How do you solve quadratic and linear simultaneous equations?

For two linear simultaneous equations, two algebraic methods can be used to find solutions.The first is elimination, which is usually a lot quicker. Elimination involves subtracting one equation form the other to eliminate one variable. For example, given equations:

(1) x+y=4

(2) 2x+3y=5

Currently, no variables can be immediately eliminated, as subtracting y from 3y will give a remainder of 2y, and subtracting x form 2x will give a remainder of x. To overcome this, equation (1) can be rewritten as:

(1) 2x+2y=8

(multiplying both sides of the equation by 2 keeps the equation true)

Our new pair of equations is 

(1) 2x+2y=8
(2) 2x+3y=5

Subtracting equation (1) from (2) we obtain:

(2x+3y)-(2x+2y)=5-8

Therefore

y=-3

as we know x+y=4 we can substitute in our y value to obtain the equation

x-3=4, giving x=7

The alternate method is substitution, the involves making one variable the subject of one equation, then substitution the expression for that value into the other equation, for example.

(1) x+y=4 

(2) 2x+3y=5
making x the subject of equation (1) x=(4-y)

This expression then can be substituted for x in equation (2) giving

2(4-y)+3y=5, which simplifies to:

y=-3, which can again be substituted into either equation to find x=7

For non-linear equations, where both variables have a different power from what they have in the other equation, substitution is the only viable algebraic method.

For instance:

(1) x²+y²=9

(2) x+y=2

Elimination does not work here, as an x or y cannot be eliminated from an x² or y² by multiplying by a constance factor. (no Ax²-Bx=0 for all x)

Therefore substitution must be used, it is easier to use the linear equation for substitution, so I will make y the subject of (2) giving

y=2-x

Substituting this expression for y into equation (1) gives:

x²+(2-x)²=9
expanding:
x²+4-4x+x²=9
collecting terms
2x²-4x-5=0
simplifying:

x²-2x-2.5=0

using the quadratic formula this gives us the solution

x=1±√(14)/2 and substituting back into (2) gives us solution y=1∓√(14)/2