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Exercises Systems of equations

Example 5 from Algebra and Trigonometry, 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables

Graphing a System of Inequalities

Graph the given system of inequalities.

\begin{array}{r} x^{2} - y \leq 0 \\ 2 x^{2} + y \leq 12 \end{array}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable $y .$ will be eliminated. Then we solve for $x .$

$\begin{array}{l} \underset{____________}{\begin{matrix} \begin{array}{l} x^{2} - y = 0 \end{array} \\ 2 x^{2} + y = 12 \end{matrix}} \\ 3 x^{2} = 12 \\ x^{2} = 4 \\ x = \pm 2 \end{array}$
Substitute the x-values into one of the equations and solve for $y .$
$\begin{array}{r} x^{2} - y = 0 \\ {(2)}^{2} - y = 0 \\ 4 - y = 0 \\ y = 4 \\ {(−2)}^{2} - y = 0 \\ 4 - y = 0 \\ y = 4 \end{array}$
The two points of intersection are $(2, 4)$ and $(−2, 4) .$ Notice that the equations can be rewritten as follows.
$\begin{array}{l} x^{2} - y \leq 0 \\ x^{2} \leq y \\ y \geq x^{2} \\ 2 x^{2} + y \leq 12 \\ y \leq −2 x^{2} + 12 \end{array}$
Graph each inequality. See Figure 10. The feasible region is the region between the two equations bounded by $2 x^{2} + y \leq 12$ on the top and $x^{2} - y \leq 0$ on the bottom.

Figure 10

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Graph the given system of inequalities.

\begin{matrix} y & \geq & x^{2} - 1 \\ x - y & \geq & - 1 \end{matrix}

How did it go?

PreviousGraphing an Inequality for a Parabola Next Decomposing a Rational Function with Distinct Linear Factors

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.3, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.