Chapter 11.3, Problem 31AYU

$\{\begin{array}{l}3x-5y=3\\ 15x+5y=21\end{array}$

To determine

To find: The solution to the system of equations: $3x-\text{ 5}y=\text{ 3},\text{ 15}x\text{ + 5}y=\text{ 21}$.

Answer to Problem 31AYU

$x=\frac{4}{3};y=\frac{1}{5}$

Explanation of Solution

Given:

$3x-\text{ 5}y=\text{ 3},\text{ 15}x\text{ + 5}y=\text{ 21}$

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:

$3x-\text{ 5}y=\text{ 3}$

$\text{15}x\text{ + 5}y=\text{ 21}$

Where we have:

$D=\left|\begin{array}{cc}3& -5\\ 15& 5\end{array}\right|=15+75=90$

$D_x=\left|\begin{array}{cc}3& -5\\ 21& 5\end{array}\right|=15+105=120$

$D_y=\left|\begin{array}{cc}3& 3\\ 15& 21\end{array}\right|=63-45=18$

So using Cramer’s rule we get,

$x=\frac{D_x}{D}=\frac{120}{90}=\frac{4}{3}$

$y=\frac{D_y}{D}=\frac{18}{90}=\frac{1}{5}$

Thus the solutions are $x=\frac{4}{3};y=\frac{1}{5}$.