Linear Equations

A linear equation is one in which the variables are raised to the power 1. Examples x = x o + vt 5x – 7y +3z = 15

The equation of a straight line is given as y = mx + b Where m is the slope and b is the intercept on the y axis. Y b m and m = dy/dx X

The earned run average (E) of a pitcher is given by the formula Ei = 9r where i is the number of innings pitched and r is the number of earned runs. Solve the equation for E Calculate the earned run average of a pitcher who allowed 8 earned runs in 24 innings.

E = 9(r/i) E = 9(8/24) = 9(1/3) = 3.0

Solve the linear equation m/n = p – 8 for n.

You can do anything to an equation as long as you do the same thing to both sides of the equation! m/n = p – 8 n/m = 1/(p - 8) n = m/(p – 8)

Fossett set a 24 hour hot air balloon record of miles on 1 July Dustin Humphrey is trying to break that record. He has already traveled 1300 miles in 12 hours. What speed must he average in the next 12 hours to tie Fossett’s record?

Distance = speed x time v = v = – v = v = /12 = miles per hour

A system of linear equations is a collection of linear equations using the same set of variables. A linear equation is an equation where each variable is raised to the 1th power, x, y, and z. For example, 3x + 2y – z = 1 2x - 2y + 4z = -2 - x + ½ y – z = 0

Solving these three simultaneous linear equations yields x = 1 y = -2 z = -2 Which satisfies all three equations.

Elementary Example Consider the two simultaneous linear equations 2x + 3y = 6(1) 4x + 9y = 15(2) One method for solving such a system is to first solve equation (1) for x in terms of y: x = 3 - (3/2)y And then substitute that expression for x into equation (2)

4(3 – (3/2)y) + 9y = – 6y + 9y = 15 3y = 3 y = 1 And then solve for x x = 3 – (3/2)y = 3 – (3/2)(1) x = 3/2

Another method is to solve both equations (1) and (2) for x in terms of y and then set them equal to one another, i.e. x = 3 – (3/2)y x = (15/4) – (9/4)y 3 – (3/2)y = (15/4) – (9/4)y And solve for y and then substitute and solve for x.

Graphing Solve both equation (1) and equation (2) for y as a function of x. y = 2 – (2/3)x from eqn (1) y = (15/9) – (4/9)x from eqn (2) Using Excel, we can put in a range of values of x and generate solutions for y.

The solution to the system of simultaneous linear equations is given by the point on the graph where the two curves intersect, i.e. y = 1 X = 3/2

If the graph of equation (1) intersects the graph of equation (2), then there is one solution. Y X

If the graph of equation (1) is parallel to the graph of equation (2), there is no solution. Example Y y = x + 1 (1) y = x + 2 (2) (2) X (1)

If the graph of equation (1) coincides with the graph of equation (2), i.e. the two equations are the same, then there is an infinite number of solutions. Y (1), (2) Example: y = 2x + 1 (1) X 2y = 4x + 2 (2)

Independence The equations of a linear system are independent if none of the equations can be derived algebraically from the others. When the equations are independent, each equation contains new information about the variables.

For example, the equations 3x + 2y = 6 6x + 4y = 12 Are not independent – they are the same equation when scaled by a factor of 2. They would produce identical graphs!

The equations x – 2y = -1 3x + 5y = 8 4x + 3y = 7 are not independent. The third equation is the sum of the first two equations.

Consistency The equations of a linear system are consistent if they possess a common solution. They are inconsistent if a contradiction is derived from the equations, e.g. 1 = 3. Consider the equations 3x + 2y = 6 3x + 2y = 12 Which means that 6 = 12!!!!

Elimination of Variables The simplest method of solving a system of linear equations is to repeatedly eliminate variables. In the first equation, solve for one of the variables in terms of the others. Substitute this expression into the remaining equations. This yields a system with one fewer equation and one fewer unknowns. Continue until you have reduced the system to a single linear equation and solve it!

Example Consider the following system: x + 3y – 2z = 5 3x + 5y + 6z = 7 2x + 4y + 3z = 8 Solving the first equation for x yields X = 5 – 3y + 2z

Substituting that solution for x into the other two equations yields -4y + 12z = -8 -2y + 7z = -2 Which gives us two equations and two unknowns.

Solving the first of the two equations for y yields y = 2 + 3z Substituting this into the second equation yields z = 2 y = 2 + (3)(2) = 8 x = 5 + (2)(2) – (3)(8) = -15