For x= 3, we have
y= 2 × 3 + 1
= 6 + 1
= 7
The two solutions of the system are therefore (x, y)= (−2,−3) and (x, y)= (3, 7).
Finally, we check
Let’s check both solutions in the original two-variable equations to be certain that we did our
algebra and arithmetic right. This exercise can also help us to see how the system “plays out.”
First, let’s check (−2,−3) in the original two-variable linear equation:
2 x−y+ 1 = 0
2 × (−2)− (−3)+ 1 = 0
− 4 + 3 + 1 = 0
− 1 + 1 = 0
0 = 0
Next, we check (3, 7) in that same equation:
2 x−y+ 1 = 0
2 × 3 − 7 + 1 = 0
6 − 7 + 1 = 0
− 1 + 1 = 0
0 = 0
Next, we check (−2,−3) in the original two-variable quadratic:
y=x^2 +x− 5
− 3 = (−2)^2 + (−2)− 5
− 3 = 4 + (−2)− 5
− 3 = 2 − 5
− 3 =− 3
To finish up, we check (3, 7) in the original quadratic:
y=x^2 +x− 5
7 = 32 + 3 − 5
7 = 9 + 3 − 5
7 = 12 − 5
7 = 7
Linear and Quadratic 449