Question

solve the following simultaneous equation using cramer's rule 3x- 4y=10 4x+3y=5​

Answer 1

Answer:

• Cramer's Rule :

⇒ a₁x + b₁y = c₁

⇒ a₂x + b₂y = c₂

value of $\bf x=\dfrac{Dx}{D}$ and $\bf y=\dfrac{Dy}{D}$

⠀

$\sf D=\left[\begin{array}{c c}\sf a_1&\sf b_1 \\\sf a_2&\sf b_2\end{array}\right]\quad D_x=\left[\begin{array}{c c}\sf c_1&\sf b_1 \\\sf c_2&\sf b_2\end{array}\right]\quad D_y=\left[\begin{array}{c c}\sf a_1&\sf c_1 \\\sf a_2&\sf c_2\end{array}\right]$

$\rule{200}{1}$

• 3x - 4y = 10
• 4x + 3y = 5

• a₁ = 3 , b₁ = - 4 , c₁ = 10
• a₂ = 4 , b₂ = 3 , c₂ = 5

$\underline{\bigstar\:\textsf{According to the given Question :}}$

⠀⠀⠀⌬ Value of x

$:\implies\sf x=\dfrac{Dx}{D}\\\\\\:\implies\sf x=\dfrac{\left[\begin{array}{c c}\sf c_1&\sf b_1 \\\sf c_2&\sf b_2\end{array}\right]}{\left[\begin{array}{c c}\sf a_1&\sf b_1 \\\sf a_2&\sf b_2\end{array}\right]}\\\\\\:\implies\sf x =\dfrac{\left[\begin{array}{c c}\sf10&\sf-4\\\sf5&\sf3\end{array}\right]}{\left[\begin{array}{c c}\sf3&\sf-4\\\sf4&\sf3\end{array}\right]}\\\\\\:\implies\sf x = \dfrac{30 - ( - 20)}{9 - ( - 16)}\\\\\\:\implies\sf x = \dfrac{50}{25}\\\\\\:\implies\sf x = 2$

⠀

⠀⠀⠀⌬ Value of y

$:\implies\sf y=\dfrac{Dy}{D}\\\\\\:\implies\sf y=\dfrac{\left[\begin{array}{c c}\sf a_1&\sf c_1 \\\sf a_2&\sf c_2\end{array}\right]}{\left[\begin{array}{c c}\sf a_1&\sf b_1 \\\sf a_2&\sf b_2\end{array}\right]}\\\\\\:\implies\sf y =\dfrac{\left[\begin{array}{c c}\sf3&\sf10\\\sf4&\sf5\end{array}\right]}{\left[\begin{array}{c c}\sf3 &\sf-4\\\sf4&\sf3\end{array}\right]}\\\\\\:\implies\sf y = \dfrac{15 - 40)}{9 - ( - 16)}\\\\\\:\implies\sf y = \dfrac{ - 25}{25}\\\\\\:\implies\sf y = - \:1$

⠀

$\therefore\:\underline{\textsf{Hence, (x , y) will be equal to \textbf{(2 , - 1)}}}.$