -2, b = 1, c = -2
We take it as: 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 1:
(i) 3x and 7x
= 3x + 7x
= (3 + 7) x
= 10x
(ii) -5xy and 9xy
= -5xy + 9xy
= (-5 + 9) xy
= 4xy
Question 2: Simplify each of the following:
(i) 7x3y +9yx3
(ii) 12a2b + 3ba2
Solution 2:
(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 3:
(i) 7abc, -5abc, 9abc, -8abc
= 7abc + (-5abc) + (9abc) + (-8abc)
= 7abc – 5abc + 9abc – 8abc
= (7 – 5 + 9 – 8) abc [abc taking common]
= (16 – 13) abc
= 3abc
(ii) 2x2y, – 4x2y, 6x2y, -5x2y
= 2x2y +(-4x2y) + (6x2y) + (-5x2y)
= 2x2y – 4x2y + 6x2y – 5x2y
= (2- 4 + 6 – 5) x2y [by taking x2 y common]
= (8 – 9) x2y
= -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 4:
(i) x3 -2x2y + 3xy2– y3, 2x3– 5xy2 + 3x2y – 4y3
Taking like terms together,
= x3 +2x3 – 2x2y + 3x2y + 3xy2 – 5xy2 – y3– 4y3
= 3x3 + x2y – 2xy2 – 5y3
(ii) a4 – 2a3b + 3ab3 + 4a2b2 + 3b4, – 2a4 – 5ab3 + 7a3b – 6a2b2 + b4
Taking like terms together,
= 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 5:
(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b
= (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)
Taking positive and negative like terms together,= 8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b
= 8a – 10a – 7ab + 2ab + 8b – 8b
= –2a – 5ab
(ii) 5x3 + 7 + 6x – 5x2, 2x2 – 8 – 9x, 4x – 2x2 + 3 x 3, 3 x 3 – 9x – x2 and x – x2 – x3 – 4
= (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)
Taking positive and negative like terms together,
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 6:
(i) x – 3y – 2z, 5x + 7y – 8z and 3x – 2y + 5z
= (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)
Taking positive and negative like terms together,
= x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z
= 9x – 5y + 7y – 10z + 5z
= 9x + 2y – 5z
(ii) Given 4ab – 5bc + 7ca, –3ab + 2bc – 3ca and 5ab – 3bc + 4ca
= (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)
Taking positive and negative like terms together,
= 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 7:
2x2 – 3x + 1, 3x2 – 2x and 3x + 7
sum of 3x2 – 2x and 3x + 7
= (3x2 – 2x) + (3x +7)
=3x2 – 2x + 3x + 7
= (3x2 + x + 7)
Now, Add the 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 8:
x2 + 2xy + y2, x2 – 3y2and 2x2 – y2 + 9.
Firstly, find the sum of x2 – 3y2 and 2x2 – y2 + 9
= (x2 – 3y2) + (2x2 – y2 + 9)
= x2 + 2x2 – 3y2 – y2+ 9
= 3x2 – 4y2 + 9
Now, Add the expression
= (x2 + 2xy + y2) + (3x2 – 4y2 + 9)
= x2 + 3x2 + 2xy + y2 – 4y2 + 9
= 4x2 + 2xy – 3y2+ 9
Question 9: Add a3+ b3 – 3 to the sum of 2a3 – 3b3 – 3ab + 7 and -a3 + b3 + 3ab – 9.
Solution 9:
a3+ b3 – 3, 2a3 – 3b3 – 3ab + 7 and -a3 + b3 + 3ab – 9.
First, find the sum of 2a3 – 3b3– 3ab + 7 and – a3 + b3 + 3ab – 9.
= (2a3 – 3b3– 3ab + 7) + (- a3 + b3 + 3ab – 9)
Taking positive and negative like terms together, we get
= 2a3 – a3– 3b3+ b3 – 3ab + 3ab + 7 – 9
= a3 – 2b3 – 2
Now, Add the expression
= (a3 + b3 – 3) + (a3 – 2b3 – 2).
= a3+ a3+ b3– 2b3 – 3 – 2
= 2a3 – b3 – 5
Question 10: Subtract:
(i) 7a2b from 3a2b
(ii) 4xy from -3xy
Solution 10:
(i) 7a2b from 3a2b
= 3a2b -7a2b
= (3 -7) a2b
= – 4a2b
(ii) 4xy from -3xy
= –3xy – 4xy
= –7xy
Question 11: Subtract:
(i) – 4x from 3y
(ii) – 2x from – 5y
Solution 11:
(i) – 4x from 3y
= (3y) – (–4x)
= 3y + 4x
(ii) – 2x from – 5y
= (-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 12:
(i) 6x3 −7x2 + 5x − 3 and 4 − 5x + 6x2 − 8x3
= (4 – 5x + 6x2 – 8x3) – (6x3 – 7x2 + 5x – 3)
= 4 – 5x + 6x2 – 8x3 – 6x3 + 7x2 – 5x + 3
= – 8x3– 6x3 + 7x2 + 6x2– 5x – 5x + 3 + 4
= – 14x3 + 13x2 – 10x +7
(ii) − x2 −3z and 5x2 – y + z + 7
= (5x2 – y + z + 7) – (- x2 – 3z)
= 5x2 – y + z + 7 + x2 + 3z
= 5x2+ x2 – y + z + 3z + 7
= 6x2 – y + 4z + 7
(iii) x3 + 2x2y + 6xy2 − y3 and y3−3xy2−4x2y
= (y3 – 3xy2 – 4x2y) – (x3 + 2x2y + 6xy2 – y3)
= y3 – 3xy2 – 4x2y – x3 – 2x2y – 6xy2 + y3
= y3 + y3– 3xy2– 6xy2– 4x2y – 2x2y – x3
= 2y3– 9xy2 – 6x2y – x3
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 13:
(i) p3 – 4 + 3p2, take away 5p2 − 3p3 + p − 6
= (p3 – 4 + 3p2) – (5p2 – 3p3 + p – 6)
= p3 – 4 + 3p2 – 5p2 + 3p3 – p + 6
= p3 + 3p3 + 3p2 – 5p2– p – 4+ 6
= 4p3 – 2p2 – p + 2
(ii) 7 + x − x2, take away 9 + x + 3x2 + 7x3
= (7 + x – x2) – (9 + x + 3x2 + 7x3)
= 7 + x – x2 – 9 – x – 3x2 – 7x3
= – 7x3– x2 – 3x2 + 7 – 9
= – 7x3 – 4x2 – 2
(iii) 1− 5y2, take away y3 + 7y2 + y + 1
= (1 – 5y2) – (y3+ 7y2 + y + 1)
= 1 – 5y2 – y3 – 7y2 – y – 1
= – y3– 5y2 – 7y2 – y
= – y3– 12y2 – y
(iv) x3 − 5x2 + 3x + 1, take away 6x2 − 4x3 + 5 + 3x
= (x3 – 5x2 + 3x + 1) – (6x2 – 4x3 + 5 +3x)
= x3 – 5x2 + 3x + 1 – 6x2 + 4x3 – 5 – 3x
= x3+ 4x3 – 5x2 – 6x2 + 1 – 5
= 5x3 – 11x2 – 4
Question 14: From the sum of 3x2 − 5x + 2 and − 5x2 − 8x + 9 subtract 4x2 − 7x + 9.
Solution 14:
Firstly, Add 3x2 − 5x + 2 and − 5x2 − 8x + 9
= {(3x2 – 5x + 2) + (- 5x2 – 8x + 9)}
= {3x2 – 5x + 2 – 5x2 – 8x + 9}
= {3x2 – 5x2 – 5x – 8x + 2 + 9}
= {- 2x2 – 13x +11}
Then, subtract the result from 4x2 − 7x + 9.
= – 2x2 – 13x + 11 – 4x2 + 7x – 9
= – 2x2 – 4x2 – 13x + 7x + 11 – 9
= – 6x2 – 6x + 2
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 15:
Firstly, find the sum of 13x – 4y + 7z and – 6z + 6x + 3y
So, 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)
Now find the 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, Subtract 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 16:
Firstly find the sum of (x2 + 3y2 – 6xy), (2x2 – y2 + 8xy), (y2 + 8) and (x2 – 3xy)
={(x2 + 3y2 – 6xy) + (2x2 – y2 + 8xy) + ( y2 + 8) + (x2 – 3xy)}
={x2 + 3y2 – 6xy + 2x2 – y2 + 8xy + y2 + 8 + x2 – 3xy}
= {x2+ 2x2+ x2 + 3y2– y2 + y2– 6xy + 8xy – 3xy + 8}
= 4x2 + 3y2 – xy + 8
Now, subtract the result from the −3x2 + 4y2 – xy + x – y + 3.
Therefore, subtract the expression
= (4x2 + 3y2 – xy + 8) – (- 3x2 + 4y2 – xy + x – y + 3)
= 4x2 + 3y2 – xy + 8 + 3x2 – 4y2 + xy – x + y – 3
= 4x2 + 3x2+ 3y2– 4y2– x + y – 3 + 8
= 7x2 – y2– x + y + 5
Question 17: What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?
Solution 17:
By subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7,
we get the required expression.
Therefore, subtract 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 18:
Let ‘E’ be the required expression. Then,
x2 – xy + y2– x + y + 3 – E = – x2 + 3y2 – 4xy + 1
Therefore, E = (x2 – xy + y2– x + y + 3) – (- x2 + 3y2 – 4xy + 1)
= x2 – xy + y2– x + y + 3 + x2 – 3y2 + 4xy – 1
Taking positive and negative like terms together, we get
= x2 + x2– xy + 4xy + y2– 3y2 – x + y + 3 – 1
= 2x2+ 3xy- 2y2– x + y + 2
Question 19: How much is x – 2y + 3z greater than 3x + 5y – 7?
Solution 19:
Subtracting x – 2y + 3z from 3x + 5y – 7 to get the required expression,
Required expression = (x – 2y + 3z) – (3x + 5y – 7)
= x – 2y + 3z – 3x – 5y + 7
Taking 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 20:
Subtracting the x2 − 2xy + 3y2 from 2x2 − 3y2 + xy to get the required expression,
Required expression = (2x2 – 3y2 + xy) – (x2 – 2xy + 3y2)
= 2x2 – 3y2 + xy – x2 + 2xy – 3y2
Taking positive and negative like terms together, we get
= 2x2– x2 – 3y2 – 3y2 + xy + 2xy
= x2 – 6y2 + 3xy
Question 21: How much does a2 − 3ab + 2b2 exceed 2a2 − 7ab + 9b2?
Solution 21:
Subtracting 2a2 − 7ab + 9b2 from a2 − 3ab + 2b2 to the required expression
Required expression = (a2– 3ab + 2b2) – (2a2 – 7ab + 9b2)
= a2– 3ab + 2b2 – 2a2 + 7ab – 9b2
Taking positive and negative like terms together, we get
= a2 – 2a2 – 3ab + 7ab + 2b2 – 9b2
= – a2 + 4ab – 7b2
Question 22: What must be added to 12x3 − 4x2 + 3x − 7 to make the sum x3 + 2x2 − 3x + 2?
Solution 22:
Let ‘E’ be the required expression. Thus,
12x3 – 4x2 + 3x – 7 + E = x3 + 2x2 – 3x + 2
Therefore, E = (x3 + 2x2 – 3x + 2) – (12x3 – 4x2 + 3x – 7)
= x3 + 2x2 – 3x + 2 – 12x3 + 4x2 – 3x + 7
Taking positive and negative like terms together,
= x3– 12x3+ 2x2 + 4x2 – 3x – 3x + 2 + 7
= – 11x3 + 6x2 – 6x + 9
Question 23: If P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2, show that P + Q + R = 0.
Solution 23:
P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2
To prove: P + Q + R = 0,
P + Q + R = (7x2 + 5xy – 9y2) + (4y2 – 3x2 – 6xy) + (- 4x2 + xy + 5y2)
= 7x2 + 5xy – 9y2 + 4y2 – 3x2 – 6xy – 4x2 + xy + 5y2
Taking positive and negative like terms together, we get
= 7x2– 3x2 – 4x2 + 5xy – 6xy + xy – 9y2 + 4y2 + 5y2
= 7x2– 7x2 + 6xy – 6xy – 9y2 + 9y2
= 0
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 24:
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
Substituting all values
= P + Q + R + S – T = {(a2 – b2 + 2ab) + (a2 + 4b2 – 6ab) + (b2 + b) + (a2 – 4ab)} – (-2a2 + b2 – ab + a)
= {a2 – b2 + 2ab + a2 + 4b2 – 6ab + b2 + b + a2 – 4ab}- (- 2a2 + b2 – ab + a)
= {3a2 + 4b2 – 8ab + b } – (-2a2 + b2 – ab + a)
= 3a2+ 4b2 – 8ab + b + 2a2 – b2 + ab – a
positive and negative like terms together, we get
3a2+ 2a2 + 4b2 – b2 – 8ab + ab – a + b
= 5a2 + 3b2– 7ab – a + b
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:
(i) x + y – 3z + y
x + y – 3z + y
= x + y – (3z – y)
(ii) 3x – 2y – 5z – 4
3x – 2y – 5z – 4
= 3x – 2y – (5z + 4)
(iii) 3a – 2b + 4c – 5
3a – 2b + 4c – 5
= 3a – 2b – (–4c + 5)
(iv) 7a + 3b + 2c + 4
7a + 3b + 2c + 4
= 7a + 3b – (–2c – 4)
(v) 2a2 – b2 – 3ab + 6
2a2 – b2 – 3ab + 6
= 2a2 – b2 – (3ab – 6)
(vi) a2 + b2 – c2 + ab – 3ac
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 2:
(i) Sum of a – b and 3a – 2b + 5 = [(a – b) + (3a – 2b + 5)].
By subtracted from 4a + 2b – 7.
Therefore, 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)]
Subtracted from 3x – 4y + 7.
Therefore, 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)}
Above equation is added to 9x2 – 3y2 – xy,
{(2x2 + y2 – 3xy) – (x2 – y2 + 4xy)} + (9x2 – 3y2– xy))
This is the required expression.
Exercise 7.4
Simplify each of the following algebraic expressions by removing grouping symbols.
Question 1: 2x + (5x – 3y)
Solution 1:
2x + (5x – 3y)
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
Question 2: 3x – (y – 2x)
Solution 2:
3x – (y – 2x)
= 3x – y + 2x
On simplifying, we get
= 5x – y
Question 3: 5a – (3b – 2a + 4c)
Solution 3:
5a – (3b – 2a + 4c)
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
Question 4: -2(x2 – y2 + xy) – 3(x2 +y2 – xy)
Solution 4:
– 2(x2 – y2 + xy) – 3(x2 +y2 – xy)
= -2x2 + 2y2 – 2xy – 3x2 – 3y2 + 3xy
On rearranging,
= -2x2 – 3x2 + 2y2 – 3y2 – 2xy + 3xy
On simplifying, we get
= -5x2 – y2 + xy
Question 5: 3x + 2y – {x – (2y – 3)}
Solution 5:
3x + 2y – {x – (2y – 3)}
First, remove the parentheses. Then, we have to remove the braces.
Then we get expression as:
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
Question 6: 5a – {3a – (2 – a) + 4}
Solution 6:
5a – {3a – (2 – a) + 4}
Firstly , remove the parentheses. Then, we have to remove the braces.
Then we get expression as:
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
Question 7: a – [b – {a – (b – 1) + 3a}]
Solution 7:
a – [b – {a – (b – 1) + 3a}]
Firstly, remove the parentheses, then the curly brackets, and then the square brackets.
Then we get the expression:
= 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]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
Question 8: a – [2b – {3a – (2b – 3c)}]
Solution 8:
a – [2b – {3a – (2b – 3c)}]
Firstly, remove the parentheses, then the braces, and then the square brackets.
Then we get the expression:
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
Question 9: -x + [5y – {2x – (3y – 5x)}]
Solution 9:
-x + [5y – {2x – (3y – 5x)}]
Firstly remove the parentheses, then remove braces, and then the square brackets.
Then we get the ecpression:
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
Question 10: 2a – [4b – {4a – 3(2a – b)}]
Solution 10:
2a – [4b – {4a – 3(2a – b)}]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
Question 11: -a – [a + {a + b – 2a – (a – 2b)} – b]
Solution 11:
-a – [a + {a + b – 2a – (a – 2b)} – b]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= – 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]
On simplifying, we get
= – a + a – 2b
= – 2b
Question 12: 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Solution 12:
2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 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]
On simplifying, we get
= 2x – 3y – 3x + 4y – z
= – x + y – z
Question 13: 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Solution 13:
5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Firstly remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 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: x2 – [3x + [2x – (x2 – 1)] + 2]
Solution 14:
x2 – [3x + [2x – (x2 – 1)] + 2]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= x2 – [3x + [2x – (x2 – 1)] + 2]
= x2 – [3x + [2x – x2 + 1] + 2]
= x2 – [3x + 2x – x2 + 1 + 2]
= x2 – [5x – x2 + 3]
On simplifying we get
= x2 – 5x + x2 – 3
= 2x2 – 5x – 3
Question 15: 20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
Solution 15:
20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 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]
On simplifying we get
= 20 – 2xy – 3x2 – 6y + 3x
= – 3x2 – 2xy – 6y + 3x + 20
Question 16: 85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
Solution 16:
85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 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]
On simplifying, we get
= 85 + 104x – 41
= 44 + 104x
Question 17: xy [yz – zx – {yx – (3y – xz) – (xy – zy)}]
Solution 17:
Given xy [yz – zx – {yx – (3y – xz) – (xy – zy)}]
Firstly, remove the parentheses, then remove braces, and then the square brackets.
Then we get the expression:
= 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]
On simplifying, we get
= xy – [- 2zx + 3y]
= xy + 2xz – 3y