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Chapter 13: Problem 4

Solve each system by the substitution method. $$\left\\{\begin{array}{l} 2 x+y=-5 \\ y=x^{2}+6 x+7 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are (-2, 3) and (-6, 7).

Step by step solution

01

Substitution

Start by substituting \(y = x^2 + 6x + 7\) from the second equation into the first equation. This will allow us to solve for x using one equation. The first equation becomes \(2x + (x^2 + 6x + 7) = -5\). Simplifying this gives us \(x^2 + 8x + 7 + 2x = -5\).
02

Simplify Equation

Combine like terms to simplify the equation from step 1. This gives us \(x^2 + 10x + 7 = -5\). Then, move the -5 over to the left side by adding 5 to both sides, which yields \(x^2 + 10x + 12 = 0\).
03

Find the Roots

Now, solve for x by finding the roots of the equation \(x^2 + 10x + 12 = 0\). This is a quadratic equation that can be factored into (x + 2)(x + 6) = 0. Setting both factors equal to zero gives the solutions x = -2 and x = -6.
04

Solve for y

With the solutions for x, replace x by -2 and -6 in the equation \(y = x^2 + 6x + 7\) to calculate the corresponding y-values. This will yield solutions as (-2,3) and (-6,7)

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Most popular questions from this chapter

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. $$x=y^{2}-6 y+8$$

How can you distinguish ellipses from circles by looking at their equations?

Multiply: \(\quad(3 x-2)\left(2 x^{2}-4 x+3\right)\)

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.

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$x-3-4 y=6 y^{2}$$
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