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.
<u><em>Answer:</em></u> (3,14) and (-2,-1) are the solutions of the system
<u><em>Explanation:</em></u> <u>The first equation given is:</u> y = x² + 2x - 1 .........> equation I <u>The second given equation is:</u> y - 3x = 5 <u>This can be rewritten as:</u> y = 3x + 5 ..........> equation II
<u>Equate equations I and II and solve for x as follows:</u> 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
<u>Now, we get the values of y by substituting in equation II as follows:</u> 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)