GSEB Class 7 Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.2

Question 1. Simplify by combining like terms:
(i) \( 21b - 32 + 7b - 20b \)
(ii) \( -z^2 + 13z^2 - 5z + 7z^3 - 15z \)
(iii) \( p - (p - q) - q - (q - p) \)
(iv) \( 3a - 2b - ab - (a - b + ab ) + 3ab + b - a \)
(v) \( 5x^2y - 5x^2 + 3yx^2 - 3y^2 + x^2 - y^2 + 8xy^2 - 3y^2 \)
(vi) \( (3y^2 + 5y - 4) - (8y - y^2 - 4) \)
Answer:
(i) We combine the similar terms given:
\( 21b - 32 + 7b - 20b \)
\( = (21b + 7b - 20b) + (-32) \)
\( = (21 + 7 - 20)b + (-32) \)
\( = (8)b + (-32) = 8b - 32 \)
(ii) We collect the similar terms here:
\( -z^2 + 13z^2 - 5z + 7z^3 - 15z \)
\( = (7z^3) + (-z^2 + 13z^2) + (-5z - 15z) \)
\( = (7z^3) + (-1 + 13)z^2 + (-5 - 15)z \)
\( = (7)z^3 + (12)z^2 + (-20)z \)
\( = 7z^3 + 12z^2 - 20z \)
(iii) We have this expression:
\( p - (p - q) - q - (q - p) \)
\( = p - p + q - q - q + p \)
Collecting the similar terms, we get:
\( = (p - p + p) + (q - q - q) \)
\( = (1 - 1 + 1)p + (1 - 1 - 1)q \)
\( = (1)p + (-1)q \)
\( = p - q \)
(iv) We have the given expression:
\( 3a - 2b - ab - (a - b + ab) + 3ab + b - a \)
\( = 3a - 2b - ab - a + b - ab + 3ab + b - a \)
Collecting the similar terms, we find:
\( = (3a - a - a) + (-2b + b + b) + (-ab - ab + 3ab) \)
\( = (3 - 1 - 1)a + (-2 + 1 + 1)b + (-1 - 1 + 3)ab \)
\( = (1)a + (0)b + (+1)ab = a + ab \)
(v) We collect the similar terms for this expression:
\( 5x^2y - 5x^2 + 3yx^2 - 3y^2 + x^2 - y^2 + 8xy^2 - 3y^2 \)
\( = (5x^2y + 3yx^2) + (8xy^2) + (-5x^2 + x^2) + (-3y^2 - y^2 - 3y^2) \)
\( = (5 + 3)x^2y + (8)xy^2 + (-5 + 1)x^2 + (-3 - 1 - 3)y^2 \)
\( = 8x^2y + 8xy^2 + (-4)x^2 + (-7)y^2 \)
\( = 8x^2y + 8xy^2 - 4x^2 - 7y^2 \)
(vi) We have the following expression:
\( (3y^2 + 5y - 4) - (8y - y^2 - 4) \)
\( = 3y^2 + 5y - 4 - 8y + y^2 + 4 \)
Collecting the similar terms, we obtain:
\( = (3y^2 + y^2) + (5y - 8y) + (-4 + 4) \)
\( = (3 + 1)y^2 + (5 - 8)y + (-4 + 4) \)
\( = (4)y^2 + (-3)y + (0) \)
\( = 4y^2 - 3y \)
In simple words: To simplify expressions, find terms that have the exact same letters and powers, then combine their numbers. For subtraction, change the signs of all terms being subtracted and then combine like terms.

Exam Tip: Pay close attention to the signs when removing parentheses, especially when a minus sign precedes the bracket. Grouping like terms before combining them helps avoid errors.

Question 2. Add:
(i) \( 3mn, -5mn, mn, -4mn \)
(ii) \( t - 8tz, 3tz - z, z - t \)
(iii) \( -7mn + 5, 12mn + 2, 9mn - 8, -2mn - 3 \)
(iv) \( a + b - 3, b - a + 3, a - b + 3 \)
(v) \( 14x + 10y - 12xy - 13, 18 - 7x - 10y + 8xy, 4xy \)
(vi) \( 5m - 7n, 3n - 4m + 2, 2m - 3mn - 5 \)
(vii) \( 4x^2y, -3xy^2, -5xy^2, 5x^2y \)
(viii) \( 3p^2q^2 - 4pq + 5, -10p^2q^2, 15 + 9pq + 7p^2q^2 \)
(ix) \( ab - 4a, 4b - ab, 4a - 4b \)
(x) \( x^2 - y^2 - 1, y^2 - 1 - x^2, 1 - x^2 - y^2 \)
Answer:
(i) We add the given terms:
\( 3mn + (-5mn) + 8mn + (-4mn) \)
\( = [3 + (-5) + 8 + (-4)]mn \)
\( = [11 + (-9)]mn \)
\( = [2]mn \)
\( = 2mn \)
(ii) We have the expressions:
\( (t - 8tz) + (3tz - z) + (z - t) \)
\( = t - 8tz + 3tz - z + z - t \)
\( = (t - t) + (-z + z) + (-8tz + 3tz) \)
\( = (1 - 1)t + (-1 + 1)z + (-8 + 3)tz \)
\( = (0)t + (0)z + (-5)tz \)
\( = -5tz \)
(iii) We have the expressions to add:
\( (-7mn + 5) + (12mn + 2) + (9mn - 8) + (-2mn - 3) \)
\( = -7mn + 5 + 12mn + 2 + 9mn - 8 - 2mn - 3 \)
\( = (-7mn + 12mn + 9mn - 2mn) + (5 + 2 - 8 - 3) \)
\( = (-7 + 12 + 9 - 2)mn + (5 + 2 - 8 - 3) \)
\( = (21 - 9)mn + (7 - 11) \)
\( = 12mn + (-4) \)
\( = 12mn - 4 \)
(iv) We have the expressions to add:
\( (a + b - 3) + (b - a + 3) + (a - b + 3) \)
\( = a + b - 3 + b - a + 3 + a - b + 3 \)
\( = (a - a + a) + (b + b - b) + (-3 + 3 + 3) \)
\( = (1 - 1 + 1)a + (1 + 1 - 1)b + (-3 + 6) \)
\( = (2 - 1)a + (2 - 1)b + (-3 + 6) \)
\( = (1)a + (1)b + (3) \)
\( = a + b + 3 \)
(v) We have these expressions:
\( (14x + 10y - 12xy - 13) + (18 - 7x - 10y + 8xy) + 4xy \)
\( = 14x + 10y - 12xy - 13 + 18 - 7x - 10y + 8xy + 4xy \)
\( = (14x - 7x) + (10y - 10y) + (-12xy + 8xy + 4xy) + (-13 + 18) \)
\( = (14 - 7)x + (10 - 10)y + (-12 + 8 + 4)xy + (5) \)
\( = (7)x + (0)y + (-12 + 12)xy + 5 \)
\( = 7x + 0y + (0)xy + 5 \)
\( = 7x + 5 \)
(vi) We have the terms:
\( (5m - 7n) + (3n - 4m + 2) + (2m - 3mn - 5) \)
\( = 5m - 7n + 3n - 4m + 2 + 2m - 3mn - 5 \)
\( = (5m - 4m + 2m) + (-7n + 3n) - 3mn + (2 - 5) \)
\( = (5 - 4 + 2)m + (-1 + 3)n - 3mn + (-3) \)
\( = (7 - 4)m + (-4)n - 3mn - 3 \)
\( = 3m - 4n - 3mn - 3 \)
(vii) We have the expressions:
\( 4x^2y + (-3xy^2) + (-5xy^2) + 5x^2y \)
\( = 4x^2y - 3xy^2 - 5xy^2 + 5x^2y \)
\( = (4x^2y + 5x^2y) + [(-3xy^2) + (-5xy^2)] \)
\( = (4 + 5)x^2y + [(-3) + (-5)]xy^2 \)
\( = (9)x^2y + (-8)xy^2 \)
\( = 9x^2y - 8xy^2 \)
(viii) We have the expressions to add:
\( (3p^2q^2 - 4pq + 5) + (-10p^2q^2) + (15 + 9pq + 7p^2q^2) \)
\( = 3p^2q^2 - 4pq + 5 - 10p^2q^2 + 15 + 9pq + 7p^2q^2 \)
\( = (3p^2q^2 - 10p^2q^2 + 7p^2q^2) + (-4pq + 9pq) + (5 + 15) \)
\( = (3 - 10 + 7)p^2q^2 + (-4 + 9)pq + 20 \)
\( = (0)p^2q^2 + 5pq + 20 \)
\( = 5pq + 20 \)
(ix) We have the expressions:
\( (ab - 4a) + (4b - ab) + (4a - 4b) \)
\( = ab - 4a + 4b - ab + 4a - 4b \)
\( = (ab - ab) + (-4a + 4a) + (4b - 4b) \)
\( = (1 - 1)ab + (-4 + 4)a + (4 - 4)b \)
\( = (0)ab + (0)a + (0)b \)
\( = 0 + 0 + 0 = 0 \)
(x) We have the expressions:
\( (x^2 - y^2 - 1) + (y^2 - 1 - x^2) + (1 - x^2 - y^2) \)
\( = x^2 - y^2 - 1 + y^2 - 1 - x^2 + 1 - x^2 - y^2 \)
\( = (x^2 - x^2 - x^2) + (-y^2 + y^2 - y^2) + (-1 - 1 + 1) \)
\( = (1 - 1 - 1)x^2 + (-1 + 1 - 1)y^2 + (-2 + 1) \)
\( = (1 - 2)x^2 + (-2 + 1)y^2 + (-1) \)
\( = (-1)x^2 + (-1)y^2 + (-1) \)
\( = -x^2 - y^2 - 1 \)
In simple words: To add algebraic expressions, first remove any brackets, then gather terms that look alike (have the same variables and powers). Finally, add their numerical coefficients.

Exam Tip: Always double-check your sign changes when removing parentheses, especially for negative terms. Organize your work by grouping like terms vertically or horizontally to minimize errors.

Question 3. Subtract:
(i) \( -5y^2 \) from \( y^2 \)
(ii) \( 6xy \) from \( -12xy \)
(iii) \( (a - b) \) from \( (a + b) \)
(iv) \( a(b - 5) \) from \( b(5 - a) \)
(v) \( -m^2 + 5mn \) from \( 4m^2 - 3mn + 8 \)
(vi) \( -x^2 + 10x - 5 \) from \( 5x - 10 \)
(vii) \( 5a^2 - 7ab + 5b^2 \) from \( 3ab - 2a^2 - 2b^2 \)
(viii) \( 4pq - 5q^2 - 3p^2 \) from \( 5p^2 + 3q^2 - pq \)
Answer:
(i) We subtract \( -5y^2 \) from \( y^2 \):
\( y^2 - (-5y^2) \)
\( = y^2 + 5y^2 \)
\( = (1 + 5)y^2 \)
\( = 6y^2 \)
(ii) We subtract \( 6xy \) from \( -12xy \):
\( -12xy - 6xy \)
\( = (-12 - 6)xy \)
\( = -18xy \)
(iii) We subtract \( (a - b) \) from \( (a + b) \):
\( (a + b) - (a - b) \)
\( = a + b - a + b \)
\( = (a - a) + (b + b) \)
\( = (1 - 1)a + (1 + 1)b \)
\( = (0)a + (2)b \)
\( = 2b \)
(iv) We subtract \( a(b - 5) \) from \( b(5 - a) \):
\( b(5 - a) - a(b - 5) \)
\( = (5b - ab) - (ab - 5a) \)
\( = 5b - ab - ab + 5a \)
\( = 5b + 5a - ab - ab \)
\( = 5a + 5b + (-ab - ab) \)
\( = 5a + 5b + (-1 - 1)ab \)
\( = 5a + 5b + (-2)ab \)
\( = 5a + 5b - 2ab \)
(v) We subtract \( -m^2 + 5mn \) from \( 4m^2 - 3mn + 8 \):
\( (4m^2 - 3mn + 8) - (-m^2 + 5mn) \)
\( = 4m^2 - 3mn + 8 + m^2 - 5mn \)
\( = (4m^2 + m^2) + (-3mn - 5mn) + 8 \)
\( = (4 + 1)m^2 + (-3 - 5)mn + 8 \)
\( = (5)m^2 + (-8)mn + 8 \)
\( = 5m^2 - 8mn + 8 \)
(vi) We subtract \( -x^2 + 10x - 5 \) from \( 5x - 10 \):
\( (5x - 10) - (-x^2 + 10x - 5) \)
\( = 5x - 10 + x^2 - 10x + 5 \)
\( = x^2 + (5x - 10x) + (-10 + 5) \)
\( = x^2 + (5 - 10)x + (-5) \)
\( = x^2 - 5x - 5 \)
(vii) We subtract \( 5a^2 - 7ab + 5b^2 \) from \( 3ab - 2a^2 - 2b^2 \):
\( (3ab - 2a^2 - 2b^2) - (5a^2 - 7ab + 5b^2) \)
\( = 3ab - 2a^2 - 2b^2 - 5a^2 + 7ab - 5b^2 \)
\( = (3ab + 7ab) + (-2a^2 - 5a^2) + (-2b^2 - 5b^2) \)
\( = (3 + 7)ab + (-2 - 5)a^2 + (-2 - 5)b^2 \)
\( = (10)ab + (-7)a^2 + (-7)b^2 \)
\( = 10ab - 7a^2 - 7b^2 \)
(viii) We subtract \( 4pq - 5q^2 - 3p^2 \) from \( 5p^2 + 3q^2 - pq \):
\( (5p^2 + 3q^2 - pq) - (4pq - 5q^2 - 3p^2) \)
\( = 5p^2 + 3q^2 - pq - 4pq + 5q^2 + 3p^2 \)
\( = (5p^2 + 3p^2) + (3q^2 + 5q^2) + (-pq - 4pq) \)
\( = (5 + 3)p^2 + (3 + 5)q^2 + (-1 - 4)pq \)
\( = (8)p^2 + (8)q^2 + (-5)pq \)
\( = 8p^2 + 8q^2 - 5pq \)
In simple words: When you subtract one algebraic expression from another, always change the sign of every term in the expression being subtracted. Then, group the similar terms and combine them by adding or subtracting their coefficients.

Exam Tip: Remember that "subtract A from B" means B - A. Be very careful with the negative signs, distributing them to every term inside the parentheses being subtracted.

Question 4.
(a) What should be added to \( x^2 + xy + y^2 \) to obtain \( 2x^2 + 3xy \)?
(b) What should be subtracted from \( 2a + 8b + 10 \) to get \( -3a + b + 16 \)?
Answer:
(a) The required expression is:
\( (2x^2 + 3xy) - (x^2 + xy + y^2) \)
\( = 2x^2 + 3xy - x^2 - xy - y^2 \)
\( = (2x^2 - x^2) - y^2 + (3xy - xy) \)
\( = (2 - 1)x^2 - y^2 + (3 - 1)xy \)
\( = (1)x^2 - y^2 + (2)xy \)
\( = x^2 - y^2 + 2xy \)
(b) The required expression is:
\( (2a + 8b + 10) - (-3a + b + 16) \)
\( = 2a + 8b + 10 + 3a - b - 16 \)
\( = (2a + 3a) + (8b - b) + (10 - 16) \)
\( = (2 + 3)a + (8 - 1)b + (-6) \)
\( = (5)a + (7)b + (-6) \)
\( = 5a + 7b - 6 \)
In simple words: For part (a), subtract the starting expression from the target expression. For part (b), subtract the target expression from the starting expression. Always combine terms with the same variables and powers.

Exam Tip: When finding what to add or subtract, remember the simple rule: if "A + x = B", then "x = B - A". If "A - x = B", then "x = A - B".

Question 5. What should be taken away from \( 3x^2 - 4y^2 + 5xy + 20 \) to obtain \( -x^2 - y^2 + 6xy + 20 \)?
Answer:
The required expression to be taken away is:
\( (3x^2 - 4y^2 + 5xy + 20) - (-x^2 - y^2 + 6xy + 20) \)
\( = 3x^2 - 4y^2 + 5xy + 20 + x^2 + y^2 - 6xy - 20 \)
\( = (3x^2 + x^2) + (-4y^2 + y^2) + (5xy - 6xy) + (20 - 20) \)
\( = (3 + 1)x^2 + (-4 + 1)y^2 + (5 - 6)xy + (0) \)
\( = 4x^2 + (-3)y^2 + (-1)xy + 0 \)
\( = 4x^2 - 3y^2 - xy \)
In simple words: To find what needs to be removed, subtract the desired result from the starting expression. Then, combine the terms that are alike to get the final answer.

Exam Tip: This type of question tests your understanding of algebraic subtraction. Always remember to change the sign of every term in the expression being subtracted.

Question 6.
(a) From the sum of \( 3x - y + 11 \) and \( -y - 11 \), subtract \( 3x - y - 11 \).
(b) From the sum of \( 4 + 3x \) and \( 5 - 4x + 2x^2 \), subtract the sum of \( 3x^2 - 5x \) and \( -x^2 + 2x + 5 \).
Answer:
(a) First, find the sum of \( 3x - y + 11 \) and \( -y - 11 \):
\( (3x - y + 11) + (-y - 11) \)
\( = 3x - y + 11 - y - 11 \)
\( = 3x - y - y + 11 - 11 \)
\( = 3x - 2y + 0 \)
\( = 3x - 2y \)
Now, subtract \( 3x - y - 11 \) from \( 3x - 2y \):
\( (3x - 2y) - (3x - y - 11) \)
\( = 3x - 2y - 3x + y + 11 \)
\( = (3x - 3x) + (-2y + y) + 11 \)
\( = 0 + (-y) + 11 \)
\( = -y + 11 \)
(b) First, find the sum of \( 4 + 3x \) and \( 5 - 4x + 2x^2 \):
\( (4 + 3x) + (5 - 4x + 2x^2) \)
\( = 4 + 3x + 5 - 4x + 2x^2 \)
\( = (4 + 5) + (3x - 4x) + 2x^2 \)
\( = 9 + (-x) + 2x^2 \)
\( = 2x^2 - x + 9 \)
Next, find the sum of \( 3x^2 - 5x \) and \( -x^2 + 2x + 5 \):
\( (3x^2 - 5x) + (-x^2 + 2x + 5) \)
\( = 3x^2 - 5x - x^2 + 2x + 5 \)
\( = (3x^2 - x^2) + (-5x + 2x) + 5 \)
\( = (3 - 1)x^2 + (-5 + 2)x + 5 \)
\( = 2x^2 + (-3)x + 5 \)
\( = 2x^2 - 3x + 5 \)
Finally, subtract the second sum from the first sum:
\( (2x^2 - x + 9) - (2x^2 - 3x + 5) \)
\( = 2x^2 - x + 9 - 2x^2 + 3x - 5 \)
\( = (2x^2 - 2x^2) + (-x + 3x) + (9 - 5) \)
\( = (0)x^2 + (2)x + (4) \)
\( = 2x + 4 \)
In simple words: This problem asks you to do two additions first, then one subtraction. Be careful with signs when combining numbers and terms. Break down the problem into smaller steps to avoid making mistakes.

Exam Tip: For multi-step problems, calculate each sum or difference separately first. Then, perform the final subtraction, ensuring all signs are handled correctly at each stage.