Exercise 11 Page 265

In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using the Substitution Method, there are three steps.

1. Isolate a variable in one of the equations.
2. Substitute the expression for that variable into the other equation and solve.
3. Substitute this solution into one of the equations and solve for the value of the other variable.

For this exercise, x is already isolated in one equation, so we can skip straight to solving!

x=y+2 & (I) - 3x+3y=6 & (II)

Substitute

(II): x= y+2

x=y+2 - 3( y+2)+3y=6

Distr

(II): Distribute -3

x=y+2 - 3y-6+3y=6

AddTerms

(II): Add terms

x=y+2 - 6≠ 6 *

Solving this system of equations resulted in a contradiction; -6 can never be equal to 6. The lines are parallel and do not have a point of intersection. Therefore, the system has no solution.