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3x + Y = 19 3x − Y = 23

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प्रश्न

3x + y = 19
3x − y = 23

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उत्तर

\[\text{ Given }: \hspace{0.167em} 3x + y = 19\] 
\[ 3x - y = 23\] 
Using Cramer's Rule, we get
\[D = \begin{vmatrix} 3 & 1 \\3 & - 1 \end{vmatrix} = - 3 - 3 = - 6\] 
\[ D_1 = \begin{vmatrix} 19 & 1\\23 & - 1 \end{vmatrix} = - 19 - 23 = - 42\]
\[ D_2 = \begin{vmatrix} 3 & 19 \\3 & 23 \end{vmatrix} = \left( 3 \times 23 \right) - \left( 3 \times 19 \right) = 3 \times 4 = 12\] 
Now,
\[x = \frac{D_1}{D} = \frac{- 42}{- 6} = 7\] 
\[y = \frac{D_2}{D} = \frac{12}{- 6} = - 2\] 
\[ \therefore x = 7 \text{ and }y = - 2\] 

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Determinants - Exercise 6.4 [पृष्ठ ८४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.4 | Q 4 | पृष्ठ ८४
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