## Introductory Algebra for College Students (7th Edition)

The solutions is $(3, 3)$.
Since we have an equation where the $y$ term is already isolated, we can use the substitution method for this system of equations. We use $y = 2x - 3$ to substitute for $y$ in the second equation: $$x + 2(2x - 3) = 9$$ Distribute first, according to order of operations: $$x + 2(2x) + (2)(-3) = 9$$ Multiply the terms to simplify: $$x + 4x - 6 = 9$$ Group like terms: $$(x + 4x) - 6 = 9$$ Combine like terms: $$5x - 6 = 9$$ Add $6$ to both sides to isolate the constants to one side of the equation: $$5x = 15$$ Divide both sides by $5$ to solve for $x$: $$x = 3$$ We can now plug $3$ in for $x$ into the first equation to solve for $y$: $$y = 2(3) - 3$$ Multiply first, according to order of operations: $$y = 6 - 3$$ Subtract to solve for $y$: $$y = 3$$ The solutions is $(3, 3)$.