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  • Explain solution RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Equation Exercise Very short answer Question 4 Maths Textbook Solution.

Explain solution RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Equation Exercise Very short answer Question 4 Maths Textbook Solution.

Answers (1)

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Answer

\rightarrow x=\frac{2}{3}, y=-2

Given
\rightarrow\left[\begin{array}{cc} 3 & -4 \\ 9 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right]

Explanation
        \begin{gathered} \rightarrow\left[\begin{array}{cc} 3 & -4 \\ 9 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right] \\\\ \quad\left[\begin{array}{c} 3 \times x-4 \times y \\ 9 \times x+2 \times y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right] \\\\ {\left[\begin{array}{c} 3 x-4 y \\ 9 x+2 y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right]} \end{gathered}

Comparing both sides we get

\begin{aligned} &\\ &3 x-4 y=10-----(1)\\ &9 x+2 y=2------(2)\\ &\text { } \end{aligned}

Multiplying eqn. 1 by (3) and then subtracting eqn. 2 from it

\begin{aligned} &9 x-12 y=30 \\ &9 x+2 y\: \: =2 \\ &-14 y\: \: \: \: =28 \\ &\: \: \: \: \; \; \; \; \; \; \; \; y=-2 \end{aligned}

Put this value in (1)

3 x-4(-2)=10 \quad \quad \quad[3 x-4 y=10]

\begin{aligned} &3 x=10-8 \\ &x=2 / 3 \end{aligned}

Hense

x=2 / 3, y=-2

 

 

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