How to solve Simultaneous Linear Equations, e.g. (4x + 5y = 17) and (3x + 2y = 4)

Explain how linear simultaneous equations with two variables ("unknowns") differ from single linear equations with two variables. Explain that a single linear equations have multiple, and indeed infinite, solutions. Take for example the equation (3x + 4y = 12) has solutions (x=0, y=3); (x=1, y=2.25); (x=2, y=1.5) etc. and present this graphically noting each of these solutions once line is plotted. Simultaneous linear equations, and "solving equations simultaneously", can have only one finite solution, and use graphical example mentioned to prove this by adding a second line and noting that there is only one crossing point, where both equations are satisfied by the same values of x and y (perhaps explain no solutions outcome if lines are parallel but likely save for another time/extension). These values at the crossing point are where the equations are solved simultaneously. Explain that there are two methods by which we can solve simultaneous equations, through "elimination" or through "substitution". This example will use the elimination method.
Explain that using the elimination method requires us to manipulate (multiply/divide) the initial equations to equate the coefficients (the number 'in front' of the x or y) of one of the variables ("make the Xs or Ys the same in both"). In this example we can multiply the first equation (4x + 5y = 17) by 3 to give us (12x + 15y = 51) and our second equation (3x + 2y = 4) by 4 to give us (12x + 8y = 16). At point explain we can do this by multiplying each term in each equation, essentially treating the equation as a whole, and that each equation still holds true, as we have done the same to both sides. Note to the student that we now have the same coefficient of x in each equation (12x). We can therefore take our second manipulated equation (12x + 8y = 16) from our first one (12x + 15y = 51), to give us (7y = 35) (write the equations and and below each other respectively with terms lined up and subtract vertically). Note that we have "eliminated" the x term as (12x - 12x =0), and can solve (7y =35) to give us (y=5). We can now substitute this value into our first initial equation (4x + 5y = 17) to find a value for x (4x + 5(5) =17). Expand and solve this to give (x= -2). We can then check that this solution is correct, and satisfies both equations simultaneously by substituting these values into our second initial equation (3x + 2y = 4) and prove that this equation still holds true with these values (3(-2) + 2(5) = 4). This could be reinforced by plotting both lines and looking for their intercept.