$5 x = 10 - 2 y; y = 3 x - 11$ . Then

D_{x} = ∣ ∣ ∣ \begin{matrix} 10 & 2 11 & - 1 \end{matrix} ∣ ∣ ∣

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D_{x} = ∣ ∣ ∣ \begin{matrix} 11 & 2 9 & - 1 \end{matrix} ∣ ∣ ∣

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D_{x} = ∣ ∣ ∣ \begin{matrix} 12 & 2 7 & - 1 \end{matrix} ∣ ∣ ∣

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D_{x} = ∣ ∣ ∣ \begin{matrix} 13 & 2 5 & - 1 \end{matrix} ∣ ∣ ∣

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $D_{x} = ∣ ∣ ∣ \begin{matrix} 10 & 2 11 & - 1 \end{matrix} ∣ ∣ ∣$
$D_{x}$ of two simulataenous linear equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$
is given by $∣ ∣ ∣ \begin{matrix} c_{1} & b_{1} c_{2} & b_{2} \end{matrix} ∣ ∣ ∣$

The given equations can be written as $5 x + 2 y = 10$ and $3 x - y = 11$

Hence, for the given equations, $D_{x} = ∣ ∣ ∣ \begin{matrix} 10 & 2 11 & - 1 \end{matrix} ∣ ∣ ∣$

Suggest Corrections

Similar questions

5 x = 10 - 2 y; y = 3 x - 11

D_{x} = ∣ ∣ ∣ \begin{matrix} 10 & 2 11 & - 1 \end{matrix} ∣ ∣ ∣

-

-

-

y = \sqrt{sin x - \sqrt{sin x - \sqrt{sin x - . . . . . \infty}}}

\frac{d y}{d x} =

\frac{d^{2} y}{d x^{2}} + \frac{6 d y}{d x} + 9 y = 6 e^{- 3 x}

y (0) = 0, y (+ 1) = 6 e^{- 3}

y (- 1)

If $2 x = y^{1 / 5} + y^{- 1 / 5}$ then $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} =$

Join BYJU'S Learning Program