DAV Class 7 Maths Chapter 6 Worksheet 3

DAV Class 7 Maths Chapter 6 Worksheet 3

Algebraic Expression Worksheet 3


1. Find the product of the following binomials.

(i) \((5x + 3) (2x + 4) \)

Answer

\[ \begin{align*} &=(5x + 3)(2x + 4) \\ &= 5x(2x + 4) + 3(2x + 4) \\ &= 5x \times 2x + 5x \times 4 + 3 \times 2x + 3 \times 4 \\ &= 10x^2 + 20x + 6x + 12 \\ &= 10x^2 + 26x + 12 \end{align*} \]

(ii) \( (7p - 3q) (2p + 5q) \)

Answer

\[ \begin{align*} &= (7p - 3q)(2p + 5q) \\ &= 7p(2p + 5p) - 3q(2p + 5q) \\ &= 7p \times 2p + 7p \times 5q - 3q \times 2p - 3q \times 5q \\ &= 14p^2 + 35pq - 6pq - 15q^2 \\ &= 14p^2 + 29pq - 15q^2 \end{align*} \]

(iii) \( \left(\frac{2}{5}a + b\right) \left(a^2 - \frac{1}{5}b^2\right) \)

Answer

\[ \begin{align*} &= \left(\frac{2}{5}a + b\right) \left(a^2 - \frac{1}{5}b^2\right) \\ \\ &= \frac{2}{5}a \left(a^2 - \frac{1}{5}b^2\right) + b \left(a^2 - \frac{1}{5}b^2\right) \\ \\ &= \frac{2}{5}a^3 - \frac{2}{25}ab^2 + a^2b - \frac{1}{5}b^3 \\ \\ &= \frac{2}{5}a^3 + a^2b - \frac{2}{25}ab^2 - \frac{1}{5}b^3 \end{align*} \]

(iv) \( (m^2n - 5) (6 - mn^2) \)

Answer

\begin{align*} &=(m^2n - 5) \times (6 - mn^2) \\ & = m^2n (6 - mn^2) - 5(6 - mn^2) \\ &= 6m^2n - m^3n^3 - 30 + 5mn^2 \\ &= 6m^2n + 5mn^2 - m^3n^3 - 30 \\ \end{align*}

(v) \((4x^2 + 7y^3) (3xy - 2y^2) \)

Answer

\begin{align*} &=(4x^2 + 7y^3) \times (3xy - 2y^2) \\ &= 4x^2 (3xy -2y^2) + 7y^3 (3xy - 2y^2) \\ &= 12x^3y - 8x^2y^2 + 21xy^4 - 14y^5 \\ \end{align*}

(vi) \( (1.1x + 2.7y) (1.1x - 2.7y) \)

Answer

\begin{align*} &=(1.1x + 2.7y) \times (1.1x - 2.7y) \\ &= 1.1x (1.1x - 2.7y) + 2.7y (1.1x - 2.7y) \\ &= 1.1x \times 1.1x - 1.1x \times 2.7y + 2.7y \times 1.1x - 2.7y \times 2.7y \\ &= 1.21x^2 - 2.97xy + 2.97xy - 7.29y^2 \\ &= 1.21x^2 - 7.29y^2 \\ \end{align*}

2. Multiply the following binomials and verify the results for the given values.

(i) \( (2x^2 - 5y) (5x + 2y^2) \, ; \, x = 2, y = -1 \)

Answer

\begin{align*} &=(2x^2 - 5y) \times (5x + 2y^2) \\ &= 2x^2(5x + 2y^2) - 5y(5x + 2y^2) \\ &= 2x^2 \times 5x + 2x^2 \times 2y^2 - 5y \times 5x - 5y \times 2y^2 \\ &= 10x^3 + 4x^2y^2 - 25xy - 10y^3 \\ \\ & { \color{magenta} \text{Verification: }} x = 2, y = -1 \\ \\ & \color{magenta} \text{L.H.S } \\ &=(2x^2 - 5y) \times (5x + 2y^2) \\ &= [2(2)^2 - 5(-1)] \times (5(2) + 2(-1)^2) \\ &=(2 \times 4 + 5) \times (10 + 2 \times 1) \\ &=(8 + 5) \times (10 + 2) \\ &= 13 \times 12 \\ &= 156 \\ \\ & \color{magenta} \text{R.H.S } \\ &= 10x^3 + 4x^2y^2 - 25xy - 10y^3 \\ &= 10(2)^3 + 4(2)^2(-1)^2 - 25(2)(-1) - 10(-1)^3 \\ &= 10 \times 8 + 4 \times 4 \times 1 - 25 \times 2 \times (-1) - 10 \times (-1) \\ &= 80 + 16 + 50 + 10 \\ &= 156 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}

(ii) \( \left(-\frac{1}{4}a + \frac{1}{5}b\right) \left(\frac{1}{4}a + \frac{1}{5}b\right) \, ; \, a = 8, b = 5 \)

Answer

\begin{align*} &=\left(-\frac{1}{4}a + \frac{1}{5}b\right) \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ \\ &= -\frac{1}{4}a \times \left(\frac{1}{4}a + \frac{1}{5}b\right) + \frac{1}{5}b \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ \\ &= \left(-\frac{1}{4}a \times \frac{1}{4}a \right) - \left(\frac{1}{4}a \times \frac{1}{5}b \right) + \left(\frac{1}{5}b \times \frac{1}{4}a\right) + \left(\frac{1}{5}b \times \frac{1}{5}b\right) \\ \\ &= -\frac{1}{16}a^2 -\cancel{ \frac{1}{20}ab} + \cancel{\frac{1}{20}ab} + \frac{1}{25}b^2 \\ \\ &= -\frac{1}{16}a^2 + \frac{1}{25}b^2 \\ \\ & { \color{magenta} \text{Verification: }} a = 8, b = 5 \\ & \color{magenta} \text{L.H.S } \\ &=\left(-\frac{1}{4}a + \frac{1}{5}b\right) \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ \\ &= \left[\left( -\frac{1}{\cancel4_{\color{green}1}} \times {\cancel8^{\color{green}2}} \right) + \left( \frac{1}{\cancel5_{\color{green}1}} \times {\cancel5^{\color{green}1}} \right) \right] \times \left[\left( \frac{1}{\cancel4_{\color{green}1}} \times {\cancel8^{\color{green}2}} \right) + \left( \frac{1}{\cancel5_{\color{green}1}} \times {\cancel5^{\color{green}1}} \right) \right] \\ \\ &= \left[(-1 \times 2) + (1 \times 1 )\right] \times [(1 \times 2) + (1 \times 1 )] \\ &=\left(-2 + 1\right) \times \left(2 + 1\right) \\ &= (-1) \times 3 \\ &= -3 \\ \\ & \color{magenta} \text{R.H.S } \\ &= -\frac{1}{16}a^2 + \frac{1}{25}b^2 \\ \\ &= -\frac{1}{16}(8)^2 + \frac{1}{25}(5)^2 \\ \\ &= -\frac{1}{\cancel{16}_1} \times \cancel{64}^4 + \frac{1}{\cancel{25}_1} \times \cancel{25}^1 \\ \\ &= -1 \times 4 + 1 \times 1 \\ &= -4 + 1 \\ &= -3 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}

(iii) \( (m + n) (2m - 3n) \, ; \, m = -2, n = 0 \)

Answer

\begin{align*} &= (m + n) \times (2m - 3n) \\ &= m(2m - 3n) + n(2m - 3n) \\ &= 2m^2 - 3mn + 2mn - 3n^2 \\ &= 2m^2 - mn - 3n^2 \\ \\ & {\color{magenta} \text{Verification: }} m = -2, n = 0 \\ & \color{magenta} \text{L.H.S} \\ &= (m + n) \times (2m - 3n) \\ &= (-2 + 0) \times [2 \times (-2) - 3(0)] \\ &= (-2) \times (-4 - 0) \\ &= (-2) \times (-4) \\ &= 8 \\ \\ & \color{magenta} \text{R.H.S} \\ &= 2m^2 - mn - 3n^2 \\ &= 2(-2)^2 - (-2)(0) - 3(0)^2 \\ &= 2 \times 4 - 0 - 0 \\ &= 8 - 0 - 0 \\ &= 8 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}

(iv) \( (0.1p + 0.2q)(0.2q - 0.1p) \, ; \, p = 10, q = 5 \)

Answer

\begin{align*} &= (0.1p + 0.2q) \times (0.2q - 0.1p) \\ &= 0.1p(0.2q - 0.1p) + 0.2q(0.2q - 0.1p) \\ &= (0.1p \times 0.2q) - ( 0.1p \times 0.1p) + (0.2q \times 0.2q) - (0.2q \times 0.1p) \\ &= \cancel{0.02pq} - 0.01p^2 + 0.04q^2 - \cancel{0.02pq} \\ &= -0.01p^2 + 0.04q^2 \\ \\ & { \color{magenta} \text{Verification: } } p = 10, q = 5\\ & \color{magenta} \text{L.H.S} \\ &= (0.1p + 0.2q) \times (0.2q - 0.1p) \\ &= [(0.1 \times 10) + (0.2 \times 5)] \times [(0.2 \times 5) - (0.1 \times 10)] \\ &= (1 + 1) \times (1 - 1) \\ &= 2 \times 0 \\ &= 0 \\ \\ & \color{magenta} \text{R.H.S} \\ &= -0.01p^2 + 0.04q^2 \\ &= -0.01(10)^2 + 0.04(5)^2 \\ &= -0.01 \times 100 + 0.04 \times 25 \\ &= -1 + 1 \\ &= 0 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}

3. Simply the following

(i) \( [(2x - 3y)(3x + y)]+ [(x + 2y) (x - y)] \)

Solution

\begin{align*} &= [(2x - 3y) \times (3x + y)]+ [(x + 2y) \times (x - y)] \\ &= [2x(3x + y) - 3y(3x + y)] + [x(x - y) + 2y(x - y)] \\ &= [2x \times 3x + 2x \times y - 3y \times 3x - 3y \times y] + [x \times x - x \times y + 2y \times x - 2y \times y] \\ &= [6x^2 {\color{magenta} + 2xy - 9xy} - 3y^2] + [x^2 {\color{magenta} - xy + 2xy} - 2y^2] \\ &= 6x^2 {\color{magenta} - 7xy} - 3y^2 + x^2 {\color{magenta} + xy} - 2y^2 \\ &= { \color{green} 6x^2 + x^2 } { \color{magenta} - 7xy + xy} {\color{blue} - 3y^2 - 2y^2} \\ &= { \color{green} 7x^2} {\color{magenta} - 6xy} {\color{blue}- 5y^2} \end{align*}

Answer \( \color{red} 7x^2 - 6xy - 5y^2 \)

(ii) \( \left(\frac{2}{3} x + 4\right) \left(\frac{3}{2} x + 6\right) - \left(\frac{1}{7} x - 1\right) \left(\frac{1}{7} x + 1\right) \)

Solution

\begin{align*} &= \left[\frac{2}{3} x \left(\frac{3}{2} x + 6\right) + 4 \left(\frac{3}{2} x + 6\right)\right] - \left[\frac{1}{7} x \left(\frac{1}{7} x + 1\right) - 1 \left(\frac{1}{7} x + 1\right)\right] \\ \\&= \left[ \left( \frac{\cancel2}{\cancel3} x \times \frac{\cancel3}{\cancel2} x \right) + \left( \frac{2}{\cancel{3}_1} x \times \cancel{6}^2 \right) + \left( \cancel{4}^2 \times \frac{3}{\cancel{2}^1} x \right) + \left( 4 \times 6 \right) \right] - \left[ \left( \frac{1}{7} x \times \frac{1}{7} x \right) + \left( \frac{1}{7} x \times 1 \right) - \left ( 1 \times \frac{1}{7} x \right) - (1 \times 1) \right] \\ \\&= \left[(x \times x) + (2x \times 2) + (2 \times 3x) + 24\right] - \left[\frac{1}{49} x^2 + \cancel{\frac{1}{7} x} - \cancel{\frac{1}{7} x} - 1\right] \\ \\&= \left[ x^2 + 4x + 6x + 24\right] - \left[\frac{1}{49} x^2 - 1\right] \\ \\&= x^2 + 10x + 24 - \frac{1}{49} x^2 + 1 \\ \\&= x^2 - \frac{1}{49} x^2 + 10x + 24 + 1 \\ \\&= \frac{48}{49} x^2 + 10x + 25 \\ \\ \end{align*}

Answer \( \color{red} \frac{48}{49} x^2 + 10x + 25 \)

(iii) \((7 abc + b^2) (7 - bc) + c(2 b^3 - 9 ab) \)

Solution

\begin{align*} &= [(7 abc + b^2) (7 - bc)] + [c(2 b^3 - 9 ab)] \\ &= [7 abc(7 - bc) + b^2 (7 - bc)] + [c \times 2 b^3 - c \times 9 ab] \\ &= (49 abc - 7 ab^2c^2 + 7 b^2 - b^3c) + (2 b^3c - 9 abc) \\ &= 49 abc - 7 ab^2c^2 + 7 b^2 - b^3c + 2 b^3c - 9 abc \\ &={ \color{magenta} 49 abc - 9 abc} - 7 ab^2c^2 + 7 b^2 { \color{green} - b^3c + 2 b^3c} \\ &= {\color{magenta} 40 abc} - 7 ab^2c^2 + 7 b^2 {\color{green} + b^3c} \end{align*}

Answer \( \color{red} 40 abc - 7 ab^2c^2 + 7 b^2 + b^3c \)

(iv) \( (p + q)(p^2 - q^2) + (p - q)(p^2 + q^2) \)

Solution

\begin{align*} &= [(p + q)(p^2 - q^2)] + [(p - q)(p^2 + q^2)] \\ &= [p(p^2 - q^2) + q(p^2 - q^2)] + [p(p^2 + q^2) - q(p^2 + q^2)] \\ &= [p^3 - pq^2 + p^2q - q^3] + [p^3 + pq^2 - p^2q - q^3] \\ &= p^3 - pq^2 + p^2q - q^3 + p^3 + pq^2 - p^2q - q^3 \\ &= p^3 - \cancel{pq^2} + \cancel{p^2q} - q^3 + p^3 + \cancel{pq^2} - \cancel{p^2q} - q^3 \\ &= p^3 + p^3 - q^3 - q^3 \\ &= 2p^3 - 2q^3 \end{align*}

Answer \( \color{red} 2p^3 - 2q^3 \)

(v) \( (x + 2y)(-3x - y) - (x + y)(x - y) + (x - 2y)(-2x + y) \)

Solution

\begin{align*} &= [(x + 2y)(-3x - y)] - [(x + y)(x - y)] + [(x - 2y)(-2x + y)] \\ &= [x(-3x - y) + 2y(-3x - y)] - [x(x - y) + y(x - y)] + [x(-2x + y) - 2y(2x - y)] \\ &= [-3x^2 - xy - 6xy - 2y^2] - [x^2 -\cancel{xy} + \cancel{xy} - y^2] + [-2x^2 + xy + 4xy - 2y^2] \\ &= [-3x^2 - 7xy - 2y^2] - [x^2 - y^2] + [-2x^2 + 5xy - 2y^2] \\ &= -3x^2 - 7xy - 2y^2 - x^2 + y^2 - 2x^2 + 5xy - 2y^2 \\ &= {\color{green} -3x^2 - x^2 - 2x^2} { \color{magenta} - 7xy + 5xy } {\color{blue} - 2y^2 + y^2 - 2y^2} \\ &= {\color{green} -6x^2} { \color{magenta} -2xy } {\color{blue} -3y^2} \\ \end{align*}

Answer \( \color{red} -6x^2 - 2xy - 3y^2 \)

(vi) \( (a^2 + b^2)(a^2 + b^2) - (a^2 - b^2)(a^2 - b^2) \)

Solution

\begin{align*} &= [(a^2 + b^2)(a^2 + b^2)] - [(a^2 - b^2)(a^2 - b^2)] \\ &= [a^2(a^2 + b^2) + b^2(a^2 + b^2)] - [a^2(a^2 - b^2) - b^2(a^2 - b^2)] \\ &= [a^4 + a^2b^2 + a^2b^2 + b^4] - [a^4 - a^2b^2 - a^2b^2 + b^4] \\ &= [a^4 + 2a^2b^2 + b^4] - [a^4 - 2a^2b^2 + b^4] \\ &= a^4 + 2a^2b^2 + b^4 - a^4 + 2a^2b^2 - b^4 \\ &= \cancel{a^4} - \cancel{a^4} + 2a^2b^2 + 2a^2b^2 + \cancel{b^4} - \cancel{b^4} \\ &= 4a^2b^2 \end{align*}

Answer \( \color{red} 4a^2b^2 \)