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Chapter 13: Problem 1

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

Short Answer

Expert verified
The solutions of the system of equations are (2, 0) and (-3, 5).

Step by step solution

01

Solving for y

Isolating y in terms of x in the first equation gives us \(y = 2 - x\).
02

Substitution

Substitute \(y\) from the first equation into the second, and thus the second equation becomes \(2 - x = x^{2} - 4\).
03

Solve the equation for x

Solving the above equation for x, First rearrange the equation: \(0 = x^{2} + x - 6\). Then, we factor the expression and solve for x. \(0 = (x + 3)(x - 2)\). Setting each factor equal to zero gives us the solutions \(x = 2, -3\).
04

Solve for y

Substitute the obtained x-values into the first equation \(y = 2 - x\) to get \[y_1 = 2 - 2 = 0\] and \[y_2 = 2 - (-3) = 5\]. So, we have two solutions: \((2, 0)\) and \((-3, 5)\).

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

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

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

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$x=y^{2}-2 y-5$$

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