Chapter 11.3, Problem 2AYU

Using Cramer’s Rule, the value of $x$ that satisfies the system of equations. $\{\begin{array}{l}2x+3y=5\\ x-4y=-3\end{array}isx=\underset{\left[\begin{array}{l}\begin{array}{cc}2& 3\end{array}\\ \begin{array}{cc}1& -4\end{array}\end{array}\right]}{\_\_\_\_\_\_}$

To determine

To fill: Using Cramer’s rule, the value of $x$ that satisfies the system of equations:

$2x+3y=5;x-4y=-3$ is $x=\frac{}{\left|\begin{array}{cc}2& 3\\ 1& -4\end{array}\right|}$.

Answer to Problem 2AYU

$x=\frac{\left|\begin{array}{cc}5& 3\\ -3& -4\end{array}\right|}{\left|\begin{array}{cc}2& 3\\ 1& -4\end{array}\right|}$

Explanation of Solution

Given:

$2x+3y=5;x-4y=-3$

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:

The system of equations is:

$2x+3y=5$

$x-4y=-3$

By the given concept we get,

$x=\frac{\left|\begin{array}{cc}5& 3\\ -3& -4\end{array}\right|}{\left|\begin{array}{cc}2& 3\\ 1& -4\end{array}\right|}$

Using Cramer’s rule, the value of $x$ that satisfies the system of equations:

$2x+3y=5;x-4y=-3$ is $x=\frac{\left|\begin{array}{cc}5& 3\\ -3& -4\end{array}\right|}{\left|\begin{array}{cc}2& 3\\ 1& -4\end{array}\right|}$.