Problem 54 - Precalculus with Limits

Question

Choosing a Solution Method In Exercises \(49-56\) , solve the system graphically or algebraically. Explain your choice of method. $$\left\\{\begin{array}{l}{x^{2}+y=4} \\ {e^{x}-y=0}\end{array}\right.$$

Step-by-Step Solution

Answer

Based on the graphical solution, the coordinates of the points of intersection represent the solution to the system. These coordinates (x, y) need to be determined specifically by calculating or estimating based on the graph.

Step 1: Create the Graph

Plot the two equations: $x^{2}+y=4$ and $e^{x}-y=0$. The first equation forms a parabola after adjusting y = 4 - x^2, and the second equation forms an exponential curve with y = e^x.

Step 2: Identify the Points of Intersection

The points where the two curves intersect provide the solutions for the system of equations.

Step 3: Obtain the Solutions

Identify the coordinates of the points of intersection, which will give the values of x and y that satisfy both equations.

Key Concepts

Graphical Solution Method

The graphical method of solving systems of equations involves visually representing each equation on a graph and finding their points of intersection. With the equations \(x^2 + y = 4\) and \(e^x - y = 0\), you begin by plotting both on a coordinate plane. The first equation, rearranged as \(y = 4 - x^2\), creates a parabola that opens downward, and the second, rearranged as \(y = e^x\), creates an exponential curve.

Through the graphing process, you visually inspect where these two graphs meet, which corresponds to the solution of the system. This method is particularly useful when dealing with nonlinear equations, as it can provide a quick estimate of the solutions without extensive algebraic manipulation. However, it's worth considering the precision of graphical solutions; they are generally less accurate than algebraic methods, especially when drawn by hand.

Remember, for precise solutions, graphing technology or software should be used. This allows for a clearer and more accurate representation, helping you identify intersection points with greater precision.

Algebraic Solution Method

The algebraic solution method for systems of equations relies on manipulating the equations using algebra to find an exact solution. Unlike the graphical method, algebraic methods such as substitution, elimination, or using matrices, provide precise answers. However, when dealing with non-linear systems like \(x^2 + y = 4\) and \(e^x - y = 0\), the process can become more complex.

For example, you could solve for \(y\) in one of the equations and substitute it into the other, leading to an equation with one variable that you can solve. If you use this approach for our system, solving for \(y\) in the second equation \(y = e^x\) and substituting it into the first \(x^2 + e^x = 4\), you would need to apply more advanced methods such as iterative numerical methods or look for graphical intersections, as algebraic manipulation may not yield a straightforward solution.

This method is the most appropriate when systems contain equations that are linear or can be easily manipulated to be linear, as well as when the highest degree of precision is required.

Systems of Non-Linear Equations

A system of non-linear equations consists of at least one equation that isn't first order. In the case of \(x^2 + y = 4\) and \(e^x - y = 0\), we have a parabola and an exponential function, respectively. Solving such systems algebraically can be tricky because traditional methods like elimination and substitution might not always work smoothly. That’s where the graphical method shines, as it can provide a visual approximation of the solution.

Sometimes, these systems may require specific methods for a solution, such as using the quadratic formula for quadratic equations or applying logarithms for exponential equations. Systems of non-linear equations can often have multiple solutions or no solutions at all, which contrasts with linear systems that have a single solution, infinitely many solutions, or no solution. Identifying the nature of the equations in the system is critical in choosing the most efficient method to solve them.