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Chapter 13: Problem 11

Solve each system by the substitution method. $$\left\\{\begin{array}{l} y^{2}=x^{2}-9 \\ 2 y=x-3 \end{array}\right.$$

Short Answer

Expert verified
The solutions for the system of equations are \((3, 0)\) and \((-5, -4)\)

Step by step solution

01

Isolate variable

The second equation should be rearranged to the form \(x = 2y + 3\). This can be achieved by adding 3 to both sides of the equation.
02

Substitute isolated variable into second equation

Now, substitute \(x\) from the second equation into the first equation. This results in \(y^{2}=(2y+3)^{2}-9\).
03

Simplify the equation

Expand the equation to get a quadratic equation. The equation becomes \(y^{2} = 4y^{2} + 12y + 9 - 9\). Simplifying this, we obtain a quadratic equation \(0 = 3y^{2}+12y\).
04

Solve the quadratic equation

Rearrange the equation to standard form \(3y^{2} +12y = 0\), then solve for \(y\) by factoring out \(3y\) giving \(3y(y+4)=0\). This gives two possible values for \(y\), either \(y = 0\) or \(y = -4\).
05

Substitute \(y\) back into isolated variable equation to solve for \(x\)

Substitute the values of \(y\) into the equation \(x = 2y + 3\). For \(y = 0\), \(x = 3\), and for \(y = -4\), \(x = -5\). Thus, the solutions are \((3, 0)\) and \((-5, -4)\).

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Most popular questions from this chapter

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$4 x^{2}=36-y^{2}$$

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. $$y=-(x+1)^{2}+4$$

Find the slope of the line passing through \((-2,-3)\) and \((1,5)\).

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$x=(y-4)^{2}-1$$

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