Question

solve by crammer's rule 5x-3y=24,-7x+11y=14​

Answer 1

Given :

• $\sf 5x-3y=-24$
• $\sf -7x+11y=14$

To Find :

• Find The values of x and y using Cramer's Rule

Formulas Applied :

$\underline{\boxed{\sf x = \dfrac{\triangle _x }{\triangle}, y = \dfrac{\triangle_y}{\triangle}}}$

Solution :

Given Equations are,

$\sf 5x-3y=-24$

$\sf -7x+11y=14$

Then the equations in the matrix form,

$\sf \begin {bmatrix}5&-3\\-7&11\end{bmatrix} \begin {bmatrix}x\\y \end{bmatrix} = \begin {pmatrix} 24\\14 \end {pmatrix}$

Calculating ∆,

$\triangle = \sf \begin {vmatrix} 5&-3\\-7&11\end{vmatrix}\\\\\triangle = \sf \Big(55-(21)\Big)\\\\\triangle = \sf 55-21\\\\ \boxed{\triangle \bf= 34}$

$\triangle \neq 0$, Hence solution exists.

$\triangle_x = \sf \begin {vmatrix} 24&-3\\14&11\end{vmatrix}\\\\\triangle_x= \sf \Big(264-(42)\Big)\\\\\triangle_x= \sf 264+42\\\\\boxed{\bf \triangle_x = 306}$

$\triangle_y = \sf \begin {vmatrix}5&24\\-7&14\end{vmatrix}\\\\\triangle_y= \sf \Big(70-(-168)\Big)\\\\\triangle_y= \sf 70+168\\\\\boxed{\bf \triangle_y = 238}$

Substituting the values,

$\longrightarrow \sf x = \dfrac{\triangle _x }{\triangle} \Rightarrow \dfrac{306}{34}\Rightarrow 9\\\\ \longrightarrow \sf y = \dfrac{\triangle _y}{\triangle} \Rightarrow \dfrac{238}{34}\Rightarrow 7$

Required Answer :

$\boxed{\boxed{\[\sf (x,y) = (9,7)\]}}$

Answer 2

GiveN :-

• Two linear equations are given
• $\sf 5x -3y = 24$
• $\sf -7x + 11y = 14$

To FinD :-

• The value of x and y by Crammer's Rule .

SolutioN :-

Crammer's Rule is a method to solve a system of linear equations using the concept of determinants . We can write the system of linear equations as in AX = B form using matrices .

$\sf\dashrightarrow\underset{\blue{\sf AX = B \ form }}{\underbrace{ \left[\begin{array}{cc}\sf 5&\sf -3\\\sf -7&\sf 11 \end{array} \right]\left[\begin{array}{c} \sf x \\\sf y \end{array}\right] =\left[\begin{array}{c} \sf 24 \\\sf 14 \end{array}\right] }}$

• On comparing them wrt AX = B :-

$\sf\to \red{A = \left[\begin{array}{cc}\sf 5 &\sf -3\\\sf -7&\sf 11 \end{array}\right] }$

$\sf \to\red{ B = \left[\begin{array}{c} \sf 24 \\\sf 14 \end{array}\right] }$

Secondly define two matrices $\sf B_1\: \& \ B_2$ such that , The first matrix is obtained by replacing first column of matrix A with column B . In similar manner second matrix is obtained by replacing column two of matrix A by the column in B .

$\sf\to \pink{B_1 = \left[\begin{array}{cc}\sf 24&\sf -3\\\sf 14 &\sf 11 \end{array}\right] }$

$\sf\to \pink{B_2 = \left[\begin{array}{cc}\sf 5 &\sf 24\\\sf -7&\sf 14 \end{array}\right] }$

$\rule{200}2$

$\red{\bigstar}\underline{\textsf{ The value of x will be determined by :- }}$

$\sf \to x =\dfrac{ det.(B_1)}{det .(A)}=\dfrac{|B_1|}{|A|}$

$\sf\to x =\dfrac{\begin{array}{|cc|} \sf 24 &\sf -3\\\sf 14 &\sf 11 \end{array}}{\begin{array}{|cc|} \sf 5 &\sf -3\\\sf -7 &\sf 11 \end{array}} \\\\\sf\to x = \dfrac{(11)(24)-[(-3)(14)]}{(14)(5)-[(-7)(-3)]}\\\\\sf\to x = \dfrac{264+42}{55-21}\\\\\sf\to x =\dfrac{306}{34}\\\\\sf \to \pink{x = 9 }\qquad\qquad\bigg\lgroup \red{\tt Required\ value \ of \ x .}\bigg\rgroup$

$\rule{200}2$

$\red{\bigstar}\underline{\textsf{ The value of y will be determined by :- }}$

$\sf \to y =\dfrac{ det.(B_2)}{det .(A)}=\dfrac{|B_2|}{|A|}$

$\sf\to y =\dfrac{\begin{array}{|cc|} \sf 5 &\sf 24\\\sf -7 &\sf 14 \end{array}}{\begin{array}{|cc|} \sf 5 &\sf -3\\\sf -7 &\sf 11 \end{array}} \\\\\sf\to y = \dfrac{(14)(5)-[(-7)(24)]}{(14)(5)-[(-7)(-3)]}\\\\\sf\to y= \dfrac{70+168}{55-21}\\\\\sf\to y=\dfrac{238}{34}\\\\\sf \to \pink{y = 7}\qquad\qquad\bigg\lgroup \red{\tt Required\ value \ of \ y .}\bigg\rgroup$

$\rule{200}2$