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anemone
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For all real $a,\,b,\,x,\,y$ such that
$ax+by=4,\\ax^2+by^2=2,\\ax^3+by^3=-1.$
Find $(2x-1)(2y-1)$.
$ax+by=4,\\ax^2+by^2=2,\\ax^3+by^3=-1.$
Find $(2x-1)(2y-1)$.
anemone said:For all real $a,\,b,\,x,\,y$ such that
$ax+by=4---(1),\\ax^2+by^2=2---(2),\\ax^3+by^3=-1---(3).$
Find $(2x-1)(2y-1)$.
Albert said:$(2)\times x :ax^3+bxy^2=2x---(4)$
$(2)\times y :ax^2y+by^3=2y---(5)$
from $(1)(3)$ we get $(4)+(5)\rightarrow -1+4xy=2x+2y\rightarrow 4xy-2x-2y+1=1+1=2=(2x-1)(2y-1)$
Albert said:$(2)\times x :ax^3+bxy^2=2x---(4)$
$(2)\times y :ax^2y+by^3=2y---(5)$
from $(1)(3)$ we get $(4)+(5)\rightarrow -1+4xy=2x+2y\rightarrow 4xy-2x-2y+1=1+1=2=(2x-1)(2y-1)$
A quadratic system is a set of two or more equations that contain at least one quadratic equation, which is an equation with one variable raised to the power of two.
To solve a quadratic system, you can use several methods such as graphing, substitution, elimination, or using the quadratic formula.
$(2x-1)(2y-1)$ represents the product of two quadratic equations in the system, which can be expanded and simplified to solve for the values of x and y.
Yes, a quadratic system can have any number of equations, as long as at least one of them is a quadratic equation.
A quadratic system can have zero, one, or two solutions, depending on the number of equations and the nature of the equations in the system.