What is the solution of the linear-quadratic system of equations?
y=x^2+5x-3
y-x=2
2 answers:
-x=-y+2
x=y-2
now substitute iin the first one
y=(y-2)^2+5 (y-2)-3
y=y^2-4y+4 +5y-10-3
y=y^2+y-9
y^2-9=0.
y^2=9
y=3
3-x=2
-x=2-3
-x=-1
x=1
the solution are x=1 and y=3
3=1^2+5 (1)-3
3=1+5-3
3=6-3
3=3
Answer:
Step-by-step explanation:
-x=-y+2
x=y-2
now substitute iin the first one
y=(y-2)^2+5 (y-2)-3
y=y^2-4y+4 +5y-10-3
y=y^2+y-9
y^2-9=0.
y^2=9
y=3
3-x=2
-x=2-3
-x=-1
x=1
the solution are x=1 and y=3
3=1^2+5 (1)-3
3=1+5-3
3=6-3
3=3
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