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Graph the line given by 2x+y=1 and the circle given by x²+y²=10.Find all solutions to the system of equations. Verify your result both algebraically and graphically.

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Answer:

(1.8, -2.6) and (-1, 3)

Step-by-step explanation:

[tex]2x+y=1[/tex]

[tex]x^2+y^2=10[/tex]

From the first equation

[tex]y=1-2x[/tex]

Applying to the second equation

[tex]x^2+(1-2x)^2=10\\\Rightarrow x^2+1+4x^2-4x=10\\\Rightarrow 5x^2-4x-8=0[/tex]

Solving the equation we get

[tex]x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:5\left(-9\right)}}{2\cdot \:5}, \frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:5\left(-9\right)}}{2\cdot \:5}\\\Rightarrow x=1.8, -1[/tex]

At x = 1.8

Applying in first equation

[tex]2\times 1.8+y=1\\\Rightarrow y=1-3.6\\\Rightarrow y=-2.6[/tex]

At x = -1

Applying in first equation

[tex]2\times -1+y=1\\\Rightarrow y=1+2\\\Rightarrow y=3[/tex]

∴ The circle and line intersect at points (1.8, -2.6) and (-1, 3)

Ver imagen boffeemadrid
ematthewssss6

Algebraically, the point (1, -5) satisfies the first inequality, but it does not satisfy the second inequality because -5 is not greater than -5. Graphically, the point (1, -5) lies in the shaded area of the first inequality but lies on the dashed line of the second inequality, which is not inclusive. Therefore (1, -5) is not a solution to the given system of inequalities.

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