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The substitution method is a useful technique when solving systems of equations where one equation can be solved for one variable in terms of another. Here's how it works in this case: You start with two equations. The second equation is already solved for one of the variables, which makes this method ideal.
For example, if you have two equations:

• \( x^2 + 4xy - 2y^2 - 6 = 0 \)
• \( x = y + 2 \)

You can take the expression for \( x \) from the second equation \((x = y + 2)\) and substitute it into the first equation. This transformation allows you to express everything in terms of a single variable \((y)\). Then, you only need to solve a simpler equation, effectively reducing the problem's complexity.

The quadratic formula is a powerful tool for finding the roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \).
Once you've applied substitution and reached a quadratic equation like \( 3y^2 + 12y - 2 = 0 \), you can use the quadratic formula to find the solutions for \( y \). The formula is: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a, b, \) and \( c \) are the coefficients from your quadratic equation. Substitute these values into the formula to solve for \( y \).
The discriminant, \( b^2 - 4ac \), determines the nature of the roots. If it's positive, as in this example where it is 168, you get two real solutions. Calculate these solutions to proceed with solving the system of equations.

Once an expression like \((y+2)^2\) appears in a substitution-reformed equation, it's time for expansion and simplification. This involves squaring \( (y + 2) \), resulting in:\[ (y+2)^2 = y^2 + 4y + 4 \]Substitute this expansion result back into your equation in place of \((y+2)^2\). Then perform algebraic manipulations such as combining like terms to further simplify.
With terms combined, as in \( y^2 + 4y + 4 + 4y^2 + 8y - 2y^2 - 6 = 0 \), you'll end up with a more straightforward quadratic equation now ready for solving. This simplification reduces complexity and often makes it easier to apply further methods like the quadratic formula.

Solving quadratic equations is a key skill in algebra, and often involves a systematic approach. After substitution, expansion, and simplification, you find yourself with a quadratic equation like \( 3y^2 + 12y - 2 = 0 \).
Here, we've already set the stage to apply the quadratic formula. But solving doesn't stop with the formula; simplifying the results is crucial.
Once you have the potential solutions \( y = -2 \pm \frac{\sqrt{42}}{3} \), it's important to substitute these back into the reformation equation \( x = y + 2 \) to find corresponding \( x \) values. Doing this will give you the complete set of solutions for the system.
Finally, check both solutions back in the original equations to ensure they satisfy the system, confirming their validity.