Ex.3.7 Q7 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Solve the following pair of linear equations.

(i) \(\begin{align} px + qy &= p - q\\qx - py &= p + q\end{align}\)

(ii) \(\begin{align} ax + by &= c\\bx + ay &= 1 + c\end{align}\)

(iii) \(\begin{align} \frac{x}{a} - \frac{y}{b} &= 0\\ax + by &= {a^2} + {b^2}\end{align}\)

(iv) \(\begin{align} \begin{bmatrix}(a - b)x + \\(a + b)y \end{bmatrix}\!&= \! {a^2} \! -\! 2ab \! -\! {b^2}  \\(a + b)(x + y) &= {a^2} + {b^2}\end{align}\)

(v) \(\begin{align} 152x - 378y &=  - 74\\- 378x + 152y &=  - 604\end{align}\)

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.7 | Question 7

Text Solution

Steps:

(i)

\(\begin{align} px + qy &= p - q \qquad  \ldots \left( 1 \right)\\qx - py &= p + q \qquad \ldots \left( 2 \right)\end{align}\)

Multiplying equation \((1)\) by \(p\) and equation \((2)\) by \(q,\) we obtain

\[\begin{align}{p^2}x + pqy &= {p^2} - pq \qquad  \ldots \left( 3 \right)\\{q^2}x - pqy &= pq + {q^2} \qquad \ldots \left( 4 \right)\end{align}\]

Adding equations \((3)\) and \((4),\) we obtain

\[\begin{align}{p^2}x + {q^2}x &= {p^2} + {q^2}\\\left( {{p^2} + {q^2}} \right)x &= {p^2} + {q^2}\\
x &= \frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}}\\x &= 1\end{align}\]

Substituting \(x = 1\) in equation \((1),\) we obtain

\[\begin{align}p \times 1 + qy &= p – q\\qy &= - q\\y &= - 1\end{align}\]

Therefore, \(x = 1\) and \(y = - 1\)

(ii)

\(\begin{align}ax + by &= c \qquad \quad\;\;  \ldots \left( 1 \right)\\
bx + ay &= 1 + c \qquad  \ldots \left( 2 \right)\end{align}\)

Multiplying equation \((1)\) by \(a\) and equation \((2)\) by \(b\), we obtain

\[\begin{align}{a^2}x + aby &= ac \qquad \quad\;\; \ldots \left( 3 \right)\\{b^2}x + aby &= b + bc \qquad   \ldots \left( 4 \right)\end{align}\]

Subtracting equation \((4)\) from equation \((3),\)

\[\begin{align}\left( {{a^2} – {b^2}} \right)x &= ac – bc – b\\x &= \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}{\rm{ }}\end{align}\]

Substituting \(\begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}} \end{align}\) in equation \((1),\) we obtain

\[\begin{align}ax + by &= c \\ a\left[ {\frac{{c(a – b) – b}}{{{a^2} – {b^2}}}} \right] + by &= c\\\frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}} + by &= c\\ \end{align} \]

\[\begin{align} by &= c - \frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{{a^2}c – {b^2}c – {a^2}c + abc + ab}}{{{a^2} – {b^2}}}\\by &= \frac{{abc – {b^2}c + ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{bc(a – b) + ab}}{{{a^2} – {b^2}}}\\by& = \frac{{b\left[ {c(a – b) + a} \right]}}{{{a^2} – {b^2}}}\\ y &= \frac{{c(a – b) + a}}{{{a^2} – {b^2}}}\end{align}\]

Therefore,

\(\begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}\end{align}\)

and 

\(\begin{align}y=\frac{c(a-b)+a}{{{a}^{2}}-{{b}^{2}}}\end{align}\)

(iii)

\(\begin{align} \frac{x}{a} - \frac{y}{b} &= 0 \qquad \qquad\;\; \ldots \left( 1 \right)\\ax + by &= {a^2} + {b^2} \qquad  \ldots \left( 2 \right)\end{align}\)

By solving equation \((1),\) we obtain

\[\begin{align}\frac{x}{a} - \frac{y}{b} &= 0\\x &= \frac{{ay}}{b} \qquad  \ldots \left( 3 \right)\end{align}\]

Substituting \(\begin{align}x = \frac{{ay}}{b} \end{align}\) in equation \((2),\) we obtain

\[\begin{align}a \times \left( {\frac{{ay}}{b}} \right) + by &= {a^2} + {b^2}\\
\frac{{{a^2}y + {b^2}y}}{b} &= {a^2} + {b^2}\\\left( {{a^2} + {b^2}} \right)y &= b\left( {{a^2} + {b^2}} \right)\\y &= b\end{align}\]

Substituting \(y = b\) in equation \((3),\) we obtain

\[\begin{align}x &= \frac{{a \times b}}{b}\\x &= a\end{align}\]

Therefore, \(x = a\) and \(y = b\)

(iv)

\(\begin{align} &\begin{bmatrix} (a - b)x +\\ (a + b)y \end{bmatrix} \! = \! {a^2} \!  - \!  2ab  \! - \!  {b^2}  \ldots \left( 1 \right) \\ \\ & (a + b)(x + y)  \! = \!  {a^2}  \! + \!  {b^2}\ldots \left( 2 \right)\end{align}\)

By solving equation \((2),\) we obtain

\[\begin{align}(a + b)(x + y) &= {a^2} + {b^2}\\ \begin{bmatrix}(a + b)x +\\  (a + b)y \end{bmatrix}&= {a^2} + {b^2} \ldots \left( 3 \right)\end{align}\]

Subtracting equation \((3)\) from \((1),\) we obtain

\[\begin{align}\!\begin{bmatrix}(a – b)x –\\ (a + b)x \end{bmatrix}&=\! \begin{bmatrix} \left( {{a^2} – 2ab – {b^2}} \right) \!-\! \\\left( {{a^2} + {b^2}} \right)\end{bmatrix} \\ \begin{bmatrix} {(a – b) – }\\{(a + b)} \end{bmatrix}x \!& =\! \begin{bmatrix}{a^2} – 2ab – {b^2} – \\{a^2} – {b^2}\end{bmatrix} \\\left[ {a – b – a – b} \right]x \!&=\! - 2ab – 2{b^2}\\ - 2bx \!&= \!- 2b\left( {a + b} \right)\\x\!& = \!\left( {a + b} \right)\end{align}\]

Substituting \(x = \left( {a + b} \right)\) in equation (1), we obtain

\[\begin{align} \begin{bmatrix}(a – b)(a + b) + \\ (a + b)y \end{bmatrix}&=  {a^2} \! –\! 2ab \! –\! {b^2}  \\ \begin{bmatrix}({a^2} – {b^2}) +\\ (a + b)y \end{bmatrix} &= {a^2} \! – \! 2ab \! – \! {b^2}  \end{align} \]

\[\begin{align} (a \! + \! b)y & \! = \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! ({a^2} \! – \! {b^2})  \\(a \! + \! b)y \! &= \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! {a^2} \! + \! {b^2}  \\y \! &= \! \frac{{ - 2ab}}{{(a \! + \! b)}}\end{align}\]

(v)

\(\begin{align}152x - 378y &=  - 74 \;\; \ldots \left( 1 \right)\\ - 378x + 152y &=  - 604 \;\;  \ldots \left( 2 \right)\end{align}\)

Adding equations \((1)\) and \((2),\) we obtain

\[\begin{align} - 226x - 226y &=  - 678\\ - 226\left( {x + y} \right) &=  - 678\\x + y &= 3 \qquad  \ldots \left( 3 \right)\end{align}\]

Subtracting equation \((2)\) from \((1),\) we obtain

\[\begin{align}530x - 530y &= 530\\530\left( {x - y} \right) &= 530\\x - y &= 1 \qquad  \ldots \left( 4 \right)\end{align}\]

Adding equations \((3)\) and \((4),\) we obtain

\[\begin{align}2x &= 4\\x& = 2\end{align}\]

Substituting \(x = 2\) in equation \((3),\) we obtain

\[\begin{align}2 + y &= 3\\y &= 1\end{align}\]

Therefore, \(x = 2\) and \(y = 1\)