Problem 63
Question
Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}-y=0\\\&x^{2}+y^{2}=2\end{aligned}$$
Step-by-Step Solution
Verified Answer
The solutions are (1, 1) and (-1, 1).
1Step 1: Graph the Equation x^2 - y = 0
First, rearrange the equation \(x^2 - y = 0\) to solve for \(y\):\[y = x^2\]. This is a parabola that opens upwards. Plot this graph by finding points such as \((0, 0)\), \((1, 1)\), \((-1, 1)\), \((2, 4)\), and \((-2, 4)\). Connect these points smoothly to complete the parabola.
2Step 2: Graph the Equation x^2 + y^2 = 2
This equation describes a circle centered at the origin with radius \(\sqrt{2}\). To graph this, find and plot intercepts such as \((\sqrt{2}, 0)\), \((0, \sqrt{2})\), \((-\sqrt{2}, 0)\), and \((0, -\sqrt{2})\). Connect these points to form a circle around the origin.
3Step 3: Identify Points of Intersection
Look for points where the parabola from Step 1 intersects with the circle from Step 2. These points are visually where the graphs overlap.
4Step 4: Check Solutions Algebraically
To verify the intersection points, solve the system algebraically. Substitute \(y = x^2\) into \(x^2 + y^2 = 2\) to get \(x^2 + (x^2)^2 = 2\). Simplify to \(x^2 + x^4 = 2\), then \(x^4 + x^2 - 2 = 0\). Factor to find \((x^2 - 1)(x^2 + 2) = 0\). Solving gives \(x^2 = 1\) (since \(x^2 + 2 = 0\) has no real solutions), so \(x = \pm 1\). Substitute back to find \(y\) as \(1\). The solutions are \((1, 1)\) and \((-1, 1)\).
5Step 5: Verify Graphically
Check that the points \((1, 1)\) and \((-1, 1)\) are indeed where the graphs intersect on your sketch. Both points satisfy \(y = x^2\) and lie on the circle \(x^2 + y^2 = 2\), confirming their correctness.
Key Concepts
System of EquationsParabolaCircle
System of Equations
A system of equations consists of two or more equations with the same set of unknowns. Solving them means finding the values of the unknowns that satisfy all the equations simultaneously. In this exercise, we look at graphical solutions where two graphs representing the equations are plotted on the same set of axes and intersections reveal possible solutions.
Notes: It's essential to ensure the accuracy of the plotted graphs for correct intersection points.
- Equation 1: \(x^2 - y = 0\), simplifies to \(y = x^2\)
- Equation 2: \(x^2 + y^2 = 2\)
Notes: It's essential to ensure the accuracy of the plotted graphs for correct intersection points.
Parabola
A parabola is a U-shaped curve that is symmetrical. It is the graph of a quadratic equation, such as \(y = x^2\). In this vertex form, the parabola opens upwards, and its vertex is at the origin.
Key points defining a parabola help in plotting it:
Key points defining a parabola help in plotting it:
- The vertex, which is the highest or lowest point depending on direction. Here it's \((0, 0)\).
- Axis of symmetry, a vertical line that splits the parabola into two mirrored halves.
- Standard points, like \((1, 1)\) and \((-1, 1)\), which help in shaping the curve.
Circle
The circle in this exercise is described by the equation \(x^2 + y^2 = 2\). This represents a perfect circle centered at the origin with a radius of \(\sqrt{2}\).
Some features of a circle include:
Some features of a circle include:
- The center point, which coordinates both axes equally. Here, it’s \((0, 0)\).
- Radius, the constant distance from the center to any point on the circle. Here it's \(\sqrt{2}\).
- Symmetry, as a circle maintains consistency in all directions.
Other exercises in this chapter
Problem 62
Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{array}{r}1.5 x+3 y=5 \\\2 x+4 y=3\end{arr
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Graph the solution set of each system of inequalities by hand. Concept Check \(\quad\) Which one of the choices that follow is a description of the solution set
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Given a square matrix \(A^{-1}\), find matrix \(A\). $$A^{-1}=\left[\begin{array}{rrr} \frac{2}{3} & -\frac{1}{3} & 0 \\ \frac{1}{3} & -\frac{5}{3} & 1 \\ \frac
View solution Problem 63
Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} x-2 y+z &=5 \\ -2 x+4 y-2 z &=2 \\ 2 x+y-z &=2 \end{aligned}$$
View solution