Question

x+ y - 8/2= x + 2y - 14/3 =3x-y/4 solve it by cramer's rule

Answer 1

In the attachment I have answered this problem.

Two Equations 1 and 2 are formed from the given equation suitably.

Equations 1 and 2 are solved by
Cramers rule.
I hope this answer help you

Answer 2

Answer:

x = 2 , y = 6

Step-by-step explanation:

Given: equations,

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

To find: Value of x & y by cramer's rule

First we find 2 linear equation in two variable by equation given 2 equations at a time

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}\\\\3\times(x+y-8)=2\times(x+2y-14)\\\\3x+3y-24=2x+4y-28\\x-y=-4\:\:................(1)\\\\\\\frac{x+y-8}{2}=\frac{3x-y}{4}\\\\4\times(x+y-8)=2\times(3x-y)\\4x+4y-32=6x-2y\\-2x+6y=32\\-x+3y=16\:\:.................(2)$

Now we write the coefficient matrix of variable from equation, say A

$A\:\:=\:\:\begin{bmatrix}1&-1\\-1&3\end{bmatrix}$

Now we write the  constant matrix from equation, say B

$B\,=\,\begin{bmatrix}-4\\16 \end{bmatrix}$

Now using cramer's rule,

we find Determinant of A.i.e., $\Delta$ and $\Delta_x\:\:,\:\:\Delta_y$

$\Delta=\begin{vmatrix}1&-1\\-1&3\end{vmatrix}=3\times1-(-1)(-1)\\\Delta=3-1=2$

$\Delta_x=\begin{vmatrix}-4&-1\\16&3\end{vmatrix}=3\times(-4)-(-1)(16)\\\Delta=-12+16=4$

$\Delta_y=\begin{vmatrix}1&-4\\-1&16\end{vmatrix}=16\times1-(-1)(-4)\\\Delta=16-4=12$

Now,

$x=\frac{\Delta_x}{\Delta}=\frac{4}{2}=2$

$y=\frac{\Delta_y}{\Delta}=\frac{12}{2}=6$

Therefore, x = 2 and y = 6