Examples (Revised) - Chapter 9 - Algebraic Expressions & Identities - Ncert Solutions class 8 - Maths

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Chapter 8 - Algebraic Expressions & Identities - NCERT Solutions for Class 8 Maths

Example 1:

Add: $7 x y+5 y z-3 z x, 4 y z+9 z x-4 y,-3 x z+5 x-2 x y$.
Solution: Writing the three expressions in separate rows, with like terms one below the other, we have

Thus, the sum of the expressions is $5 x y+9 y z+3 z x+5 x-4 y$. Note how the terms, $-4 y$ in the second expression and $5 x$ in the third expression, are carried over as they are, since they have no like terms in the other expressions.

Example 2:

Subtract $5 x^2-4 y^2+6 y-3$ from $7 x^2-4 x y+8 y^2+5 x-3 y$.
Solution:

```
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting - -3 is the same as adding +3. Similarly, subtracting 6y is the same as
```
adding $-6 y$; subtracting $-4 y^2$ is the same as adding $4 y^2$ and so on. The signs in the third row written below each term in the second row help us in knowing which operation has to be performed.

Example 3:

Complete the table for area of a rectangle with given length and breadth.
Solution:

Example 4:

Find the volume of each rectangular box with given length, breadth and height.

Solution:
(i)
$
\begin{aligned}
& x(x-3)+2=x^2-3 x+2 \\
& \text { For } \quad x=1, x^2-3 x+2=(1)^2-3(1)+2 \\
&=1-3+2=3-3=0
\end{aligned}
$

For
(ii)
$
\begin{aligned}
3 y(2 y-7)-3(y-4)-63 & =6 y^2-21 y-3 y+12-63 \\
& =6 y^2-24 y-51 \\
\text { For } y=-2,6 y^2-24 y-51 & =6(-2)^2-24(-2)-51 \\
& =6 \times 4+24 \times 2-51 \\
& =24+48-51=72-51=21
\end{aligned}
$

$
=24+48-51=72-51=21
$

Example 6:

Add
(i) $5 m(3-m)$ and $6 m^2-13 m$
(ii) $4 y\left(3 y^2+5 y-7\right)$ and $2\left(y^3-4 y^2+5\right)$

Solution:
(i) First expression $=5 m(3-m)=(5 m \times 3)-(5 m \times m)=15 m-5 m^2$
Now adding the second expression to it, $15 m-5 m^2+6 m^2-13 m=m^2+2 m$
(ii) The first expression $=4 y\left(3 y^2+5 y-7\right)=\left(4 y \times 3 y^2\right)+(4 y \times 5 y)+(4 y \times(-7))$
$
=12 y^3+20 y^2-28 y
$

The second expression $=2\left(y^3-4 y^2+5\right)=2 y^3+2 \times\left(-4 y^2\right)+2 \times 5$
$
=2 y^3-8 y^2+10
$

Adding the two expressions,

Example 7:

Subtract $3 p q(p-q)$ from $2 p q(p+q)$.
Solution:

We have $3 p q(p-q)=3 p^2 q-3 p q^2$ and $2 p q(p+q)=2 p^2 q+2 p q^2$

Example 8:

Multiply
(i) $(x-4)$ and $(2 x+3)$
(ii) $(x-y)$ and $(3 x+5 y)$

Solution:
(i)
$
\begin{aligned}
(x-4) \times(2 x+3) & =x \times(2 x+3)-4 \times(2 x+3) \\
& =(x \times 2 x)+(x \times 3)-(4 \times 2 x)-(4 \times 3)=2 x^2+3 x-8 x-12 \\
& =2 x^2-5 x-12
\end{aligned}
$
(Adding like terms)
(ii)
$
\begin{aligned}
(x-y) \times(3 x+5 y) & =x \times(3 x+5 y)-y \times(3 x+5 y) \\
& =(x \times 3 x)+(x \times 5 y)-(y \times 3 x)-(y \times 5 y) \\
& =3 x^2+5 x y-3 y x-5 y^2=3 x^2+2 x y-5 y^2 \text { (Adding like terms) }
\end{aligned}
$

Example 9:

Multiply
(i) $(a+7)$ and $(b-5)$
(ii) $\left(a^2+2 b^2\right)$ and $(5 a-3 b)$

Solution:
(i)
$
\begin{aligned}
(a+7) \times(b-5) & =a \times(b-5)+7 \times(b-5) \\
& =a b-5 a+7 b-35
\end{aligned}
$

Note that there are no like terms involved in this multiplication.
(ii)
$
\begin{aligned}
\left(a^2+2 \mathrm{~b}^2\right) \times(5 a-3 b) & =a^2(5 a-3 b)+2 b^2 \times(5 a-3 b) \\
& =5 a^3-3 a^2 b+10 a b^2-6 b^3
\end{aligned}
$

Example 10:

Simplify $(a+b)(2 a-3 b+c)-(2 a-3 b) c$.
Solution:

We have
$
\begin{aligned}
(a+b)(2 a-3 b+c) & =a(2 a-3 b+c)+b(2 a-3 b+c) \\
& =2 a^2-3 a b+a c+2 a b-3 b^2+b c \\
& =2 a^2-a b-3 b^2+b c+a c
\end{aligned}
$
(Note, $-3 a b$ and $2 a b$ are like terms)
and $\quad(2 a-3 b) c=2 a c-3 b c$
Therefore,
$
\begin{aligned}
(a+b)(2 a-3 b+c)-(2 a-3 b) c & =2 a^2-a b-3 b^2+b c+a c-(2 a c-3 b c) \\
& =2 a^2-a b-3 b^2+b c+a c-2 a c+3 b c \\
& =2 a^2-a b-3 b^2+(b c+3 b c)+(a c-2 a c) \\
& =2 a^2-3 b^2-a b+4 b c-a c
\end{aligned}
$