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Chapter 13: Problem 18

Solve each system by the substitution method. $$\left\\{\begin{array}{l} 2 x+y=4 \\ (x+1)^{2}+(y-2)^{2}=4 \end{array}\right.$$

Short Answer

Expert verified
The solutions of the system of equations are \( (1 + sqrt(2), -2sqrt(2))\) and \( (1 - sqrt(2), 2sqrt(2)) \).

Step by step solution

01

Isolate the Variable in the First Equation

From the first linear equation \(2x+y=4\), isolate the variable \(y\) by subtracting \(2x\) from both sides to get \(y = 4 - 2x\).
02

Substitute y into the Second Equation

Substitute \(y\) from the first equation to the second circular equation \( (x+1)^{2}+(y-2)^{2}=4 \), we will have \( (x+1)^{2}+[(4-2x)-2]^{2}=4 \). Simplify to get a quadratic equation \( x^2 - x -1 = 0\).
03

Solve the Quadratic Equation

Solving the quadratic equation \( x^2 - x -1 = 0\) will give the solution for the variable \(x\). Using the formula for the quadratic equation \(x = [-b \pm sqrt(b^2 - 4ac)] / 2a\), the solutions for \(x\) are \(x = 1 \pm sqrt(2)\).
04

Find y Corresponding to each x Solution

Substitute the obtained \(x\)-values into the first equation \(y = 4 - 2x\) to find corresponding \(y\)-values. For \(x = 1 + sqrt(2)\), we get \(y = 4 - 2(1 + sqrt(2)) = -2sqrt(2)\). For \(x = 1 - sqrt(2)\) we get \(y = 4 - 2(1 - sqrt(2)) = 2sqrt(2)\).

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Most popular questions from this chapter

The equation of a parabola is given. Determine: a. if the parabola is horizontal or vertical. b. the way the parabola opens. c. the vertex. $$x=2(y-1)^{2}+2$$

Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. $$\left\\{\begin{array}{l}x=y^{2}-5 \\ x^{2}+y^{2}=25\end{array}\right.$$

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$(x-2)^{2}+(y+1)^{2}=16$$

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. I'm graphing an equation that contains neither an \(x^{2}\) -term nor a \(y^{2}\) -term, so the graph cannot be a conic section.

Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. $$\left\\{\begin{array}{l}x=(y-2)^{2}-4 \\\y=-\frac{1}{2} x\end{array}\right.$$
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