Q. 59 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $57-64$, use a graphing utility to solve each system of equations. Express the solution(s) rounded to two decimal places.

$\left\{\begin{array}{l}{x}^2+y^3=2\\ x^{3}y=4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is, $\left(-1.65,-0.89\right)$.

Step 1. Given information

$\left\{\begin{array}{l}{x}^2+y^3=2\\ x^{3}y=4\end{array}\right.$

We need to solve each equation for $y$.

Let us consider the first equation,

$$\begin{gathered} x^2+y^3=2 \\ {y}^3=2-x^2 \\ \ln \left(y^3\right)=\ln \left(2-x^2\right) \\ 3 \ln \left(y\right)=\ln \left(2-x^2\right) \\ \ln \left(y\right)=\frac{\ln \left(2-x^2\right)}{3} \\ y=e^{\frac{\ln \left(2-x^2\right)}{3}} \end{gathered}$$

Now let us consider the second equation,

$$\begin{gathered} x^{3}y=4 \\ y=\frac{4}{x^3} \end{gathered}$$

Step 2 Now sketch a graph for the obtained equations.

From the graph, the two equations are intersecting at a point $\left(-1.65,-0.89\right)$.So, the system has only one equation.