Chapter 11.3, Problem 16AYU

$\{\begin{array}{l}x+2y=5\\ x-y=3\end{array}$

To determine

To find: The solution to the system of equations: $x+2y=5$, $x-y=3$.

Answer to Problem 16AYU

$x=\frac{11}{3},y=\frac{2}{3}$

Explanation of Solution

Given:

$x+2y=5$, $x-y=3$

Formula used:

To find the value of a second order $2\times 2$ determinant:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=a\times d-b\times c=ad-bc$

Cramer’s rule states that:

The solution of a system of two linear equations $ax+by=m$ and $cx+dy=n$ is given by,

$x=\frac{D_x}{D};y=\frac{D_y}{D}$

Where we have,

$D=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|;D_x=\left|\begin{array}{cc}m& b\\ n& d\end{array}\right|$ and $D_y=\left|\begin{array}{cc}a& m\\ c& n\end{array}\right|$; provided $D=ad-bc\ne 0$.

When $D=0$, the equations are inconsistent or dependent.

Calculation:

$x+2y=5$

$x-y=3$

The above system of equations is equivalent to:

$1x+2y=5$

$1x-1y=3$

Where we have:

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

$D_x=\left|\begin{array}{cc}5& 2\\ 3& -1\end{array}\right|=-5-6=-11$

$D_y=\left|\begin{array}{cc}1& 5\\ 1& 3\end{array}\right|=3-5=-2$

So using Cramer’s rule we get,

$x=\frac{D_x}{D}=\frac{-11}{-3}=\frac{11}{3}$

$y=\frac{D_y}{D}=\frac{-2}{-3}=\frac{2}{3}$

Thus the solutions are $x=\frac{11}{3},y=\frac{2}{3}$.