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Chapter 13: Problem 5

Solve each system by the substitution method. $$\left\\{\begin{array}{l} y=x^{2}-4 x-10 \\ y=-x^{2}-2 x+14 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is (4, -6) and (-3, 5).

Step by step solution

01

Substitute

Substitute the second equation into the first equation in place of \(y\), so the equations become: \(x^2-4x-10=-x^2-2x+14\).
02

Combine like terms

Bring all terms to one side to make the equation equal to zero: \(2x^2-2x-24=0\). Then simplify by dividing all sides by 2 we have: \(x^2-x-12=0\).
03

Solve the Quadratic equation

The equation \(x^2-x-12=0\) is a quadratic equation and can be factored as \((x-4)(x+3)=0\). Setting each factor equal to zero gives the solutions: \(x=4\) and \(x=-3\).
04

Find the corresponding y-values

Substitute \(x=4\) into either of the original equations, for this example, use the first one: \(y=4^2-4*4-10 = -6\). Do the same for \(x=-3\): \(y=(-3)^2-4*(-3)-10=5\).

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Most popular questions from this chapter

(GRAPH CANT COPY) Find the coordinates of the vertex for the horizontal parabola defined by the given equation. $$x=-2 y^{2}+4 y+6$$

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

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. $$x=-y^{2}-4 y+5$$

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. $$\left\\{\begin{array}{l}x=(y-3)^{2}+2 \\ x+y=5\end{array}\right.$$

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. $$x=2(y-1)^{2}+2$$
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