Question

Solve the following simultaneous equations by Cramer's rule. 6x - 3y = -10 ; 3x + 5y - 8 = 0​

Answer 1

Solution:-

Given equation

6x - 3y = -10                       ...(i) eq

3x + 5y = 8                         ....(ii) eq

So

a₁ = 6 , b₁ = -3 and c₁ = -10

a₂ = 3 , b₂ = 5 and c₂ = 8

so

$\rm x=\dfrac{D_1}{D} \:and \: y=\dfrac{D_2}{D}$

Now find D

$\rm D= \begin{vmatrix} \sf a_1 & \sf b_1 \\ \sf a_2 &\sf b_2 \end{vmatrix}$

$\rm D= \begin{vmatrix} \sf 6 & \sf -3\\ \sf 3 &\sf 5 \end{vmatrix}$

$\rm D= 30-(-9)$

$\rm D=30+9$

$\rm D= 39$

Now find D₁

$\rm D_1= \begin{vmatrix} \sf c_1 & \sf b_1 \\ \sf c_2 &\sf b_2 \end{vmatrix}$

$\rm D_1= \begin{vmatrix} \sf -10 & \sf -3 \\ \sf 8 &\sf 5 \end{vmatrix}$

$\rm D_1= -50-(-24)$

$\rm D_1 = -26$

Now find D₂

$\rm D_2= \begin{vmatrix} \sf a_1 & \sf c_1 \\ \sf a_2 &\sf c_2 \end{vmatrix}$

$\rm D_2= \begin{vmatrix} \sf 6 & \sf -10 \\ \sf 3 &\sf 8 \end{vmatrix}$

$\rm D_2=48-(-30)$

$\rm D_2=78$

Now we find x and y

$\rm X=\dfrac{D_1}{D} =\dfrac{-26}{39} =\dfrac{-2}{3}$

$\rm Y=\dfrac{D_2}{D} =\dfrac{78}{39} =2$

Now we get value of x and y

$\boxed{\rm X=\dfrac{-2}{3} \: and \: Y= 2}$

Answer 2

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