NCERT Solutions for Class 7 Maths Chapter 12: Algebraic Expressions - Exercise 12.3

Exercise 12.3

1. If $\mathbf{m = 2}$, find the value of:

(i)$\mathbf{m - 2}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow m - 2 = 2 - 2 \\ $

$ \Rightarrow m - 2 = 0  \\ $

The value is $0$.

(ii) $\mathbf{3m - 5}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3m - 5 = 3\left( 2 \right) - 5 \\$

$ \Rightarrow 3m - 5 = 6 - 5 \\ $

$ \Rightarrow 3m - 5 = 1 \\ $

The value is $1$.

(iii) $\mathbf{9 - 5m}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 9 - 5m = 9 - 5\left( 2 \right) \\ $

$ \Rightarrow 9 - 5m = 9 - 10  \\ $

$ \Rightarrow 9 - 5m =  - 1 \\ $

The value is $ - 1$.

(iv)$\mathbf{3{m^2} - 2m - 7}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3{m^2} - 2m - 7 = 3{\left( 2 \right)^2} - 2\left( 2 \right) - 7 \\ $

$ \Rightarrow 3{m^2} - 2m - 7 = 3\left( 4 \right) - 4 - 7  \\ $

 $ \Rightarrow 3{m^2} - 2m - 7 = 12 - 11  \\ $

  $ \Rightarrow 3{m^2} - 2m - 7 = 1 \\  $

The value is $1$.

(v) $\mathbf{\dfrac{{5m}}{2} - 4}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow \dfrac{{5m}}{2} - 4 = \dfrac{{5 \times 2}}{2} - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 5 - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 1 \\ $

The value is $1$.

2. If $\mathbf{p =  - 2}$, find the value of:

(i) $\mathbf{4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow 4p + 7 = 4\left( { - 2} \right) + 7 \\ $

$ \Rightarrow 4p + 7 =  - 8 + 7 \\ $

 $ \Rightarrow 4p + 7 =  - 1 \\ $

 The value is $ - 1$.

(ii) $\mathbf{ - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$\Rightarrow  - 3{p^2} + 4p + 7 =  - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 3\left( 4 \right) - 8 + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 12 - 1 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 13 \\ $

The value is $ - 13$.

(iii) $\mathbf{ - 2{p^3} - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2{\left( { - 2} \right)^3} - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

 $ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2 \times \left( { - 8} \right) - 3\left( 4 \right) - 8 + 7 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 16 - 12 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 4 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 3  \\ $

The value is $3$.

3. Find the value of the following expressions, when $\mathbf{x =  - 1}$:

(i) $\mathbf{2x - 7}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2x - 7 = 2\left( { - 1} \right) - 7 \\ $

$ \Rightarrow 2x - 7 =  - 2 - 7 \\ $

$ \Rightarrow 2x - 7 =  - 9 \\ $

The value of expression $2x - 7$ is $ - 9$.

(ii) $ \mathbf{- x + 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow  - x + 2 =  - \left( { - 1} \right) + 2  \\ $

$ \Rightarrow  - x + 2 = 1 + 2 \\ $

 $ \Rightarrow  - x + 2 = 3  \\ $ 

The value of expression $ - x + 2$ is $3$.

(iii) $\mathbf{{x^2} + 2x + 1}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow {x^2} + 2x + 1 = {\left( { - 1} \right)^2} + 2\left( { - 1} \right) + 1 \\ $

$ \Rightarrow {x^2} + 2x + 1 = 1 - 2 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 =  - 1 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 = 0 \\ $

The value of expression ${x^2} + 2x + 1$ is $0$.

(iv)$\mathbf{2{x^2} - x - 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2{x^2} - x - 2 = 2{\left( { - 1} \right)^2} - \left( { - 1} \right) - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 \times 1 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 3 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 1 \\ $

The value of expression $2{x^2} - x - 2$ is $1$.

4. If $\mathbf{a = 2$, $b =  - 2}$, find the value of:

(i) $\mathbf{{a^2} + {b^2}}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} + {b^2} = {\left( 2 \right)^2} + {\left( { - 2} \right)^2} \\ $

$ \Rightarrow {a^2} + {b^2} = 4 + 4  \\ $

$ \Rightarrow {a^2} + {b^2} = 8 \\ $

The value is $8$.

(ii) ${a^2} + ab + {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} + ab + {b^2} = {\left( 2 \right)^2} + \left( 2 \right)\left( { - 2} \right) + {\left( { - 2} \right)^2} \\ $

$ \Rightarrow {a^2} + ab + {b^2} = 4 - 4 + 4 \\ $

$ \Rightarrow {a^2} + ab + {b^2} = 4 \\ $

The value is $4$.

(iii) ${a^2} - {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} - {b^2} = {\left( 2 \right)^2} - {\left( { - 2} \right)^2}  \\ $

$ \Rightarrow {a^2} - {b^2} = 4 - 4 \\ $

$ \Rightarrow {a^2} - {b^2} = 0  \\ $

The value is $0$.

5. When $\mathbf{a = 0$, $b =  - 1}$, find the value of the given expressions:

(i) $\mathbf{2a + 2b}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2a + 2b = 2\left( 0 \right) + 2\left( { - 1} \right)  \\ $

$ \Rightarrow 2a + 2b = 0 - 2  \\ $

$ \Rightarrow 2a + 2b =  - 2  \\ $

The value of expression $2a + 2b$ is $ - 2$.

(ii) $\mathbf{2{a^2} + {b^2} + 1}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2{\left( 0 \right)^2} + {\left( { - 1} \right)^2} + 1  \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \times 0 + 1 + 1 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 0 + 2 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \\ $

The value of expression $2{a^2} + {b^2} + 1$ is $2$.

(iii) $\mathbf{2{a^2}b + 2a{b^2} + ab}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 2{\left( 0 \right)^2}\left( { - 1} \right) + 2\left( 0 \right){\left( { - 1} \right)^2} + \left( 0 \right)\left( { - 1} \right) \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 + 0 + 0 \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 \\ $

The value of expression $2{a^2}b + 2a{b^2} + ab$ is $0$.

(iv)$\mathbf{{a^2} + ab + 2}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow {a^2} + ab + 2 = {\left( 0 \right)^2} + \left( 0 \right)\left( { - 1} \right) + 2 \\$

$ \Rightarrow {a^2} + ab + 2 = 0 + 0 + 2 \\ $

$ \Rightarrow {a^2} + ab + 2 = 2 \\ $

The value of expression ${a^2} + ab + 2$ is $2$.

6. Simplify the expressions and find the value if $\mathbf{x}$ is equal to $\mathbf{2}$:

(i) $\mathbf{x + 7 + 4\left( {x - 5} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 7 + 4x - 20  \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 4x + 7 - 20 \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5x - 13 \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5\left( 2 \right) - 13  \\ $

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 10 - 13 \\ $

$  \Rightarrow x + 7 + 4\left( {x - 5} \right) =  - 3 \\  $

 The determined value is $ - 3$.

(ii) $\mathbf{3\left( {x + 2} \right) + 5x - 7}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 6 + 5x - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 5x + 6 - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8x - 1 \\  $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8\left( 2 \right) - 1 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 16 - 1  \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 15 \\ $

The determined value is $15$.

(iii) $\mathbf{6x + 5\left( {x - 2} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 6x + 5x - 10 \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11x - 10  \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11 \times 2 - 10  \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 22 - 10   \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 12  \\  $

The determined value is $12$.

(iv)$\mathbf{4\left( {2x - 1} \right) + 3x + 11}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x - 4 + 3x + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x + 3x - 4 + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11x + 7   \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11 \times 2 + 7 \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 22 + 7  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 29  \\ $

The determined value is $29$.

7. Simplify these expressions and find their values if $\mathbf{x = 3$, $a =  - 1$, $b =  - 2}$:

(i) $\mathbf{3x - 5 - x + 9}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3x - 5 - x + 9 = 3x - x - 5 + 9 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 2x + 4 \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 3x - 5 - x + 9 = 2 \times 3 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 6 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 10 \\ $

The determined value is $10$.

(ii) $\mathbf{2 - 8x + 4x + 4}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 8x + 4x + 2 + 4 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4x + 6  \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4 \times 3 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 12 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 6 \\ $

The determined value is $ - 6$.

(iii) $\mathbf{3a + 5 - 8a + 1}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 \\ $

 $ \Rightarrow 3a + 5 - 8a + 1 =  - 5a + 6 \\ $

Substitute $ - 1$ for $a$ and then solve.

$ \Rightarrow 3a + 5 - 8a + 1 =  - 5\left( { - 1} \right) + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 5 + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 11 \\ $

The determined value is $11$.

(iv)$\mathbf{10 - 3b - 4 - 5b}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 3b - 5b + 10 - 4 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8b + 6 \\  $

Substitute $ - 2$ for $b$ and then solve.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8\left( { - 2} \right) + 6  \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 16 + 6 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 22 \\ $

The determined value is $22$.

(v) $\mathbf{2a - 2b - 4 - 5 + a}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5  \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3a - 2b - 9 \\  $

 Substitute $ - 2$ for $b$ and $ - 1$ for $a$ and then solve.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3\left( { - 1} \right) - 2\left( { - 2} \right) - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 3 + 4 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 1 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 8 \\  $

The determined value is $ - 8$.

8. (i) If $\mathbf{z = 10}$, find the value of $\mathbf{{z^3} - 3\left( {z - 10} \right)}$.

Ans: Substitute the value of $z$ as $10$ then find a cube of $10$ to solve.

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = {\left( {10} \right)^3} - 3\left( {10 - 10} \right) \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 - 3 \times 0 \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 \\ $

The value of ${z^3} - 3\left( {z - 10} \right)$ is $1000$.

8. (ii) If $\mathbf{p =  - 10}$, find the value of $\mathbf{{p^2} - 2p - 100}$

Ans: Substitute the value of $p$ as $ - 10$ then find a square of $10$ to solve.

$  \Rightarrow {p^2} - 2p - 100 = {\left( {10} \right)^2} - 2\left( { - 10} \right) - 100  \\ $

$ \Rightarrow {p^2} - 2p - 100 = 100 + 20 - 100 \\ $

$ \Rightarrow {p^2} - 2p - 100 = 20  \\  $

The value of ${p^2} - 2p - 100$ is $20$.

9. What should be the value of $a$ if the value of $\mathbf{2{x^2} + x - a}$ equals to $5$, when $\mathbf{x = 0}$?

Ans: Substitute $0$ for $x$ in the expression $2{x^2} + x - a$ and equate it to $5$ in order to solve for $a$.

$ \Rightarrow 2{x^2} + x - a = 5 \\ $

$ \Rightarrow 2{\left( 0 \right)^2} + 0 - a = 5 \\ $

$ \Rightarrow 0 + 0 - a = 5 \\ $

$ \Rightarrow  - a = 5 \\ $

$ \Rightarrow a =  - 5 \\ $

So, the value of $a$ is $ - 5$.

10. Simplify the expression and find its value when $\mathbf{a = 5}$ and $\mathbf{b =  - 3}$:

$\mathbf{2\left( {{a^2} + ab} \right) + 3 - ab}$

Ans: Solve the brackets and combine the like terms with the variable $ab$. 

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab + 3 - ab  \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab - ab + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + ab + 3 \\  $

Now substitute $5$ for $a$ and $ - 3$ for $b$ to solve.

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{\left( 5 \right)^2} + \left( 5 \right)\left( { - 3} \right) + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2 \times 25 - 15 + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 50 - 12 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 38  \\ $

The value of expression $2\left( {{a^2} + ab} \right) + 3 - ab$ is $38$.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.3

Opting for the NCERT solutions for Ex 12.3 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.3 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 12 Exercise 12.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 12 Exercise 12.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 12 Exercise 12.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.