Examples (Revised) - Chapter 12 - Algebraic Expressions - Ncert Solutions class 7 - Maths

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Chapter 10 - Algebraic Expressions | NCERT Solutions for Class 7 Maths

Example 1

Identify, in the following expressions, terms which are not constants. Give their numerical coefficients:
$
x y+4,13-y^2, 13-y+5 y^2, 4 p^2 q-3 p q^2+5
$

Solution

Example 2
(a) What are the coefficients of $x$ in the following expressions?
$
4 x-3 y, 8-x+y, y^2 x-y, 2 z-5 x z
$
(b) What are the coefficients of $y$ in the following expressions?
$
4 x-3 y, 8+y z, y z^2+5, m y+m
$

Solution
(a) In each expression we look for a term with $x$ as a factor. The remaining part of that term is the coefficient of $x$.

(b) The method is similar to that in (a) above. 

Example 3

State with reasons, which of the following pairs of terms are of like terms and which are of unlike terms:
(i) $7 x, 12 y$
(ii) $15 x,-21 x$
(iii) $-4 a b, 7 b a$
(iv) $3 x y, 3 x$
(v) $6 x y^2, 9 x^2 y$
(vi) $p q^2,-4 p q^2$
(vii) $m n^2, 10 m n$

Solution

Following simple steps will help you to decide whether the given terms are like or unlike terms:
(i) Ignore the numerical coefficients. Concentrate on the algebraic part of the terms.
(ii) Check the variables in the terms. They must be the same.
(iii) Next, check the powers of each variable in the terms. They must be the same.

Note that in deciding like terms, two things do not matter (1) the numerical coefficients of the terms and (2) the order in which the variables are multiplied in the terms.

Example 4

Find the values of the following expressions for $x=2$.
(i) $x+4$
(ii) $4 x-3$
(iii) $19-5 x^2$
(iv) $100-10 x^3$

Solution

Putting $x=2$
(i) In $x+4$, we get the value of $x+4$, i.e., $x+4=2+4=6$
(ii) In $4 x-3$, we get $4 x-3=(4 \times 2)-3=8-3=5$
(iii) In $19-5 x^2$, we get $19-5 x^2=19-\left(5 \times 2^2\right)=19-(5 \times 4)=19-20=-1$
(iv) In $100-10 x^3$, we get
$
\begin{aligned}
& 100-10 x^3=100-\left(10 \times 2^3\right)=100-(10 \times 8)\left(\text { Note } 2^3=8\right) \\
& =100-80=20
\end{aligned}
$

Example 5

Find the value of the following expressions when $n=-2$.
(i) $5 n-2$
(ii) $5 n^2+5 n-2$
(iii) $n^3+5 n^2+5 n-2$

Solution
(i) Putting the value of $n=-2$, in $5 n-2$, we get,
$
5(-2)-2=-10-2=-12
$
(ii) In $5 n^2+5 n-2$, we have, for $n=-2,5 n-2=-12$ and $5 n^2=5 \times(-2)^2=5 \times 4=20$ [as $\left.(-2)^2=4\right]$
Combining,
$
5 n^2+5 n-2=20-12=8
$
(iii) Now, for $n=-2$,

$5 n^2+5 n-2=8$ and
$
n^3=(-2)^3=(-2) \times(-2) \times(-2)=-8
$

Combining,
$
n^3+5 n^2+5 n-2=-8+8=0
$

We shall now consider expressions of two variables, for example, $x+y, x y$. To work out the numerical value of an expression of two variables, we need to give the values of both variables. For example, the value of $(x+y)$, for $x=3$ and $y=5$, is $3+5=8$.

Example 6

Find the value of the following expressions for $a=3, b=2$.
(i) $a+b$
(ii) $7 a-4 b$
(iii) $a^2+2 a b+b^2$
(iv) $a^3-b^3$

Solution

Substituting $a=3$ and $b=2$ in
(i) $a+b$, we get
$
a+b=3+2=5
$
(ii) $7 a-4 b$, we get
$
7 a-4 b=7 \times 3-4 \times 2=21-8=13 .
$
(iii) $a^2+2 a b+b^2$, we get
$
a^2+2 a b+b^2=3^2+2 \times 3 \times 2+2^2=9+2 \times 6+4=9+12+4=25
$
(iv) $a^3-b^3$, we get
$
a^3-b^3=3^3-2^3=3 \times 3 \times 3-2 \times 2 \times 2=9 \times 3-4 \times 2=27-8=19
$