## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The solution set is $\{(a,b)\ |\ a=2b+3, b\in \mathbb{R}\}$
Divide the second equation with 3 to isolate $y$ $a=2b+3\qquad$ (*) Replace $a$ with $2b+3$ in the other equation: $(2b+3)-2b=3\qquad$ ... and solve for $x$. $2b-2b=3-3$ $0=0\qquad$ ... always true, there will be infinitely many solutions. This is the case where both equations have graphs that coincide. All points on that common line have coordinates that satisfy both equations. Letting $b\in \mathbb{R}$ (be any real number), from equation 1 it follows: $a=2b+3$ The solution set is $\{(a,b)\ |\ a=2b+3, b\in \mathbb{R}\}$