Chapter 10.8, Problem 20E

To determine

To find: the solutions of the system of equations

Answer to Problem 20E

The two solutions are $(0,0)$ and $(1,1)$

Explanation of Solution

Given:

  $x-y^2=0$

  $y-x^2=0$

Calculation:

Consider the system of equations

  $\begin{array}{l}x-y^2=0 Equation1\\ y-x^2=0 Equation\text{ 2}\end{array}$

Need to find all the solutions to the system of equations using substitution method.

Solve equation $1$ for $x$

  $x-y^2=0$

  $x=y^2$ Adding $y^2$ to both sides

Substitute:

Now substitute $x=y^2$ in $x-y^2=0$ and solve for $y.$

  $y-{(y^2)}^2=0$ substitute $x=y^2$ into equation $\text{2}$

  $y-y^4=0$ Simplify

  $y(1-y^3)=0$ Taking out the common factor $y$

  $y=0$ or $1-y^3=0$

  $y=0$ or $y=1,$

  $-i,i$ Taking square roots

Discard the imaginary roots.

Therefore, $y=0,1$

Back-substitute:

Now substitute the values of $y$ into the equation $x=y^2$

For $y=0$ : $x={(0)}^2=0$ Back substitution

For $y=1$ : $x={(1)}^2=1$ Back substitution

Thus, the two solutions are $\boxed{(0,0)}$ and $\boxed{(1,1)}$ .

Conclusion: The two solutions are $(0,0)$ and $(1,1)$