# Ex.3.6 Q1 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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## Question

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) \begin{align} \quad \frac{1}{{2x}} + \frac{1}{{3y}} &= 2\\\frac{1}{{3x}} + \frac{1}{{2y}} &= \frac{{13}}{6}\end{align}

(ii) \begin{align} \quad \frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} &= 2\\\frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} &= - 1\end{align}

(iii) \begin{align}\quad \frac{4}{x} + 3y &= 14\\\frac{3}{x} - 4y &= 23\end{align}

(iv) \begin{align}\quad \frac{5}{{x - 1}} + \frac{1}{{y - 2}} &= 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 1}}& = 2\end{align}

(v) \begin{align}\quad \frac{{7x - 2y}}{{xy}} &= 5\\\frac{{8x + 7y}}{{xy}} &= 15\end{align}

(vi) \begin{align} \quad 6x + 3y &= 6xy\\2x + 4y &= 5xy\end{align}

(vii) \begin{align}\quad \frac{{10}}{{x + y}} + \frac{2}{{x - y}} &= 4\\\frac{{15}}{{x + y}} - \frac{5}{{x - y}}& = - 2\end{align}

(viii) \begin{align} \quad \frac{1}{{3x + y}} + \frac{1}{{3x - y}} &= \frac{3}{4}\\\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} &= \frac{{ - 1}}{8}\end{align}

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.6 | Question 1

## Text Solution

Reasoning:

When the variable is in denominator, consider the reciprocal of variable as new variable.

Steps:

(i)

\begin{align} \quad \frac{1}{{2x}} + \frac{1}{{3y}} &= 2\\\frac{1}{{3x}} + \frac{1}{{2y}} &= \frac{{13}}{6}\end{align}

Let $$\frac{1}{x} = p\; {\rm{ and }}\;\frac{1}{y} = q$$, then the equations change as follows:

\begin{align} & \frac{1}{{2x}} + \frac{1}{{3y}} = 2 \\ & \Rightarrow \frac{p}{2} + \frac{q}{3} = 2 \\ \\ & \Rightarrow \,\,3p\, + 2q - 12\, = 0 \qquad (1)\\ \\ & \frac{1}{{3x}} + \frac{1}{{2y}} = \frac{{13}}{6} \\ & \Rightarrow \frac{p}{3} + \frac{q}{2} = \frac{{13}}{6} \\ \\ & \Rightarrow \,\,2p + 3q - 13\, = 0 \quad (2) \end{align}

Using cross-multiplication method, we obtain

\begin{align}\frac{p}{{ - 26 - ( - 36)}}& = \frac{q}{{ - 24 - ( - 39)}} \\ & = \frac{1}{{9 - 4}}\\ \frac{p}{{10}} &= \frac{q}{{15}} = \frac{1}{5}\\\frac{p}{{10}} &= \frac{1}{5}\,\,{\rm{and}}\,\,\frac{q}{{15}} = \frac{1}{5}\\p &= 2\,\,{\rm{and}}\,\,q = 3\\\end{align}

Therefore, \begin{align}\frac{1}{x} = 2 \end{align} and \begin{align}\frac{1}{y} = 3 \end{align}
Hence, \begin{align}x = \frac{1}{2}\end{align} and  \begin{align}y = \frac{1}{3}\end{align}

(ii)

\begin{align} \quad \frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} &= 2\\\frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} &= - 1\end{align}

Substituting \begin{align}\frac{1}{{\sqrt x }} = p\end{align}  and \begin{align}\frac{1}{{\sqrt y }} = q \end{align} in the given equations, we obtain

\begin{align}&\frac{2}{{\sqrt x }} + \frac{3}{{\sqrt y }} = 2 \\ \Rightarrow \, & 2p + 3q = 2 \ldots \left( {\rm{1}} \right)\\ & \frac{4}{{\sqrt x }} - \frac{9}{{\sqrt y }} = - 14 \\ \Rightarrow \, & 4p - 9q = - 14 \ldots \left( {\rm{2}} \right)\end{align}

Multiplying equation (1) by 3, we obtain

$6p + 9q = 6 \ldots \left( {\rm{3}} \right)$

Adding equation $$(2)$$ and $$(3)$$, we obtain

\begin{align}10p &= 5\\p &= \,\frac{1}{2}\end{align}

Putting \begin{align}p = \,\frac{1}{2}\end{align} in equation $$(1)$$, we obtain

\begin{align}2 \times \frac{1}{2} + 3q\, &= \,2\\3q &= 2 - 1\\q &= \frac{1}{3}\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{{\sqrt x }} = \frac{1}{2} & \\ &\Rightarrow \sqrt x \,\, = \,\,2\\& \Rightarrow x\,\, = \,\,4\\{\text{And }}q = \frac{1}{{\sqrt y }} &= \frac{1}{3} & \\ &\Rightarrow \sqrt y = 3\\ &\Rightarrow y = 9\\{\text{Hence, }}\,\,x &= 4\,{\text{ and }}\,y = 9\end{align}

\begin{align}{\rm{(iii) }}\quad \frac{4}{x} + 3y &= 14\\\frac{3}{x} - 4y &= 23\end{align}

Substituting \begin{align}\frac{1}{x} = p \end{align} in the given equations, we obtain

\begin{align}&4p+3y=14 \\ \Rightarrow \, & 4p+3y-14=0 \ldots \left( 1 \right) \\ \\ &3p-4y=23 \\ \Rightarrow \, &3p-4y-23=0 \ldots \left( 2 \right)\end{align}

By cross-multiplication, we obtain

\begin{align}\frac{p}{\begin{Bmatrix} - 69 \\ - 56\end{Bmatrix}} &= \frac{y}{\begin{Bmatrix} - 42 \\- ( - 92) \end{Bmatrix}} \\ &= \frac{1}{\begin{Bmatrix} - 16 \\- 9\end{Bmatrix}}\\ \frac{p}{{ - 125}} &= \frac{y}{{50}} = \frac{1}{{ - 25}}\\ \frac{p}{{ - 125}} &= \frac{1}{{ - 25}}{\text{ and }}\frac{y}{{50}} = \frac{1}{{ - 25}}\\ p &= 5{\text{ and }}y = - 2\end{align}

\begin{align}{\text{Therefore, }}p &= \,\,\frac{1}{x}\,\,\, = \,\,5\\ \Rightarrow x &= \frac{1}{5}\\{\text{Hence, }}x &= \frac{1}{5}{\text{ and }}y = - 2 \end{align}

\begin{align}{\rm{ (iv)}}\quad \frac{5}{{x - 1}} + \frac{1}{{y - 2}} &= 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 1}}& = 2\end{align}

Putting \begin{align}\frac{1}{{x - 1}} = p\end{align} and \begin{align}\frac{1}{{y - 2}} = q\end{align} in the given equation, we obtain

\begin{align} & \frac{5}{{x - 1}} + \frac{1}{{y - 2}} = 2 \\ \Rightarrow \, & 5p + q = 2 \cdots \cdots \left( 1 \right)\\ \\ & \frac{6}{{x - 1}} - \frac{3}{{y - 1}} = 2 \\ \Rightarrow \, & 6p - 3q = 1 \cdots \cdots \left( 2 \right) \end{align}

Multiplying equation $$(1)$$ by $$3$$, we obtain

$15p + 3q = 6\qquad \left( 3 \right)$

Adding $$(2)$$ and $$(3)$$, we obtain

\begin{align} 21p&=7 \\ p &=\frac{1}{3} \\\end{align}

Putting \begin{align}p = \frac{1}{3} \end {align} in equation ($$1$$), we obtain

\begin{align}5 \times \frac{1}{3} + q &= 2\\\,q &= 2 - \frac{5}{3}\\q &= \frac{1}{3}\\\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{{x - 1}} = \frac{1}{3}\\ &\Rightarrow x - 1 = 3\\ &\Rightarrow \,\,x = 4\\{\text{and }}q &= \frac{1}{{y - 2}} = \frac{1}{3}\\ &\Rightarrow y - 2 = 3\\ &\Rightarrow y = 5\\{\text{Hence, }}x &= 4{\text{ and }}y = 5\end{align}

\begin{align}{\rm{(v)}} \quad \frac{{7x - 2y}}{{xy}} &= 5\\\frac{{8x + 7y}}{{xy}} &= 15\end{align}

\begin{align}&\frac{{7x - 2y}}{{xy}} = 5\\ \Rightarrow\, & \frac{{7x}}{{xy}} - \frac{{2y}}{{xy}} = 5 \\ \Rightarrow \, & \frac{7}{y} - \frac{2}{x} = 5 \cdots \left( 1 \right)\\ \\ &\frac{{8x + 7y}}{{xy}} = 15 \\ \Rightarrow \, & \frac{{8x}}{{xy}} + \frac{{7y}}{{xy}} = 15\\ \Rightarrow\, & \frac{8}{y} + \frac{7}{x} = 15 \cdots \left( 2 \right)\end{align}

Putting \begin{align} \frac{1}{x} = p \end{align} and  \begin{align}\frac{1}{y} = q \end{align}in the equations $$(1)$$ and $$(2)$$, we obtain

\begin{align} & \frac{7}{y} - \frac{2}{x} = 5 \\ \Rightarrow \, & - 2p + 7q - 5 = 0 \cdots \left( 3 \right)\\ \\ &\frac{8}{y} + \frac{7}{x} = 15 \\ \Rightarrow \, & 7p + 8q - 15 = 0 \cdots \left( 4 \right)\end{align}

By cross-multiplication method, we obtain

\begin{align}\frac{p}{\begin{Bmatrix} - 105 \\- ( - 40) \end{Bmatrix}} &= \frac{q}{\begin{Bmatrix} - 35\\ - 30\end{Bmatrix}} \\ &= \frac{1}{{\begin{Bmatrix} - 16 \\ - 49\end{Bmatrix}}}\\ \frac{p}{{ - 65}} &= \frac{q}{{ - 65}} = \frac{1}{{ - 65}}\\ \\ \frac{p}{{ - 65}} &= \frac{1}{{ - 65}} {\text{ and }} \\ \frac{q}{{ - 65}} &= \frac{1}{{ - 65}}\\ \\ p& = 1\quad {\text{ and }}\quad q = 1 \end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = 1\\& \Rightarrow x = 1\\{\text{ and, }}q &= \frac{1}{y} = 1\\& \Rightarrow y = 1\\\\{\text{Hence, }}x &= 1{\text{ and }}y = 1\end{align}

\begin{align}{\rm{(vi)}} \quad 6x + 3y &= 6xy\\2x + 4y &= 5xy\end{align}

By dividing both the given equations by $$(xy),$$ we obtain

\begin{align} & 6x + 3y = 6xy \\ \Rightarrow \, & \frac{6}{y} + \frac{3}{x} = 6 \cdots \left( 1 \right)\\ \\ & 2x + 4y = 5xy \\ \Rightarrow \, & \frac{2}{y} + \frac{4}{x} = 5 \cdots \left( 2 \right)\end{align}

Substituting \begin{align}\frac{1}{x} = p\end{align} and \begin{align}\frac{1}{y} = q\end{align} in the equations $$(1)$$ and $$(2)$$, we obtain

\begin{align}3p + 6q - 6 &= 0 \cdots \left( 3 \right)\\4p + 2q - 5 &= 0 \cdots \left( 4 \right) \end{align}

By cross-multiplication method, we obtain

\begin{align}\frac{p}{{ - 30 - ( - 12)}} &= \frac{q}{{ - 24 - ( - 15)}}\\& = \frac{1}{{6 - 24}}\\ \\ \frac{p}{{ - 18}} &= \frac{q}{{ - 9}} = \frac{1}{{ - 18}}\\ \\ \frac{p}{{ - 18}} &= \frac{1}{{ - 18}}{\text{ and }} \\ \frac{q}{{ - 9}} & = \frac{1}{{ - 18}}\\p &= 1 \quad {\text{ and }}\quad q = \frac{1}{2}\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = 1\\& \Rightarrow x = 1\\{\text{and, }}q &= \frac{1}{y} = \frac{1}{2}\\& \Rightarrow y = 2\\\\{\text{Hence, }}x &= 1{\text{ and }}y = 2\end{align}

\begin{align}{\rm{(vii)}}\quad \frac{{10}}{{x + y}} + \frac{2}{{x - y}} &= 4\\\frac{{15}}{{x + y}} - \frac{5}{{x - y}}& = - 2\end{align}

Substituting \begin{align}\frac{1}{{x + y}} = p \end{align} and  \begin{align}\frac{1}{{x - y}} = q \end{align} in the given equations, we obtain

\begin{align} & \frac{{10}}{{x + y}} + \frac{2}{{x - y}} = 4 \\ \Rightarrow \, &10p + 2q = 4 \\ \Rightarrow \, & 5p + q - 2 = 0 \cdots\left( 1 \right)\\ \\ &\frac{{15}}{{x + y}} - \frac{5}{{x - y}} = - 2 \\ \Rightarrow \, & 15p - 5q = - 2 \\ \Rightarrow \, &15p - 5q + 2 = 0 \cdots \left( 2 \right) \end{align}

Using cross-multiplication method, we obtain

\begin{align}\frac{p}{{2 - 10}} &= \frac{q}{{ - 30 - 10}} \\ &= \frac{1}{{ - 25 - 15}}\\\frac{p}{{ - 8}} &= \frac{q}{{ - 40}} = \frac{1}{{ - 40}}\\ \\ \frac{p}{{ - 8}} &= \frac{1}{{ - 40}}{\text{ and }} \\ \frac{q}{{ - 40}}& = \frac{1}{{ - 40}}\\ p& = \frac{1}{5} \quad {\text{ and }}\quad q = 1\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{{x + y}} = \frac{1}{5}\\ &\Rightarrow x + y = 5 \qquad \left( 3 \right)\\{\text{and, }}q &= \frac{1}{{x - y}} = 1\\ &\Rightarrow x - y = 1 \qquad \left( 4 \right)\end{align}

Adding equation $$(3)$$ and $$(4)$$, we obtain

\begin{align}2x &= 6\\x &= 3\end{align}

Substituting$$x = 3$$ in equation (3), we obtain

\begin{align}3 + y &= 5\\y &= 2\end{align}

Hence, $$x = 3$$ and $$y = 2$$

\begin{align}{\rm{(viii)}} \quad \frac{1}{{3x + y}} + \frac{1}{{3x - y}} &= \frac{3}{4}\\\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} &= \frac{{ - 1}}{8}\end{align}

Substituting \begin{align} \frac{1}{{3x + y}} = p \end{align} and  \begin{align}\frac{1}{{3x - y}} = q \end{align} in these equations, we obtain

\begin{align}& \frac{1}{{3x + y}} + \frac{1}{{3x - y}} = \frac{3}{4}\\ \Rightarrow \, & p + q = \frac{3}{4} \cdots \left( 1 \right)\\ \\ &\frac{1}{{2(3x + y)}} - \frac{1}{{2(3x - y)}} = \frac{{ - 1}}{8} \\ \Rightarrow \, &\frac{p}{2} - \frac{q}{2} = - \frac{1}{8} \\ \Rightarrow \, &p - q = - \frac{1}{4} \cdots \left( 2 \right) \end{align}

Adding $$(1)$$ and $$(2)$$, we obtain

\begin{align}2p &= \frac{3}{4} - \frac{1}{4}\\2p &= \frac{1}{2}\\p &= \frac{1}{4}\end{align}

Substituting \begin{align} p = \frac{1}{4} \end{align} in $$(2)$$, we obtain

\begin{align}\frac{1}{4} – q &= - \frac{1}{4}\\q &= \frac{1}{4} + \frac{1}{4}\\q& = \frac{1}{2} \end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{{3x + y}} = \frac{1}{4}\\ &\Rightarrow 3x + y = 4 \cdots \left( 3 \right)\\{\text{and, }}q &= \frac{1}{{3x - y}} = \frac{1}{2}\\ &\Rightarrow 3x - y = 2 \cdots \left( 4 \right)\end{align}

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

\begin{align}6x &= 6\\x &= 1\end{align}

Substituting$$x = 1$$ in $$(3)$$, we obtain

\begin{align}3 \times 1 + y &= 4\\y& = 1\end{align}

Hence, $$x = 1$$ and $$y = 1$$

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