91Ó°ÊÓ

The substitution method is useful for simplifying complex equations by replacing a part of the equation with a single variable.
In our example, we start with the quadratic equation: $$ (3y + 2)^2 + (3y + 2) - 6 = 0 $$
We set up a substitution where: $$u = 3y + 2$$
Substituting this into the original equation replaces all instances of $$3y + 2$$ with $$u$$, leading to a simpler form: $$ u^2 + u - 6 = 0 $$
This step-by-step approach of using substitution can transform a more complicated equation into one that is easier to solve.

Factoring quadratic equations is a method used to find the roots, or solutions, of the quadratic equation.
In our simplified equation from the substitution method, we have: $$ u^2 + u - 6 = 0 $$
To factor this, we look for two numbers that multiply to -6 and add up to 1 (the coefficient of $$u$$).
These numbers are 3 and -2. So we rewrite our equation as: $$ (u + 3)(u - 2) = 0 $$
Factoring allows us to break the quadratic equation into simpler binomial expressions.

Algebraic manipulation is the process of rearranging and simplifying equations to isolate variables and find solutions.
Once the quadratic equation is factored, we must solve for $$u$$. Setting each factor to zero gives us: $$ u + 3 = 0 $$ and $$ u - 2 = 0 $$ Solving these, we find: $$ u = -3 $$ and $$ u = 2 $$
Next, we substitute back to the original variable $$y$$ using the initial substitution: $$ u = 3y + 2 $$
With $$u = -3$$: $$ 3y + 2 = -3 $$
solving for $$y$$ gives $$ y = \frac{-5}{3} $$
With $$u = 2$$: $$ 3y + 2 = 2 $$
solving for $$y$$ gives $$ y = 0 $$
Finally, correctly manipulating and substituting back ensures we have the solutions: $$ y = \frac{-5}{3} $$ and $$ y = 0 $$.