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ElenaW [278]
3 years ago
6

Give all solutions if the nonlinear system of equations, including those with no real complex components.

Mathematics
2 answers:
Ksivusya [100]3 years ago
6 0
Y=x^2+6x
Apply the value of y into the equation below:
4x-y=-24
4x-(x^2+6x) = -24
4x-x^2-6x = -24
x^2+2x-24=0
(x+6)(x-4)=0
Therefore x can be -6 or 4
If x = -6
Then y = (-6)^2+6(-6) = 36 -36=0
If x = 4
Then y = (4)^2 + 6(4) = 16 + 24 = 40<span>
</span>
konstantin123 [22]3 years ago
4 0
\left\{\begin{array}{ccc}y=x^2+6x\\4x-y=-24\end{array}\right\\\\substitute:\\\\4x-(x^2+6x)=-24\\\\4x-x^2-6x+24=0\\\\-x^2-2x+24=0\\\\a=-1;\ b=-2;\ c=24\\\\\Delta=b^2-4ac

\Delta=(-2)^2-4\cdot(-1)\cdot24=4+96=100\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\x_1=\frac{2-10}{2\cdot(-1)}=\frac{-8}{-2}=4;\ x_2=\frac{2+10}{2\cdot(-1)}=\frac{12}{-2}=-6\\\\y_1=4^2+6\cdot4=16+24=40;\ y_2=(-6)^2+6\cdot(-6)=36-36=0\\\\Answer:\\x=4\ and\ y=40\ or\ x=-6\ and\ y=0
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