## Elementary Linear Algebra 7th Edition

The solution of this system is $$x=\frac{1}{5}\qquad y= \frac{1}{2}.$$
Follow the steps bellow: Step 1: Subtract from the second equation the first one multiplied by $3/2$ to eliminate $x$: $$3x-\frac{3}{2}2x+2y-\frac{3}{2}(-y) =1.6-\frac{3}{2}(-0.1)\Rightarrow 3x-3x+2y+\frac{3}{2}y = 1.6+0.15$$ which becomes $$\frac{7}{2}y = 1.75 = \frac{7}{4}$$ and this gives $$y=\frac{1}{2}.$$ Step 2: Use back substitution to find $x$ from the 1st equation: $$2x-\frac{1}{2} = -0.1\Rightarrow 2x=\frac{1}{2}-0.1\Rightarrow 2x = 0.5-0.1 = 0.4=\frac{2}{5}.$$ this gives $$x=\frac{1}{5}.$$