Mathematics

Solve the system of equations

x^2-y=0
-3x+y=-2


Answers

yassautumn

4 years ago Comment

[latex]x^2-y=0\\ -3x+y=-2\\ -------\\ x^2-3x=-2\\ x^2-3x+2=0\\ x^2-2x-x+2=0\\ x(x-2)-1(x-2)=0\\ (x-1)(x-2)=0\\ x=1 \vee x=2\\ 1^2-y=0 \vee 2^2-y=0\\ y=1 \vee y=4\\\\ (x=1 \wedge y=1) \vee (x=2 \wedge y=4)[/latex]

lyssie11234

4 years ago Comment

[latex]\left\{\begin{array}{ccc}x^2-y=0&\to y=x^2\\-3x+y=-2\end{array}\right\\\\substitute:\\\\-3x+x^2=-2\\x^2-3x+2=0\\\\a=1;\ b=-3;\ c=2\\\Delta=b^2-4ac\to\Delta=(-3)^2-4\cdot1\cdot2=9-8=1\\\\x_1=\frac{-b-\sqrt\Delta}{2a}\to x_1=\frac{3-\sqrt1}{2\cdot1}=\frac{3-1}{2}=\frac{2}{2}=1\\\\x_2=\frac{-b+\sqrt\Delta}{2a}\to x_2=\frac{3+\sqrt1}{2\cdot1}=\frac{3+1}{2}=\frac{4}{2}=2\\\\y_1=x_1^2\to y_1=1^2=1\\\\y_2=x_2^2\to y_2=2^2=6\\\\Answer:x=1\ and\ y=1\ or\ x=2\ and\ y=4.[/latex]