The drop down menus are not provided. However, I will help get the solution of the system of equation which is the requirement of the question.
Answer:
(3,14) and (-2,-1) are the solutions of the system
Step-by-step explanation:
The first equation given is:
y = x² + 2x - 1 .........> equation I
The second given equation is:
y - 3x = 5
This can be rewritten as:
y = 3x + 5 ..........> equation II
Equate equations I and II and solve for x as follows:
x² + 2x - 1 = 3x + 5
x² + 2x - 1 - 3x - 5 = 0
x² - x - 6 = 0
(x-3)(x+2) = 0
either x-3 = 0 .........> x = 3
or x+2 = 0 ...........> x = -2
Now, we get the values of y by substituting in equation II as follows:
At x = 3:
y = 3x + 5
y = 3(3) + 5
y = 14
The first solution is (3,14)
At x = -2:
y = 3x + 5
y = 3(-2) + 5
y = -1
The second solution is (-2,-1)
Hope this helps :)