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5.(a) \(Add: p(p – q), q(q – r) and r(r – p)\)
(b) \(Add: 2x(z – x – y) and 2y(z – y – x)\)

(c) \(Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)\)

(d) \(Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)\)


Answer :

(a) \(p(p – q) + q(q – r) + r(r – p)\)

=\( (p \times p) – (p \times q) + (q \times q) – (q \times r) + (r \times r) – (r \times p)\)[Opening the brackets]

= \(p^2 – pq + q^2 – qr + r^2 – rp\)

=\( p^2 + q^2 + r^2 – pq – qr – rp\)

(b) \(2x(z – x – y) + 2y(z – y – x)\)

= \((2x \times z) – (2x \times x) – (2x \times y) + (2y \times z) – (2y \times y) – (2y \times x)\)[Opening the brackets]

= \(2xz – 2x^2 – 2xy + 2yz – 2y^2 – 2xy\)

\(= -2x^2 – 2y^2 + 2xz + 2yz – 4xy\)

\(= -2x^2 – 2y^2 – 4xy + 2yz + 2xz\)

(c) \(4l(10n – 3m + 2l) – 3l(l – 4m + 5n)\)

\(= (4l \times 10n) – (4l \times 3m) + (4l \times 2l) – (3l \times l) – (3l \times -4m) – (3l ×\times 5n)\)[Opening the brackets]

\(= 40ln – 12lm + 8l^2 – 3l^2 + 12lm – 15ln\)

\(= (40ln – 15ln) + (-12lm + 12lm) + (8l^2 – 3l^2)\)

\(= 25ln + 0 + 5l^2\)

\(= 25ln + 5l^2\)

\(= 5l^2 + 25ln\)

(d)\( [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]\)

\(= (-4ac + 4bc + 4c^2) – (3a^2 + 3ab + 3ac – 2ab + 2b^2 – 2bc)\)[Opening the brackets]

\(= -4ac + 4bc + 4c^2 – 3a^2 – 3ab – 3ac + 2ab – 2b^2 + 2bc\)

\(= -3a^2 – 2b^2 + 4c^2 – ab + 6bc – 7ac\)

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