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Simplify combining like terms:
(i) \(21b – 32 + 7b – 20b\)
Solution:-
\(= (21b + 7b – 20b) – 32\)
\(= b (21 + 7 – 20) – 32\)
\(= b (28 – 20) – 32\)
\(= b (8) – 32\)
\(= 8b – 32\)
(ii) \(– z^2 + 13z^2 – 5z + 7z^3 – 15z\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 7z^3 + (-z^2 + 13z^2) + (-5z – 15z)\)
\(= 7z^3 + z^2 (-1 + 13) + z (-5 – 15)\)
\(= 7z^3 + z2 (12) + z (-20)\)
\(= 7z^3 + 12z^2 – 20z\)
(iii) \(p – (p – q) – q – (q – p)\)
Solution:-
When term have the same algebraic factors, they are like terms.
When term have the same algebraic factors, they are like terms.
\(= p – p + q – q – q + p\)
\(= p – q\)
(iv) \(3a – 2b – ab – (a – b + ab) + 3ab + b – a\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 3a – 2b – ab – a + b – ab + 3ab + b – a\)
\(= 3a – a – a – 2b + b + b – ab – ab + 3ab\)
\(= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)\)
\(= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)\)
\(= a (1) + b (0) + ab (1)\)
\(= a + ab\)
(v) \(5x^2y – 5x^2 + 3yx^2 – 3y^2 + x^2 – y^2 + 8xy^2 – 3y^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 5x^2y + 3yx^2 – 5x^2 + x^2 – 3y^2 – y^2 – 3y^2\)
\(= x^2y (5 + 3) + x^2 (- 5 + 1) + y^2 (-3 – 1 -3) + 8xy^2\)
\(= x^2y (8) + x^2 (-4) + y^2 (-7) + 8xy^2\)
\(= 8x^2y – 4x^2 – 7y2 + 8xy^2\)
(vi) \((3y^2 + 5y – 4) – (8y – y^2 – 4)\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 3y^2 + 5y – 4 – 8y + y^2 + 4\)
\(= 3y^2 + y^2 + 5y – 8y – 4 + 4\)
\(= y^2 (3 + 1) + y (5 – 8) + (-4 + 4)\)
\(= y^2 (4) + y (-3) + (0)\)
\(= 4y^2 – 3y\)
2. Add:
(i) \(3mn, – 5mn, 8mn, – 4mn\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 3mn + (-5mn) + 8mn + (- 4mn)\)
\(= 3mn – 5mn + 8mn – 4mn\)
\(= mn (3 – 5 + 8 – 4)\)
\(= mn (11 – 9)\)
\(= mn (2)\)
\(= 2mn\)
(ii) \(t – 8tz, 3tz – z, z – t\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we hav e to add the like terms
\(= t – 8tz + (3tz – z) + (z – t)\)
\(= t – 8tz + 3tz – z + z – t\)
\(= t – t – 8tz + 3tz – z + z\)
\(= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)\)
\(= t (0) + tz (- 5) + z (0)\)
\(= – 5tz\)
\(= p – q\)
(iv) \(3a – 2b – ab – (a – b + ab) + 3ab + b – a\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 3a – 2b – ab – a + b – ab + 3ab + b – a\)
\(= 3a – a – a – 2b + b + b – ab – ab + 3ab\)
\(= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)\)
\(= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)\)
\(= a (1) + b (0) + ab (1)\)
\(= a + ab\)
(v) \(5x^2y – 5x^2 + 3yx^2 – 3y^2 + x^2 – y^2 + 8xy^2 – 3y^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 5x^2y + 3yx^2 – 5x^2 + x^2 – 3y^2 – y^2 – 3y^2\)
\(= x^2y (5 + 3) + x^2 (- 5 + 1) + y^2 (-3 – 1 -3) + 8xy^2\)
\(= x^2y (8) + x^2 (-4) + y^2 (-7) + 8xy^2\)
\(= 8x^2y – 4x^2 – 7y2 + 8xy^2\)
(vi) \((3y^2 + 5y – 4) – (8y – y^2 – 4)\)
Solution:-
When term have the same algebraic factors, they are like terms.
\(= 3y^2 + 5y – 4 – 8y + y^2 + 4\)
\(= 3y^2 + y^2 + 5y – 8y – 4 + 4\)
\(= y^2 (3 + 1) + y (5 – 8) + (-4 + 4)\)
\(= y^2 (4) + y (-3) + (0)\)
\(= 4y^2 – 3y\)
2. Add:
(i) \(3mn, – 5mn, 8mn, – 4mn\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 3mn + (-5mn) + 8mn + (- 4mn)\)
\(= 3mn – 5mn + 8mn – 4mn\)
\(= mn (3 – 5 + 8 – 4)\)
\(= mn (11 – 9)\)
\(= mn (2)\)
\(= 2mn\)
(ii) \(t – 8tz, 3tz – z, z – t\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we hav e to add the like terms
\(= t – 8tz + (3tz – z) + (z – t)\)
\(= t – 8tz + 3tz – z + z – t\)
\(= t – t – 8tz + 3tz – z + z\)
\(= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)\)
\(= t (0) + tz (- 5) + z (0)\)
\(= – 5tz\)
(iii) \(– 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)\)
\(= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3\)
\(= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3\)
\(= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)\)
\(= mn (- 9 + 21) + (7 – 11)\)
\(= mn (12) – 4\)
\(= 12mn – 4\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)\)
\(= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3\)
\(= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3\)
\(= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)\)
\(= mn (- 9 + 21) + (7 – 11)\)
\(= mn (12) – 4\)
\(= 12mn – 4\)
(iv) \(a + b – 3, b – a + 3, a – b + 3\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= a + b – 3 + (b – a + 3) + (a – b + 3)\)
\(= a + b – 3 + b – a + 3 + a – b + 3\)
\(= a – a + a + b + b – b – 3 + 3 + 3\)
\(= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)\)
\(= a (2 -1) + b (2 -1) + (-3 + 6)\)
\(= a (1) + b (1) + (3)\)
\(= a + b + 3\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= a + b – 3 + (b – a + 3) + (a – b + 3)\)
\(= a + b – 3 + b – a + 3 + a – b + 3\)
\(= a – a + a + b + b – b – 3 + 3 + 3\)
\(= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)\)
\(= a (2 -1) + b (2 -1) + (-3 + 6)\)
\(= a (1) + b (1) + (3)\)
\(= a + b + 3\)
(v) \(14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy\)
\(= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy\)
\(= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18\)
\(= x (14 – 7) + y (10 – 10) + xy(-12 + 8 + 4) + (-13 + 18)\)
\(= x (7) + y (0) + xy(0) + (5)\)
\(= 7x + 5\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy\)
\(= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy\)
\(= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18\)
\(= x (14 – 7) + y (10 – 10) + xy(-12 + 8 + 4) + (-13 + 18)\)
\(= x (7) + y (0) + xy(0) + (5)\)
\(= 7x + 5\)
(vi) \(5m – 7n, 3n – 4m + 2, 2m – 3mn – 5\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 5m – 7n + (3n – 4m + 2) + (2m – 3mn – 5)\)
\(= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5\)
\(= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5\)
\(= m (5 – 4 + 2) + n (-7 + 3) – 3mn + (2 – 5)\)
\(= m (3) + n (-4) – 3mn + (-3)\)
\(= 3m – 4n – 3mn – 3\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 5m – 7n + (3n – 4m + 2) + (2m – 3mn – 5)\)
\(= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5\)
\(= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5\)
\(= m (5 – 4 + 2) + n (-7 + 3) – 3mn + (2 – 5)\)
\(= m (3) + n (-4) – 3mn + (-3)\)
\(= 3m – 4n – 3mn – 3\)
(vii) \(4x^2y, – 3xy^2, –5xy^2, 5x^2y\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 4x^2y + (-3xy^2) + (-5xy^2) + 5x^2y\)
\(= 4x^2y + 5x^2y – 3xy^2 – 5xy^2\)
\(= x^2y (4 + 5) + xy^2 (-3 – 5)\)
\(= x^2y (9) + xy^2 (- 8)\)
\(= 9x^2y – 8xy^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 4x^2y + (-3xy^2) + (-5xy^2) + 5x^2y\)
\(= 4x^2y + 5x^2y – 3xy^2 – 5xy^2\)
\(= x^2y (4 + 5) + xy^2 (-3 – 5)\)
\(= x^2y (9) + xy^2 (- 8)\)
\(= 9x^2y – 8xy^2\)
(viii) \(3p^2q^2 – 4pq + 5, – 10 p^2q^2, 15 + 9pq + 7p^2q^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 3p^2q^2 – 4pq + 5 + (- 10p^2q^2) + 15 + 9pq + 7p^2q^2\)
\(= 3p^2q^2 – 10p^2q^2 + 7p^2q^2 – 4pq + 9pq + 5 + 15\)
\(= p^2q^2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)\)
\(= p^2q^2 (0) + pq (5) + 20\)
\(= 5pq + 20\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= 3p^2q^2 – 4pq + 5 + (- 10p^2q^2) + 15 + 9pq + 7p^2q^2\)
\(= 3p^2q^2 – 10p^2q^2 + 7p^2q^2 – 4pq + 9pq + 5 + 15\)
\(= p^2q^2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)\)
\(= p^2q^2 (0) + pq (5) + 20\)
\(= 5pq + 20\)
(ix) \(ab – 4a, 4b – ab, 4a – 4b\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= ab – 4a + (4b – ab) + (4a – 4b)\)
\(= ab – 4a + 4b – ab + 4a – 4b\)
\(= ab – ab – 4a + 4a + 4b – 4b\)
\(= ab (1 -1) + a (4 – 4) + b (4 – 4)\)
\(= ab (0) + a (0) + b (0)\)
\(= 0\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= ab – 4a + (4b – ab) + (4a – 4b)\)
\(= ab – 4a + 4b – ab + 4a – 4b\)
\(= ab – ab – 4a + 4a + 4b – 4b\)
\(= ab (1 -1) + a (4 – 4) + b (4 – 4)\)
\(= ab (0) + a (0) + b (0)\)
\(= 0\)
(x) \(x^2 – y^2 – 1, y^2 – 1 – x^2, 1 – x^2 – y^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= x^2 – y^2 – 1 + (y^2 – 1 – x^2) + (1 – x^2 – y^2)\)
\(= x^2 – y^2 – 1 + y^2 – 1 – x^2 + 1 – x^2 – y^2\)
\(= x^2 – x^2 – x^2 – y^2 + y^2 – y^2 – 1 – 1 + 1\)
\(= x^2 (1 – 1- 1) + y^2 (-1 + 1 – 1) + (-1 -1 + 1)\)
\(= x^2 (1 – 2) + y^2 (-2 +1) + (-2 + 1)\)
\(= x^2 (-1) + y^2 (-1) + (-1)\)
\(= -x^2 – y^2 -1\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
\(= x^2 – y^2 – 1 + (y^2 – 1 – x^2) + (1 – x^2 – y^2)\)
\(= x^2 – y^2 – 1 + y^2 – 1 – x^2 + 1 – x^2 – y^2\)
\(= x^2 – x^2 – x^2 – y^2 + y^2 – y^2 – 1 – 1 + 1\)
\(= x^2 (1 – 1- 1) + y^2 (-1 + 1 – 1) + (-1 -1 + 1)\)
\(= x^2 (1 – 2) + y^2 (-2 +1) + (-2 + 1)\)
\(= x^2 (-1) + y^2 (-1) + (-1)\)
\(= -x^2 – y^2 -1\)
3. Subtract:
(i) \(–5y^2\) from y^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= y^2 – (-5y^2)\)
\(= y^2 + 5y^2\)
\(= 6y^2\)
(i) \(–5y^2\) from y^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= y^2 – (-5y^2)\)
\(= y^2 + 5y^2\)
\(= 6y^2\)
(ii) \(6xy\) from \(–12xy\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= -12xy – 6xy
\(= – 18xy
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= -12xy – 6xy
\(= – 18xy
(iii) \((a – b)\) from \((a + b)\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= (a + b) – (a – b)\)
\(= a + b – a + b\)
\(= a – a + b + b\)
\(= a (1 – 1) + b (1 + 1)\)
\(= a (0) + b (2)\)
\(= 2b\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= (a + b) – (a – b)\)
\(= a + b – a + b\)
\(= a – a + b + b\)
\(= a (1 – 1) + b (1 + 1)\)
\(= a (0) + b (2)\)
\(= 2b\)
(iv) \(a (b – 5)\) from \(b (5 – a)\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= b (5 -a) – a (b – 5)\)
\(= 5b – ab – ab + 5a\)
\(= 5b + ab (-1 -1) + 5a\)
\(= 5a + 5b – 2ab\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= b (5 -a) – a (b – 5)\)
\(= 5b – ab – ab + 5a\)
\(= 5b + ab (-1 -1) + 5a\)
\(= 5a + 5b – 2ab\)
(v) \(–m^2 + 5mn\) from \(4m^2 – 3mn + 8\)
Solution:-
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 4m^2 – 3mn + 8 – (- m^2 + 5mn)\)
\(= 4m^2 – 3mn + 8 + m^2 – 5mn\)
\(= 4m^2 + m^2 – 3mn – 5mn + 8\)
\(= 5m^2 – 8mn + 8\)
\(= 4m^2 – 3mn + 8 – (- m^2 + 5mn)\)
\(= 4m^2 – 3mn + 8 + m^2 – 5mn\)
\(= 4m^2 + m^2 – 3mn – 5mn + 8\)
\(= 5m^2 – 8mn + 8\)
(vi) \(– x^2 + 10x – 5\) from \(5x – 10\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 5x – 10 – (-x^2 + 10x – 5)\)
\(= 5x – 10 + x^2 – 10x + 5\)
\(= x^2 + 5x – 10x – 10 + 5\)
\(= x^2 – 5x – 5\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 5x – 10 – (-x^2 + 10x – 5)\)
\(= 5x – 10 + x^2 – 10x + 5\)
\(= x^2 + 5x – 10x – 10 + 5\)
\(= x^2 – 5x – 5\)
(vii) \(5a^2 – 7ab + 5b^2\) from \(3ab – 2a^2 – 2b^2\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 3ab – 2a^2 – 2b^2 – (5a^2 – 7ab + 5b^2)\)
\(= 3ab – 2a^2 – 2b^2 – 5a^2 + 7ab – 5b^2\)
\(= 3ab + 7ab – 2a^2 – 5a^2 – 2b^2 – 5b^2\)
\(= 10ab – 7a^2 – 7b^2\)
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 3ab – 2a^2 – 2b^2 – (5a^2 – 7ab + 5b^2)\)
\(= 3ab – 2a^2 – 2b^2 – 5a^2 + 7ab – 5b^2\)
\(= 3ab + 7ab – 2a^2 – 5a^2 – 2b^2 – 5b^2\)
\(= 10ab – 7a^2 – 7b^2\)
(viii) \(4pq – 5q^2 – 3p^2\) from \(5p^2 + 3q^2 – pq\)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
\(= 5p^2 + 3q^2 – pq – (4pq – 5q^2 – 3p^2)\)
\(= 5p^2 + 3q^2 – pq – 4pq + 5q^2 + 3p^2\)
\(= 5p^2 + 3p^2 + 3q^2 + 5q^2 – pq – 4pq\)
\(= 8p^2 + 8q^2 – 5pq\)
4. (a) What should be added to \(x^2 + xy + y^2\) to obtain \(2x^2 + 3xy\)?
Solution:-
Let us assume \(p\) be the required term
\(p + (x^2 + xy + y^2) = 2x^2 + 3xy\)
\(p = (2x^2 + 3xy) – (x^2 + xy + y^2)\)
\(p = 2x^2 + 3xy – x^2 – xy – y^2\)
\(p = 2x^2 – x^2 + 3xy – xy – y^2\)
\(p = x^2 + 2xy – y^2\)
(b) What should be subtracted from \(2a + 8b + 10\) to get \(– 3a + 7b + 16\)?
Solution:-
Let us assume \(x\) be the required term
\(2a + 8b + 10 – x = -3a + 7b + 16\)
\(x = (2a + 8b + 10) – (-3a + 7b + 16)\)
\(x = 2a + 8b + 10 + 3a – 7b – 16\)
\(x = 2a + 3a + 8b – 7b + 10 – 16\)
\(x = 5a + b – 6 \)
5. What should be taken away from \(3x^2 – 4y^2 + 5xy + 20\) to obtain \(-x^2 – y^2 + 6xy + 20\)?
Solution:-
Let us assume a be the required term \(3x^2 – 4y^2 + 5xy + 20 – a = -x^2 – y^2 + 6xy + 20\)
\(a = 3x^2 – 4y^2 + 5xy + 20 – (-x^2 – y^2 + 6xy + 20)\)
\(a = 3x^2 – 4y^2 + 5xy + 20 + x^2 + y^2 – 6xy – 20\)
\(a = 3x^2 + x^2 – 4y^2 + y^2 + 5xy – 6xy + 20 – 20\)
\(a = 4x^2 – 3y^2 – xy\)
Solution:-
Let us assume \(p\) be the required term
\(p + (x^2 + xy + y^2) = 2x^2 + 3xy\)
\(p = (2x^2 + 3xy) – (x^2 + xy + y^2)\)
\(p = 2x^2 + 3xy – x^2 – xy – y^2\)
\(p = 2x^2 – x^2 + 3xy – xy – y^2\)
\(p = x^2 + 2xy – y^2\)
(b) What should be subtracted from \(2a + 8b + 10\) to get \(– 3a + 7b + 16\)?
Solution:-
Let us assume \(x\) be the required term
\(2a + 8b + 10 – x = -3a + 7b + 16\)
\(x = (2a + 8b + 10) – (-3a + 7b + 16)\)
\(x = 2a + 8b + 10 + 3a – 7b – 16\)
\(x = 2a + 3a + 8b – 7b + 10 – 16\)
\(x = 5a + b – 6 \)
5. What should be taken away from \(3x^2 – 4y^2 + 5xy + 20\) to obtain \(-x^2 – y^2 + 6xy + 20\)?
Solution:-
Let us assume a be the required term \(3x^2 – 4y^2 + 5xy + 20 – a = -x^2 – y^2 + 6xy + 20\)
\(a = 3x^2 – 4y^2 + 5xy + 20 – (-x^2 – y^2 + 6xy + 20)\)
\(a = 3x^2 – 4y^2 + 5xy + 20 + x^2 + y^2 – 6xy – 20\)
\(a = 3x^2 + x^2 – 4y^2 + y^2 + 5xy – 6xy + 20 – 20\)
\(a = 4x^2 – 3y^2 – xy\)
6. (a) From the sum of \(3x – y + 11\) and \(– y – 11\), subtract \(3x – y – 11\).
Solution:-
First we have to find out the sum of \(3x – y + 11\) and \(– y – 11\)
First we have to find out the sum of \(3x – y + 11\) and \(– y – 11\)
\(= 3x – y + 11 + (-y – 11)\)
\(= 3x – y + 11 – y – 11\)
\(= 3x – y – y + 11 – 11\)
\(= 3x – 2y\)
Now, subtract \(3x – y – 11\) from \(3x – 2y\)
\(= 3x – y + 11 – y – 11\)
\(= 3x – y – y + 11 – 11\)
\(= 3x – 2y\)
Now, subtract \(3x – y – 11\) from \(3x – 2y\)
\(= 3x – 2y – (3x – y – 11)\)
\(= 3x – 2y – 3x + y + 11\)
\(= 3x – 3x – 2y + y + 11\)
\(= -y + 11\)
\(= 3x – 2y – 3x + y + 11\)
\(= 3x – 3x – 2y + y + 11\)
\(= -y + 11\)
(b) From the sum of \(4 + 3x\) and \(5 – 4x + 2x^2\), subtract the sum of \(3x^2 – 5x\) and \(–x^2 + 2x + 5\).
Solution:-
First we have to find out the sum of \(4 + 3x\) and \(5 – 4x + 2x^2\)
\(= 4 + 3x + (5 – 4x + 2x^2)\)
\(= 4 + 3x + 5 – 4x + 2x^2\)
\(= 4 + 5 + 3x – 4x + 2x^2\)
\(= 9 – x + 2x^2\)
\(= 2x^2 – x + 9\) … [equation 1]
Then, we have to find out the sum of \(3x^2 – 5x \) and \(– x^2 + 2x + 5\)
\(= 3x^2 – 5x + (-x^2 + 2x + 5)\)
\(= 3x^2 – 5x – x^2 + 2x + 5\)
\(= 3x^2 – x^2 – 5x + 2x + 5\)
\(= 2x^2 – 3x + 5\) … [equation 2]
Now, we have to subtract equation (2) from equation (1)
\(= 2x^2 – x + 9 – (2x^2 – 3x + 5)\)
\(= 2x^2 – x + 9 – 2x^2 + 3x – 5\)
\(= 2x^2 – 2x^2 – x + 3x + 9 – 5\)
\(= 2x + 4\)
Solution:-
First we have to find out the sum of \(4 + 3x\) and \(5 – 4x + 2x^2\)
\(= 4 + 3x + (5 – 4x + 2x^2)\)
\(= 4 + 3x + 5 – 4x + 2x^2\)
\(= 4 + 5 + 3x – 4x + 2x^2\)
\(= 9 – x + 2x^2\)
\(= 2x^2 – x + 9\) … [equation 1]
Then, we have to find out the sum of \(3x^2 – 5x \) and \(– x^2 + 2x + 5\)
\(= 3x^2 – 5x + (-x^2 + 2x + 5)\)
\(= 3x^2 – 5x – x^2 + 2x + 5\)
\(= 3x^2 – x^2 – 5x + 2x + 5\)
\(= 2x^2 – 3x + 5\) … [equation 2]
Now, we have to subtract equation (2) from equation (1)
\(= 2x^2 – x + 9 – (2x^2 – 3x + 5)\)
\(= 2x^2 – x + 9 – 2x^2 + 3x – 5\)
\(= 2x^2 – 2x^2 – x + 3x + 9 – 5\)
\(= 2x + 4\)
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