How do I solve simultaneous equations by substitution?

Often, substitution is used when dealing with quadratic simultaneous equations as the other method, elimination, is more straightforward and can be used to solve linear equations but it cannot be used for quadratic equations. Here is an example of a set of quadratic simultaneous equations:  x² + y² = 4    y = 3x -2 . In order to solve simultaneous equations by substitution, you substitute one of the squared terms in the quadratic equation using the linear equation.  So for this example, we would substitue y² for (3x -2)² as the linear equation tells us y = 3x-2 .  Our equation is now x² + (3x -2)² = 4 . Now we expand to solve for our variable (x in this case) as normal x² + 9x² - 12x + 4 = 4 becomes 10x² - 12x = 0  becomes 5x - 6 = 0  becomes 5x = 6 means x = 6/5 . Then once we have our first variable, in this case x, we put that back into the linear equation to find the other variable, in this case y. So  y = 3x -2 becomes y = 3(6/5) -2 means y =  8/5 . And there we have our solution!