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Chapter 13: Problem 3

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

Short Answer

Expert verified
The solution to the system of equations is (1, 1) and (2, 0).

Step by step solution

01

Substitute \(x\) in the first equation

Substitute \(x\) from the second equation into the first equation. This will give the equation \(x + (x^{2} - 4x + 4) = 2\)
02

Simplify the equation

Simplify this equation by combining like terms. Simplification will result in the quadratic equation \(x^{2} - 3x + 4 - 2 = 0\), i.e., \(x^{2} - 3x + 2 = 0\)
03

Solve for \(x\)

Solve for \(x\) in the quadratic equation. This can be factored to \((x - 1)(x - 2) = 0\), which gives us \(x = 1\) or \(x = 2\)
04

Solve for \(y\)

Substitute the found \(x\) values into the original first equation to solve for \(y\). When substituting \(x = 1\), the result is \(y = 2 - 1 = 1\), and when substituting \(x = 2\), the result is \(y = 2 - 2 = 0\)

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Most popular questions from this chapter

Will help you prepare for the material covered in the next section. Solve: \(x^{2}=2(3 x-9)+10\)

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$7 x^{2}+7 y^{2}=28$$

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. $$x=-(y+3)^{2}+4$$

Use the vertex and intercepts to sketch the graph of each equation. If needed, find additional points on the parabola by choosing values of y on each side of the axis of symmetry. $$x=-2(y+6)^{2}+2$$

Solve: \((x+1)^{2}+(x+3)^{2}=4 .\) (Section 6.6, Example 6)
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