Solve the following simultaneous equations using Cramer's rule.
$4 x + 3 y - 4 = 0; 6 x = 8 - 5 y$

x = - 3, y = 4

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

x = - 2, y = 4

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

x = - 2, y = 3

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

x = - 4, y = 4

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

Open in App

Solution

The correct option is A $x = - 2, y = 4$
Given, $4 x + 3 y = 4, 6 x + 5 y = 8$
$D = ∣ ∣ ∣ \begin{matrix} 4 & 3 6 & 5 \end{matrix} ∣ ∣ ∣ = 20 - 18 = 2 D_{x} = ∣ ∣ ∣ \begin{matrix} 4 & 3 8 & 5 \end{matrix} ∣ ∣ ∣ = 20 - 24 = - 4 D_{y} = ∣ ∣ ∣ \begin{matrix} 4 & 4 6 & 8 \end{matrix} ∣ ∣ ∣ = 32 - 24 = 8 x = \frac{D_{x}}{D} = \frac{- 4}{2} = - 2 y = \frac{D_{y}}{D} = \frac{8}{2} = 4$

Suggest Corrections

Similar questions

-

y = \frac{5 x - 10}{2}

-

2 x + 3 y = 2; x - \frac{y}{2} = \frac{1}{2}

Solve the following system using Cramer’s Rule.

$- 4 x + 3 y = - 4$

- 8 x + 5 y = 0

2 x + 3 y = 4

x + y = 1

Join BYJU'S Learning Program