Given:
The equations of a system of equations are
[tex]y=-2x^2+3x-1[/tex]
[tex]x+y=-1[/tex]
To find:
The solution of the given system of equations.
Solution:
We have,
[tex]y=-2x^2+3x-1[/tex] ...(i)
[tex]x+y=-1[/tex] ...(ii)
Using (i) and (ii), we get
[tex]x+(-2x^2+3x-1)=-1[/tex]
[tex]x-2x^2+3x-1+1=0[/tex]
[tex]-2x^2+4x=0[/tex]
Taking out the common factors.
[tex]2x(-x+2)=0[/tex]
Using zero product property,we get
[tex]2x=0[/tex] and [tex]-x+2=0[/tex]
[tex]x=0[/tex] and [tex]2=x[/tex]
For x=0, we have
[tex]0+y=-1[/tex]
[tex]y=-1[/tex]
For x=2, we have
[tex]2+y=-1[/tex]
[tex]y=-1-2[/tex]
[tex]y=-3[/tex]
Therefore, the two solutions of the given system of equations are (0,-1) and (2,-3).
