Systems of Equations: E.g. 2x + 3y = 1 x − 3y = 5 To solve this system of two equations with two unknowns what we are looking for are values of (x, y) that would solve both equations. For example, the 1 ordered pair ( , 0) solves the first equation, but does not solve the second one, 2 so it is not a solution to the system. The ordered pair (2, −1) solves both the top and bottom equation, so it is a solution. Geometrically, a solution to an equation is a point on the graph of the equation, so a solution to the first equation is a point on the graph of the first equation, and a solution to the second equation is a point on the second equation. Therefore, a solution to the system of equation is a point of intersection of the two graphs. To solve systems of equations, geometrically, is to find point(s) of intersection of the graphs. If two lines intersect at one point the two equations has one solution. We have an independent system. If two lines are parallel we have no solution. The system of equations is inconsistent. If two lines are the same line we have infinitely many solutions. The system is dependent. Algebraically, we may use the method of substitution or the method of addition (sometimes called the method of elimination) to solve the system. Method of Substitution: 2x + 3y = 1 x − 3y = 5 Solve for one of the variable in terms of the other using one of the equations, then substitute for that variable in the other equation. Using the second equation: x = 3y + 5 Substitute the expression for x (in terms of y) in the first equation: 2(3y + 5) + 3y = 1 6y + 10 + 3y = 1 9y + 10 = 1 9y = −9 y = −1 Once we found the value of y, we may substitute it into any of the two equations to solve for x: 2x + 3(−1) = 1 2x − 3 = 1 2x = 4 x=2 The solution is (2, −1), notice that the solution is an ordered pair. Method of Addition: Example: Solve the system of equation: 4x − 3y = 6 3x − 2y = 5 We add the two equations with one of the variables having coefficients of the same magnitude but opposite sign (for example, one maybe 4x and the other −4x) so we may eliminate one of the variables. If the two equations do not have any variables with coefficients of opposite sign, we multiply by the LCM of the coefficient of one of the variables. Multiply first equation by −3, multiply second equation by 4: −3[4x − 3y = 6] 4[3x − 2y = 5] We get a new system of equations: −12x + 9y = −18 12x − 8y = 20 We now add the two equations and this eliminates the x variable: y=2 Knowing the value of y we may once again substitute the value into one of the two equations to solve for the value of x: 4x − 3(2) = 6 4x − 6 = 6 4x = 12 x=3 The solution is (3, 2)