Since the coefficient of the quadratic variable is missing, I obtained it from a similar question.
The system is:
- y = (1/2)x² + 2²x - 1 and 3x - y = 1
Answer:
- The two solutions are (0, -1) and (2, 5)
Explanation:
1. Write the system
[tex]y=(1/2)x^2+2x-1\\ \\ 3x-y=1[/tex]
2. Clear y from the second equation to solve by substitution
[tex]y= 3x-1[/tex]
3. Substitute in the first equation and solve for x
[tex]3x-1=(1/2)x^2+2x-1\\ \\ (1/2)x^2+2x-3x=0\\ \\ (1/2)x^2-x=0\\ \\ x(x/2-1)=0[/tex]
[tex]x=0[/tex]
[tex]x/2-1=0\\ \\ x/2=1\\ \\ x=2[/tex]
4. Subsitute both values into the equation y= 3x - 1
- y = 3(0) - 1 = - 1 ⇒ solution (0, -1)
- y = 3(2) - 1 = 5 ⇒ solution (2, 5)
