Question

Activity 1: To solve the simultaneous equations by determinant method, fill
in the blanks
y + 2x - 19 = 0 ; 2x - 3y + 3=0








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Answer 1

Determinant method (Cramer's Rule):

Let the two linear equations be

$\quad\quad a_{1}x+b_{1}y=c_{1}$

$\quad\quad a_{2}x+b_{2}y=c_{2}$

By Cramer's rule, the required solution be

$\quad x=\frac{D_{1}}{D},\:y=\frac{D_{2}}{D}$

where $D=\left|\begin{array}{cc}a_{1}&b_{1}\\a_{2}&b_{2}\end{array}\right|,\:D_{1}=\left|\begin{array}{cc}c_{1}&d_{2}\\c_{2}&d_{2}\end{array}\right|$ and $D_{2}=\left|\begin{array}{cc}a_{1}&c_{1}\\a_{2}&c_{2}\end{array}\right|$, where $D\neq 0$.

Solution:

The given equations are

$\quad\quad 2x+y=19$

$\quad\quad 2x-3y=-3$

Comparing these equations with the above equations, we get

$\quad\quad a_{1}=2,\:b_{1}=1,\:c_{1}=19$

$\quad\quad a_{2}=2,\:b_{2}=-3,\:c_{2}=-3$

Then, $D=\left|\begin{array}{cc}2&1\\2&-3\end{array}\right|=-8$,

$\quad\quad D_{1}=\left|\begin{array}{cc}19&-3\\1&-3\end{array}\right|=-54$ and

$\quad\quad D_{2}=\left|\begin{array}{cc}2&19\\2&-3\end{array}\right|=-44$.

$\therefore x=\frac{D_{1}}{D}=\frac{-54}{-8}=\frac{27}{4}$

$\quad y=\frac{D_{2}}{D}=\frac{-44}{-8}=\frac{11}{2}$

Therefore the required solution is

$\quad\quad \boxed{x=\frac{27}{4},\:y=\frac{11}{2}}$.

Answer 2

Answer:

here is yur answer mate...

Step-by-step explanation:

x=27/4,y=11/2

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