Question

by cramer's rule 6x-4y=-12;8x-3y=-2​

Answer 1

Question :

1. 6x - 4y = -12 ; 8x - 3y = -2

Lesson :- Linear equations in two variable

Solution :

• Refer the attachment

   

Linear equation in two variable :-

2x + 5y = 12

x - y = 4

2x + 5y - 12 = 0

ax + by = c

  OR

ax + by + c = 0

a, b and c → Constants

x and y → Variables

(i) a, b and c → Real numbers

(ii) a, b  → At least one non-zero $\sf{(a^{2}+b^{2} \neq 0)}$

2x + 5y = 12

(a) When x = 1 and y = 2  (✔)

LHS = 2(1) + 5(2)

        = 2 + 10 = 12 = RHS

(b) When x = 3 and y = 1  (X)

     LHS = 2(3) + 5(1)

             = 6 + 5 = 11 ≠ RHS

Answer 2

Solve the equations :-

• 6x - 4y = 12

• 8x - 3y = 2

Solution :-

The given equations in matrix form can be represented as

$\rm :\longmapsto\:\begin{bmatrix} 6 & - 4\\ 8 & - 3\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 12 \\ 2\end{bmatrix}$

where,

$\rm :\longmapsto\:D \: = \: \begin{array}{|cc|}\sf 6 &\sf - 4 \\ \sf 8 &\sf - 3 \\\end{array}$

$\rm :\longmapsto\:D = - 18 + 32$

$\rm :\longmapsto\:D = 14$

$\rm :\implies\:system \: of \: equations \: have \: unique \: solution$

Consider,

$\rm :\longmapsto\:D_1 \: = \: \begin{array}{|cc|}\sf 12 &\sf - 4 \\ \sf 2 &\sf - 3 \\\end{array}$

$\rm :\longmapsto\:D_1 = - 36 + 8$

$\rm :\longmapsto\:D_1 = - 28$

Consider,

$\rm :\longmapsto\:D_2 \: = \: \begin{array}{|cc|}\sf 6 &\sf 12 \\ \sf 8 &\sf 2 \\\end{array}$

$\rm :\longmapsto\:D_2 = 12 - 96$

$\rm :\longmapsto\:D_2 = - 84$

So,

$\rm :\longmapsto\:x = \dfrac{D_1}{D} = - \dfrac{28}{14} = - 2$

$\rm :\longmapsto\:y = \dfrac{D_2}{D} = - \dfrac{84}{14} = - 6$

$\begin{gathered}\begin{gathered}\bf \: Hence - \begin{cases} &\sf{x \: = \: - \: 2} \\ &\sf{y \: = \: - \: 6} \end{cases}\end{gathered}\end{gathered}$