Our second solution is (x,y)= (−2,−5). Plugging in x=−3, we havey= 3 x+ 1
= 3 × (−3)+ 1
=− 9 + 1
=− 8Our third solution is (x,y)= (−3,−8).Finally, we check
There are six arithmetic exercises to do! It’s tedious, but if we want to be sure our solutions
are right, it’s mandatory. We’d better be careful with the signs, using negative additions rather
than subtractions as much as possible! We check (−1,−2) in the first original equation:x^3 + 6 x^2 + 14 x−y=− 7
(−1)^3 + 6 × (−1)^2 + 14 × (−1)− (−2)=− 7
− 1 + 6 × 1 + (−14)+ 2 =− 7
− 1 + 6 + (−14)+ 2 =− 7
− 7 =− 7Next, we check (−2,−5) in the first original equation:x^3 + 6 x^2 + 14 x−y=− 7
(−2)^3 + 6 × (−2)^2 + 14 × (−2)− (−5)=− 7
− 8 + 6 × 4 + (−28)+ 5 =− 7
− 8 + 24 + (−28)+ 5 =− 7
− 7 =− 7Next, we check (−3,−8) in the first original equation:x^3 + 6 x^2 + 14 x−y=− 7
(−3)^3 + 6 × (−3)^2 + 14 × (−3)− (−8)=− 7
− 27 + 6 × 9 + (−42)+ 8 =− 7
− 27 + 54 + (−42)+ 8 =− 7
− 7 =− 7That completes the check for the original cubic. Now we plug (−1,−2) into the second origi-
nal equation:− 6 x+ 2 y= 2
− 6 × (−1)+ 2 × (−2)= 2
6 + (−4)= 2
2 = 2458 More Two-by-Two Systems