Ex.9.3 Q5 Algebraic Expressions and Identities - NCERT Maths Class 8

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Question

(a) Add: $$p\left( {p - q} \right),q\left( {q - r} \right)$$ and $$r\left( {r - p} \right)$$

(b) Add: $$2x\left( {z - x - y} \right)$$ and $$2y\left( {z - y - x} \right)$$

(c) Subtract: $$3l\left( {l - 4m + 5n} \right)$$ from $$4l\left( {10n - 3m + 2l} \right)$$

(d) Subtract: $$3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)$$ from $$4c\left( { - a + b + c} \right)$$

Video Solution
Algebraic Expressions & Identities
Ex 9.3 | Question 5

Text Solution

What is known?

Expressions

What is unknown?

Simplification

Reasoning:

Addition and Subtraction takes place between like terms.

Steps:

(a)

First expression

$$=p\left( {p{\rm{ }} - {\rm{ }}q} \right) = {p^2} - pq$$

Second expression

$$=q\left( {q - r} \right) = {q^2} - qr$$

Third expression

$$=r\left( {r - p} \right) = {r^2} - pr$$

Adding the three expressions, we obtain

\frac{\begin{align}{p^2} - pq&\\+\;\; &{q^2} - qr\\+& \;\;{r^2} - pq\end{align}}{{{p^2} - pq + {q^2} - qr + {r^2} - pq}}

Therefore, the sum of the given expressions is $${p^2} + {q^2} + {r^2} - pq - qr - rp.$$

(b)

First expression

\begin{align}&=2x\left( {z - x - y} \right) \\ &= 2xz - 2{x^2} - 2xy\end{align}

Second expression

\begin{align}&=2y\left( {z - y - x} \right) \\ &= 2yz - 2{y^2} - 2yx\end{align}

Adding the two expressions, we obtain

\begin{align}\frac{\begin{align}& 2xz-2{{x}^{2}}-2xy \\& +\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2xy+2yz-2{{y}^{2}} \\\end{align}}{2xz-2{{x}^{2}}-4xy+2yz-2{{y}^{2}}}\end{align}

Therefore, the sum of the given expressions is $$- 2{x^2} - 2{y^2} - 4xy + 2yz + 2zx.$$

(c)

\begin{align}3l\left( {l \!- \!4m \!+ \!5n} \right) &= 3{l^2} \!-\! 12lm \!+ \!15ln \\4l\left( {10n\! -\! 3m \!+\! 2l} \right) &=40ln \!-\! 12lm \!+\! 8{l^2} \end{align}

Subtracting these expressions, we obtain

\frac{\begin{align}40ln - 12lm + 8{l^2}\\15l - 12lm + 3{l^2}\\\left(-\right)\,\,\,\,\,\,\,\left(+\right)\,\,\,\,\,\,\,\,\,\left(-\right)\end{align}}{{ + 25ln\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 5{l^2}}}

Therefore, the result is $$5{l^2} + 25ln.$$

(d)

\begin{align}&3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)\\&=\begin{bmatrix} 3{a^2} + 3ab + 3ac- \\ 2ba + 2{b^2} - 2bc \end{bmatrix}\\&= \begin{bmatrix} 3{a^{2}} + 2{b^2} +\\ ab + 3ac - 2bc\end{bmatrix} \\&4c\left( \! - a + b + c \right) \!= - 4ac + 4bc + 4{c^2} \end{align}

Subtracting these expressions, we obtain

\dfrac{\begin{align}& -4ac+4bc+4{{c}^{2}} \\& \,\,\,\,\,3ac-2bc\,\,\,\,\,\,\,\,\,\,\,+3{{a}^{2}}+2{{b}^{2}}+ab \\& \left( - \right)\,\,\,\,\,\,\,\,\,\,\,\left( + \right)\,\,\,\,\,\,\,\,\,\,\,\left( - \right)\,\,\,\,\,\,\,\,\left( -\right)\,\,\,\,\,\left( - \right) \\\end{align}}{-7ac+6bc+4{{c}^{2}}\!-\!3{{a}^{2}}\!-2{{b}^{2}}\!-ab}

Therefore, the result is $$- 3{a^2} - 2{b^2} +\! 4{c^2} - ab \!+ 6bc - 7ac$$

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