Question

Solve the following:
$
  {\text{A}}{\text{. Add: p}}\left( {{\text{p - q}}} \right),{\text{ q}}\left( {{\text{q - r}}} \right){\text{ and r}}\left( {{\text{r - p}}} \right) \\
  {\text{B}}{\text{. Add: 2x}}\left( {{\text{z - x - y}}} \right){\text{ and 27}}\left( {{\text{z - y - x}}} \right) \\
  {\text{C}}{\text{. Subtract: 3l}}\left( {{\text{l - 4m + 5n}}} \right){\text{ from 4l}}\left( {10{\text{n - 3m + 2l}}} \right) \\
  {\text{D}}{\text{. Subtract: 3a}}\left( {{\text{a + b + c}}} \right) - 2{\text{b}}\left( {{\text{a - b + c}}} \right){\text{ from 4c}}\left( { - {\text{a + b + c}}} \right) \\
$

Answer 1

Hint: To solve, we use the relation mentioned before the terms to perform the operation. To add is to find the sum of all the terms, to subtract b from a means ‘a – b’.Simplify and multiply the term which is outside the bracket and add or subtract the coefficients having the same variables.

Complete step-by-step answer:
(A). Add p(p – q), q(q – r) and r(r – p)
⟹$p(p – q) + q(q – r) + r(r – p)$
⟹${{\text{p}}^2} - {\text{pq + }}{{\text{q}}^2} - {\text{qr + }}{{\text{r}}^2} - {\text{rp}}$
⟹${{\text{p}}^2} + {{\text{q}}^2} + {{\text{r}}^2} - {\text{pq - qr - rp}}$

(B). Add 2x(z – x – y) and 27(z – y –x)
⟹ $2x(z – x – y) + 27(z – y –x)$
⟹ ${\text{2xz - 2}}{{\text{x}}^2} - {\text{2xy + 27z - 27y - 27x}}$

(C). Subtract 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
⟹$4l(10n – 3m + 2l) - 3l(l – 4m + 5n)$
⟹${\text{40ln - 12lm + 8}}{{\text{l}}^2} - 3{{\text{l}}^2} + 12{\text{lm - 15ln}}$
⟹$25{\text{ln + 5}}{{\text{l}}^2}$

(D). Subtract 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
⟹$4c(-a + b + c) – [3a(a + b + c) – 2b(a – b + c)]$
⟹$ - {\text{4ac + 4bc + 4}}{{\text{c}}^2} - \left[ {3{{\text{a}}^2} + 3{\text{ab + 3ac - 2ab + 2}}{{\text{b}}^2} - 2{\text{bc}}} \right]$
⟹$ - {\text{4ac + 4bc + 4}}{{\text{c}}^2} - 3{{\text{a}}^2} - 3{\text{ab - 3ac + 2ab - 2}}{{\text{b}}^2} + 2{\text{bc}}$
⟹${\text{ - 3}}{{\text{a}}^2} - 2{{\text{b}}^2} + 4{{\text{c}}^2} - 7{\text{ac + 6bc - ab}}$

Note: In order to solve questions of this type the key is to watch out for + and – signs denoted for addition and subtraction respectively, as they play a key role in determining the answer respective answers. If we are not given brackets to separate the terms and if more than operation is given, we use the PEMDAS rule which says the computation should take place in order of parenthesis, exponents, multiplication, division, addition, subtraction.