Remember that a quadratic equation is a parabola. The equation is of the type y = Ax^2 + Bx + C
A linear equation is a straight line. The equation is of the type y = MX + N
The soluction of that system is Ax^2 + Bx + C = MX + N
=> Ax^2 + (B-M)x + (C-N) = 0
That is a quadratic equation.
A quadratic equation may have 0, 1 or 2 real solutions. Those are all the possibilitis.
So you must select 0, 1 and 2.
You can also get to that conclusion if you draw a parabola and figure out now many point of it you can intersect with a straight line.
You will realize that depending of the straight line position it can intersect the parabola in none point, or one point or two points.
The answer to the question
Answer:
400
Step-by-step explanation:
Solve your system of equations step by step.
x−2y=14;x+3y=9
Solve x−2y=14 for x:
x−2y+2y=14+2y (Add 2y to both sides)
x=2y+14
Substitute (2y+14) for x in x+3y=9:
x+3y=9
2y+14+3y=9
5y+14=9(Simplify both sides of the equation)
5y+14+−14=9+−14 (Add -14 to both sides)
5y=−5
5y5=−55 (Divide both sides by 5)
y=−1
Substitute (−1) for y in x=2y+14:
x=2y+14
x=(2)(−1)+14
x=12(Simplify both sides of the equation)
So,
y=-1
x=12
Answer is C)
