Q. 32 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 15 – 42, solve each system of equations using Cramer’s Rule if it is applicable. If Cramer’s Rule is not applicable, say so.

$\left\{\begin{array}{l}2x-y=-1\\ x+\frac{1}{2}y=\frac{3}{2}\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations $\left\{\begin{array}{l}2x-y=-1\\ x+\frac{1}{2}y=\frac{3}{2}\end{array}\right.$is $\left(\frac{1}{2},2\right).$

Step 1. Given information

The given system of equations is 

$\left\{\begin{array}{l}2x-y=-1\\ x+\frac{1}{2}y=\frac{3}{2}\end{array}\right.$

Step 2. Determinant

The determinant D of the coefficients of the variables 

$$\begin{gathered} D=\left|\begin{array}{cc}2& -1\\ 1& \frac{1}{2}\end{array}\right| \\ D=\left(2\right)\left(\frac{1}{2}\right)-(-1)\left(1\right) \\ D=1+1 \\ D=2 \end{gathered}$$

$D\ne 0,$so Cramer's rule can be used to determine the solution of the system of equations 

Step 3. Solution for x

According to Cramer's rule $x=\frac{D_x}{D}$

Determine $D_x$

$$\begin{gathered} D_x=\left|\begin{array}{cc}-1& -1\\ \frac{3}{2}& \frac{1}{2}\end{array}\right| \\ {D}_x=(-1)\left(\frac{1}{2}\right)-(-1)\left(\frac{3}{2}\right) \\ {D}_x=-\frac{1}{2}+\frac{3}{2} \\ {D}_x=1 \end{gathered}$$

Then 

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{1}{2} \end{gathered}$$

Step 4. Solution for y

According to Cramer's rule $y=\frac{D_y}{D}$

Determine $D_y$

$$\begin{gathered} D_y=\left|\begin{array}{cc}2& -1\\ 1& \frac{3}{2}\end{array}\right| \\ {D}_y=\left(2\right)\left(\frac{3}{2}\right)-(-1)\left(1\right) \\ {D}_y=3+1 \\ {D}_y=4 \end{gathered}$$

Then $$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{4}{2} \\ y=2 \end{gathered}$$

So the solution of the system of equations is$\left(\frac{1}{2},2\right)$