Ex.3.5 Q3 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

\[\begin{align}8x + 5y &= 9\\3x + 2y &= 4\end{align}\]

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.5 | Question 3

Text Solution

Steps:

\[\begin{align} 8x + 5y &= 9 \qquad \ldots (1)\\ 3x + 2y &= 4 \qquad \ldots (2)\end{align}\]

From equation \((2)\), we obtain

\[\begin{align}3x + 2y &= 4\\3x &= 4 - 2y\\x &= \frac{{4 - 2y}}{3}\, \qquad \ldots (3)\end{align}\]

Substituting \(\begin{align}x = \frac{{4 - 2y}}{3}\,\end{align}\) in equation \((1),\) we obtain

\[\begin{align}8\left( {\frac{{4 - 2y}}{3}} \right) + 5y &= 9\\\frac{{32 - 16y + 15y}}{3} &= 9\\32 - y &= 27\\
y &= 32 - 27\\y &= 5\end{align}\]

Substituting \(y=5\) in equation \((3),\) we obtain

\[\begin{align}x &= \frac{{4 - 2 \times 5}}{3}\\x &= \frac{{ - 6}}{3}\\x &= - 2\\\end{align}\]

Hence, \(x = - 2, \;y = 5\)

Again, by cross-multiplication method

\[\begin{align}8x + 5y &= 9\\3x + 2y &= 4\end{align}\]

\[\begin{align}8x + 5y-9 &= 0\\3x + 2y-4 &= 0\\{a_1} &= 8,\quad{b_1} = 5,\quad{c_1} = - 9\\{a_2} &= 3,\quad{b_2} = 2,\quad{c_2} = - 4\\\frac{{x}}{{{b_1}{c_2} - {b_2}{c_1}}} &= \frac{{y}}{{{c_1}{a_2} - {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\\\frac{x}{{ - 20 - ( - 18)}} &= \frac{y}{{ - 27 - ( - 32)}} = \frac{1}{{16 - 15}}\\\frac{x}{{ - 2}} &= \frac{y}{5} = 1\\\frac{x}{{ - 2}}& = 1\,\quad {\rm{and}}\quad \frac{y}{5} = 1\\x &= - 2\,\quad {\rm{and}}\quad y = 5\end{align}\]

  
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