Question

plz answer...................​

Answer 1

Question :

Solve the following simultaneous equations with cramer's rule

2x+3y= 2

$x-\frac{y}{2}=\frac{1}{2}$

Theory ;

Cramer's rule :

consider a system of simulateneous linear equations :

$a_{1}x+b_{1}y=d_{1}$

$a_{2}x+b_{2}y=d_{2}$

Δ=$\left|\begin{array}{cc}a_{1}&b_{1}\\a_{2}&b_{2}\end{array}\right|$

$\triangle_{1}$ =$\left|\begin{array}{cc}d_{1}&b_{1}\\d_{2}&b_{2}\end{array}\right|$

$\triangle_{2}$ =$\left|\begin{array}{cc}a_{1}&d_{1}\\a_{2}&d_{2}\end{array}\right|$

Case 1 :

If Δ≠0 ,then solution is unique

x=$\frac{\triangle_{1}}{\triangle}$

y=$\frac{\triangle_{2}}{\triangle}$

Case 2:

If Δ= 0 , and alteast of $\triangle_{1}$ ,$\triangle_{2}$ is not equal to 0.

then given system of equations is inconsistent. No solution .

Case 3:

If Δ=0 and $\triangle_{1}$=$\triangle_{1}$=0

then ,infinite solution or No solution .

Solution:

2x+3y= 2

$x-\frac{y}{2}=\frac{1}{2}$

Here Δ≠0, then solution is unique solution .

x=$\frac{\triangle_{1}}{\triangle}$=$\dfrac{5}{8}$

y=$\frac{\triangle_{2}}{\triangle}$=$\dfrac{1}{4}$

refer to the attachment.