To solve the system of equations using the substitution method, we need to solve one equation for a particular variable and substitute that into the other equation. Let's go step by step.

The system of equations is:

$1. \ 2x^3 + 3y^2 = 5z$ $2. \ 6x^3 + 7y^2 = 8z$

Step 1: Solve the first equation for $z$

From the first equation:

$2x^3 + 3y^2 = 5z$

Solve for $z$:

$z = \frac{2x^3 + 3y^2}{5}$

Step 2: Substitute this expression for $z$ in the second equation

Substitute $z = \frac{2x^3 + 3y^2}{5}$ into the second equation $6x^3 + 7y^2 = 8z$:

$6x^3 + 7y^2 = 8 \times \frac{2x^3 + 3y^2}{5}$

Now simplify the right-hand side:

$6x^3 + 7y^2 = \frac{8(2x^3 + 3y^2)}{5}$ $6x^3 + 7y^2 = \frac{16x^3 + 24y^2}{5}$

Step 3: Eliminate the denominator

To eliminate the fraction, multiply both sides of the equation by 5:

$5(6x^3 + 7y^2) = 16x^3 + 24y^2$ $30x^3 + 35y^2 = 16x^3 + 24y^2$

Step 4: Simplify the equation

Now, bring like terms together:

$30x^3 - 16x^3 = 24y^2 - 35y^2$ $14x^3 = -11y^2$

Step 5: Express the relationship between $x^3$ and $y^2$

Now, solve for $x^3$ in terms of $y^2$:

$x^3 = \frac{-11y^2}{14}$

Step 6: Substitute $x^3 = \frac{-11y^2}{14}$ into the equation for $z$

Now, substitute this into the expression for $z = \frac{2x^3 + 3y^2}{5}$:

$z = \frac{2\left(\frac{-11y^2}{14}\right) + 3y^2}{5}$

Simplify:

$z = \frac{\frac{-22y^2}{14} + 3y^2}{5}$

Convert $\frac{-22y^2}{14}$ to simplify:

$z = \frac{-\frac{11y^2}{7} + 3y^2}{5}$

Now combine the terms:

$z = \frac{-\frac{11y^2}{7} + \frac{21y^2}{7}}{5}$ $z = \frac{\frac{10y^2}{7}}{5}$

Finally:

$z = \frac{2y^2}{7}$

Final Result

Thus, the relationship between $x$, $y$, and $z$ is:

$x^3 = \frac{-11y^2}{14}, \quad z = \frac{2y^2}{7}$

Would you like further details or have any questions?

Expanding Further:

1. How do you solve nonlinear equations using the elimination method?
2. What if we solve this system by graphing instead of substitution?
3. What would be the result if $x = 1$ or $y = 1$?
4. How would you approach this problem with a different substitution variable?
5. What is the significance of having a negative term in these equations?

Tip: When solving by substitution, always ensure that your final expressions are fully simplified before concluding the solution.