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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