704 Worked-Out Solutions to Exercises: Chapters 21 to 29
x-value into either of the original quadratic functions. Let’s use the first one:y=x^2 +x+ 1
= (−1)^2 + (−1)+ 1
= 1 + (−1)+ 1
= 1The system has the single solution (x,y)= (−1,1).- First, let’s check (−1,1) in the first original quadratic:
x^2 +x−y=− 1
(−1)^2 + (−1)− 1 =− 1
1 + (−1)− 1 =− 1
0 − 1 =− 1
− 1 =− 1Next, let’s check (−1,1) in the second original quadratic:x^2 − 2 x−y= 2
(−1)^2 − 2 × (−1)− 1 = 2
1 − (−2)− 1 = 2
1 + 2 − 1 = 2
3 − 1 = 2
2 = 2- Here are the two equations in their original forms:
x^2 +y= 0and2 x^3 −y= 0We can morph these into functions of x, gettingy=−x^2andy= 2 x^3Mixing the right sides of these equations gives us−x^2 = 2 x^3