Answer:
No solutions
Step-by-step explanation:
Given the following systems of linear equations, 2x + 6y = - 18, and y = - ⅓x + 2:
Let, Equation 1: 2x + 6y = - 18
Equation 2: y = - ⅓x + 2
Use the substitution method to solve for the values of x and y.
Start by substituting the value of y from Equation 2 into Equation 1:
2x + 6y = - 18
2x + 6(- ⅓x + 2) = - 18
Distribute 6 into the parenthesis:
2x - 2x + 12 = - 18
0 + 12 = -18
12 = -18 (False statement).
Therefore, the given systems of linear equations have no solutions. In fact, when you transform Equation 1 into its slope-intercept form:
2x + 6y = - 18 (Currently in standard form, Ax + By = C)
Subtract 2x from both sides:
2x - 2x + 6y = - 2x - 18
6y = -2x - 18
Divide both sides by 6:
[tex]\frac{6y}{6} = \frac{-2x - 18}{6}[/tex]
y = -⅓x - 3 ⇒ This is the equivalent linear equation of 2x + 6y = - 18 in its slope-intercept form, y = mx + b.
Since both equations in the given system have the same slope of -⅓, then it means that their lines are parallel, and will never have a point of intersection that will provide a solution to the system.
Hence, this proves the given systems of linear equations have no solutions.
Attached is the screenshot of the graphed systems of linear equations for your reference.
