Systems of Nonlinear Equations and Inequalities: Two Variables

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Graph the parabola \(y=-{x}^{2}+4x-3\)

   Bonisa impendulo

   Standard form	\(y=-{x}^{2}+4x-3a=-1,b=4,c=-3\)
   Step 1	Since \(a=-1\) the parabola opens downward
   Step 2	The axis of symmetry is given by \(x=-\frac{b}{2a}\) , \(x=-\frac{4}{2(-1)}=2\) The axis of symmetry is \(x=2\)
   Step 3	The vertex is on the line \(x=2\) Let's substitute \(x=2\) into \(y=-{x}^{2}+4x-3\) \(\begin{array}{l}y=-{(2)}^{2}+4(2)-3 \\ y=-4+8-3 \\ y=1\end{array}\) The vertex is the point (2, 1)
   Step 4	To find y-intercept, substitute \(x=0\) into \(y=-{x}^{2}+4x-3\) \(\begin{array}{l}y=-{(0)}^{2}+4(0)-3 \\ y=-3\end{array}\) The y-intercept is the point (0, -3) and is it 2 units to the left of the vertex. The symmetric point is 2 units to the right of the vertex and is (4, -3)
   Step 5	To find the x-intercepts, substitute \(y=0\) into \(y=-{x}^{2}+4x-3\) and solve for \(x\) \(\begin{array}{l}0=-{x}^{2}+4x-3 \\ 0=-({x}^{2}-4x+3) \\ 0={x}^{2}-4x+3 \\ 0=(x-3)(x-1) \\ x=3orx=1\end{array}\) The x-intercepts are (3, 0) and (1, 0)
   Step 6	Graph the parabola Axis of symmetry \(x=2\) Vertex: (2,1) y-intercept: (0, -3), symmetric point: (4, -3) x-intercepts: (3, 0) and (1, 0)

2. Graph the parabola \(y=2{x}^{2}+4x+6\)
3. 1. ⓐ Find the center and radius and then graph the circle \({(x+3)}^{2}+{(y+4)}^{2}=4\)
   2. ⓑ Find the center and radius and then graph the circle \({x}^{2}+{y}^{2}-6x-8y+9=0\)
   Bonisa impendulo

   1. ⓐ

      Use the standard form of the equation of a circle. Identify the center, (h,k) and radius, r.	\(\begin{array}{l}{(x-h)}^{2}+{(y-k)}^{2}={r}^{2} \\ {(x+3)}^{2}+{(y+4)}^{2}=4 \\ (x-{(-3))}^{2}+(y-{(-4))}^{2}={2}^{2}\end{array}\) Center: (-3, -4) Radius: 2
      Graph the circle

   2. ⓑ

      We need to rewrite this general form into standard form in order to find the center and radius.

      Step 1 Group the x-terms and y-terms. Collect the constants on the right side.	\(\begin{array}{l}{x}^{2}+{y}^{2}-6x-8y+9=0 \\ {x}^{2}-6x+{y}^{2}-8y=-9\end{array}\)
      Step 2 Complete the squares	\(\begin{array}{l}{x}^{2}-6x+{(\frac{6}{2})}^{2}+{y}^{2}-8y+{(\frac{8}{2})}^{2}=-9+{(\frac{6}{2})}^{2}+{(\frac{8}{2})}^{2} \\ {x}^{2}-6x+9+{y}^{2}-8y+16=-9+9+16\end{array}\)
      Step 3 Rewrite as binomial squares.	\({(x-3)}^{2}+{(y-4)}^{2}=16\)
      Step 4 Identify the center and radius.	Center: (3,4) Radius: 4
      Step 5 Graph the circle.

4. Find the center and radius and then graph the circle \({(x-3)}^{2}+{(y+4)}^{2}=25\)

   Use the standard form of the equation of a circle. Identify the center, (h, k) and radius, r.
   Graph the circle

5. Find the center and radius and then graph the circle \({x}^{2}+{y}^{2}+12x-14y+21=0\)

   Step 1 Group the x-terms and y-terms. Collect the constants on the right side.
   Step 2 Complete the squares	________________________________________
   Step 3 Rewrite as binomial squares.	________________________________________
   Step 4 Identify the center and radius.	Center: ________ Radius: ________
   Step 5 Graph the circle.

6. Solve the system of equations.

   \[\begin{array}{l}x-y=-1 \\ y={x}^{2}+1\end{array}\]
   Bonisa impendulo

   Solve the first equation for \(x\) and then substitute the resulting expression into the second equation.

   \[\begin{array}{llll}x-y=-1 & & & \\ x=y-1 & & & \text{Solve for }x. \\ & & & \\ y={x}^{2}+1 & & & \\ y={(y-1)}^{2}+1 & & & \text{Substitute expression for }x.\end{array}\]

   Expand the equation and set it equal to zero.

   \[\begin{array}{l}y={(y-1)}^{2}+1 \\ =({y}^{2}-2y+1)+1 \\ ={y}^{2}-2y+2 \\ 0={y}^{2}-3y+2 \\ =(y-2)(y-1)\end{array}\]

   Solving for \(y\) gives \(y=2\) and \(y=1.\) Next, substitute each value for \(y\) into the first equation to solve for \(x.\) Always substitute the value into the linear equation to check for extraneous solutions.

   \[\begin{array}{l}\ x-y=-1 \\ x-(2)=-1 \\ x=1 \\ \\ x-(1)=-1 \\ x=0\end{array}\]

   The solutions are \((1,2)\) and \((0,1),\) which can be verified by substituting these \((x,y)\) values into both of the original equations. See .

7. Solve the given system of equations by substitution.

   \[\begin{array}{lll}3x-y & = & -2 \\ 2{x}^{2}-y & = & 0\end{array}\]
   Bonisa impendulo

   \((-\frac{1}{2},\frac{1}{2})\) and \((2,8)\)

8. Find the intersection of the given circle and the given line by substitution.

   \[\begin{array}{l}{x}^{2}+{y}^{2}=5 \\ y=3x-5\end{array}\]
   Bonisa impendulo

   One of the equations has already been solved for \(y.\) We will substitute \(y=3x-5\) into the equation for the circle.

   \[\begin{array}{l}{x}^{2}+{(3x-5)}^{2}=5 \\ {x}^{2}+9{x}^{2}-30x+25=5 \\ 10{x}^{2}-30x+20=0\end{array}\]

   Now, we factor and solve for \(x.\)

   \[\begin{array}{l}10({x}^{2}-3x+2)=0 \\ 10(x-2)(x-1)=0 \\ x=2 \\ x=1\end{array}\]

   Substitute the two x-values into the original linear equation to solve for \(y.\)

   \[\begin{array}{l}y=3(2)-5 \\ =1 \\ \\ \\ \\ y=3(1)-5 \\ =-2\end{array}\]

   The line intersects the circle at \((2,1)\) and \((1,-2),\) which can be verified by substituting these \((x,y)\) values into both of the original equations. See .

9. Solve the system of nonlinear equations.

   \[\begin{array}{lll}{x}^{2}+{y}^{2} & = & 10 \\ x-3y & = & -10\end{array}\]
   Bonisa impendulo

   \((-1,3)\)

10. Solve the system of nonlinear equations.

    \[\begin{array}{ll}{x}^{2}+{y}^{2}=26 & (1) \\ 3{x}^{2}+25{y}^{2}=100 & (2)\end{array}\]
    Bonisa impendulo

    Let’s begin by multiplying equation (1) by \(-3,\) and adding it to equation (2).

    \[\frac{\begin{array}{l}\begin{array}{l} \\ (-3)({x}^{2}+{y}^{2})=(-3)(26) \\ \ -3{x}^{2}-3{y}^{2}=-78\end{array} \\ \ 3{x}^{2}+25{y}^{2}=100\end{array}}{\ 22{y}^{2}=22}\]

    After we add the two equations together, we solve for \(y.\)

    \[\begin{array}{l}{y}^{2}=1 \\ y=\pm \sqrt{1}=\pm 1\end{array}\]

    Substitute \(y=\pm 1\) into one of the equations and solve for \(x.\)

    \[\begin{array}{l}\ {x}^{2}+{(1)}^{2}=26 \\ {x}^{2}+1=26 \\ {x}^{2}=25 \\ x=\pm \sqrt{25}=\pm 5 \\ \\ {x}^{2}+{(-1)}^{2}=26 \\ {x}^{2}+1=26 \\ {x}^{2}=25=\pm 5\end{array}\]

    There are four solutions: \((5,1),(-5,1),(5,-1),\text{and}\ (-5,-1).\) See .

11. Find the solution set for the given system of nonlinear equations.

    \[\begin{array}{lll}4{x}^{2}+{y}^{2} & = & 13 \\ {x}^{2}+{y}^{2} & = & 10\end{array}\]
    Bonisa impendulo

    \(\{(1,3),(1,-3),(-1,3),(-1,-3)\}\)

12. Graph the inequality \(y>{x}^{2}+1.\)

    Bonisa impendulo

    First, graph the corresponding equation \(y={x}^{2}+1.\) Since \(y>{x}^{2}+1\) has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points
    \((0,2)\) and \((2,0).\) One point is clearly inside the parabola and the other point is clearly outside.

    \[\begin{array}{ll}y>{x}^{2}+1 & \\ 2>{(0)}^{2}+1 & \\ 2>1 & \text{True} \\ & \\ & \\ & \\ 0>{(2)}^{2}+1 & \\ 0>5 & \text{False}\end{array}\]

    The graph is shown in . We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

13. Graph the given system of inequalities.

    \[\begin{array}{l}{x}^{2}-y\le 0 \\ 2{x}^{2}+y\le 12\end{array}\]
    Bonisa impendulo

    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 \(x.\)

    \[\begin{array}{l}\underset{____________}{\begin{array}{l}\begin{array}{l} \\ {x}^{2}-y=0\end{array} \\ 2{x}^{2}+y=12\end{array}} \\ 3{x}^{2}=12 \\ {x}^{2}=4 \\ x=\pm 2\end{array}\]

    Substitute the x-values into one of the equations and solve for \(y.\)

    \[\begin{array}{l}{x}^{2}-y=0 \\ {(2)}^{2}-y=0 \\ 4-y=0 \\ y=4 \\ \\ {(-2)}^{2}-y=0 \\ 4-y=0 \\ y=4\end{array}\]

    The two points of intersection are \((2,4)\) and \((-2,4).\) Notice that the equations can be rewritten as follows.

    \[\begin{array}{l}{x}^{2}-y\le 0 \\ {x}^{2}\le y \\ y\ge {x}^{2} \\ \\ 2{x}^{2}+y\le 12 \\ y\le -2{x}^{2}+12\end{array}\]

    Graph each inequality. See . The feasible region is the region between the two equations bounded by \(2{x}^{2}+y\le 12\) on the top and \({x}^{2}-y\le 0\) on the bottom.

14. Graph the given system of inequalities.
    

    \[\begin{array}{lll}y & \ge & {x}^{2}-1 \\ x-y & \ge & -1\end{array}\]

15. Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers.

    Bonisa impendulo

    A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches also intersect the circle, hence three solutions.

16. When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution?

17. When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?

    Bonisa impendulo

    No. There does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region.

18. If you graph a revenue and cost function, explain how to determine in what regions there is profit.

19. If you perform your break-even analysis and there is more than one solution, explain how you would determine which x-values are profit and which are not.

    Bonisa impendulo

    Choose any number between each solution and plug into \(C(x)\) and \(R(x).\) If \(C(x)

20. \(\begin{array}{l}\ x+y=4 \\ {x}^{2}+{y}^{2}=9\end{array}\)

21. \(\begin{array}{l}\ y=x-3 \\ {x}^{2}+{y}^{2}=9\end{array}\)

    Bonisa impendulo

    \((0,-3),(3,0)\)

22. \(\begin{array}{l}\ y=x \\ {x}^{2}+{y}^{2}=9\end{array}\)

23. \(\begin{array}{l}\ y=-x \\ {x}^{2}+{y}^{2}=9\end{array}\)

    Bonisa impendulo

    \((-\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}),(\frac{3\sqrt{2}}{2},-\frac{3\sqrt{2}}{2})\)

24. \(\begin{array}{l}\ x=2 \\ {x}^{2}-{y}^{2}=9\end{array}\)

25. \(\begin{array}{l}4{x}^{2}-9{y}^{2}=36 \\ 4{x}^{2}+9{y}^{2}=36\end{array}\)

    Bonisa impendulo

    \((-3,0),(3,0)\)

26. \(\begin{array}{l}{x}^{2}+{y}^{2}=25 \\ {x}^{2}-{y}^{2}=1\end{array}\)

27. \(\begin{array}{l}2{x}^{2}+4{y}^{2}=4 \\ 2{x}^{2}-4{y}^{2}=25x-10\end{array}\)

    Bonisa impendulo

    \((\frac{1}{4},-\frac{\sqrt{62}}{8}),(\frac{1}{4},\frac{\sqrt{62}}{8})\)

28. \(\begin{array}{l}{y}^{2}-{x}^{2}=9 \\ 3{x}^{2}+2{y}^{2}=8\end{array}\)

29. \(\begin{array}{l}{x}^{2}+{y}^{2}+\frac{1}{16}=2500 \\ y=2{x}^{2}\end{array}\)

    Bonisa impendulo

    \((-\frac{\sqrt{398}}{4},\frac{199}{4}),(\frac{\sqrt{398}}{4},\frac{199}{4})\)

30. \(\begin{array}{l}-2{x}^{2}+y=-5 \\ 6x-y=9\end{array}\)

31. \(\begin{array}{l}-{x}^{2}+y=2 \\ -x+y=2\end{array}\)

    Bonisa impendulo

    \((0,2),(1,3)\)

32. \(\begin{array}{l}{x}^{2}+{y}^{2}=1 \\ y=20{x}^{2}-1\end{array}\)

33. \(\begin{array}{l}{x}^{2}+{y}^{2}=1 \\ y=-{x}^{2}\end{array}\)

    Bonisa impendulo

    \((-\sqrt{\frac{1}{2}(\sqrt{5}-1)},\frac{1}{2}(1-\sqrt{5})),(\sqrt{\frac{1}{2}(\sqrt{5}-1)},\frac{1}{2}(1-\sqrt{5}))\)

34. \(\begin{array}{l}2{x}^{3}-{x}^{2}=y \\ y=\frac{1}{2}-x\end{array}\)

35. \(\begin{array}{l}9{x}^{2}+25{y}^{2}=225 \\ {(x-6)}^{2}+{y}^{2}=1\end{array}\)

    Bonisa impendulo

    \((5,0)\)

36. \(\begin{array}{l}{x}^{4}-{x}^{2}=y \\ {x}^{2}+y=0\end{array}\)

37. \(\begin{array}{l}2{x}^{3}-{x}^{2}=y \\ {x}^{2}+y=0\end{array}\)

    Bonisa impendulo

    \((0,0)\)

38. \(\begin{array}{l}{x}^{2}+{y}^{2}=9 \\ y=3-{x}^{2}\end{array}\)

39. \(\begin{array}{l}{x}^{2}-{y}^{2}=9 \\ x=3\end{array}\)

    Bonisa impendulo

    \((3,0)\)

40. \(\begin{array}{l}{x}^{2}-{y}^{2}=9 \\ y=3\end{array}\)

Zama ngokwakho

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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