Systems of Linear Equations

Systems of Linear Equatiosn A system of linear equations is two or more linear equations that are being solved simultaneously. In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true.

Solutions of a System The solution of a system is where two graphs intersect or what they have in common. If an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system. A consistent system is a system that has at least one solution. An inconsistent system is a system that has no solution.

Solution of a System The equation of a system is dependent if ALL the solutions of one equation are also solutions of the other equation. In other words, they end up being the same line. The equations of a system are independent if they do not share ALL solutions. In other words, they have one point in common.

Solution of a System There are three possible outcomes: One solution No solution Infinitely many solutions

One Solution If the system in two variables has one solution, it is an ordered pair that is a solution to BOTH equations. The graph here illustrates a system of two equations and two unknowns that has one solution:

No Solution If the two lines are parallel to each other, they will never intersect. This means they do not have any points in common. In this system, you would have no solution. This graph illustrates a system of two equations and two unknowns that has no solutions.

Infinite Solutions If two lines end up lying on top of each other, then there is an infinite number of solutions. They end up being the same line, so any solution that would work in one equation is going to work in the other. This graph illustrates a system of two equations and two unknowns that has an infinite number of solutions.

Practice Determine whether each ordered pair is a solution of the system.           (3, -1) and (0, 2)

Solve by Graphing

Solve by Graphing

Solve by Graphing

Substitution Method To solve a system using substitution, solve one equation for either x or y. Then substitute this equation into the other equation for the variable you solved for in the first.

Practice 3y - 2x = 11  y + 2x = 9

Practice −5x + y = −3 3x − 8y = 24

Practice −2x + 6y = 6 −7x + 8y = −5

Elimination Method First, be sure that the variables are "lined up" under one another. In this problem, they are already "lined up“ x - 2y = 14 x + 3y = 9 Decide which variable ("x" or "y") will be easier to eliminate. In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another. Looks like "x" is the easier variable to eliminate in this problem since the x's already have the same coefficients.

Elimination Method Now, in this problem we need to subtract to eliminate the "x" variable. Subtract ALL of the sets of lined up terms. x - 2y = 14 -x - 3y = - 9 - 5y = 5 Solve this simple equation. -5y = 5 y = -1

Elimination Method Substitute "y = -1" into either of the ORIGINAL equations to get the value for "x". x - 2y = 14 x - 2(-1) = 14 x + 2 = 14 x = 12 Check: substitute x = 12 and y = -1 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! x - 2y = 14 x + 3y = 9 12 + 3(-1) = 9 12 - 2(-1) = 14 12 - 3 = 9 12 + 2 = 14 9 = 9 (check!) 14 = 14 (check!)

More Challenging You can probably see the dilemma with this problem right away. Neither of the variables have the same (or negative) coefficients to eliminate. 4x + 3y = -1 5x + 4y = 1

Challenge! In this type of situation, we must MAKE the coefficients the same (or negatives) by multiplication. You can MAKE either the "x" or the "y" coefficients the same. In this problem, the "y" variables will be changed to the same coefficient by multiplying the top equation by 4 and the bottom equation by 3. Remember: * you can multiply the two differing coefficients to obtain the new coefficient if you cannot think of another smaller value that will work. * multiply EVERY element in each equation by your adjustment numbers. 4(4x + 3y = -1) 16x + 12y = -4 3(5x + 4y = 1) 15x + 12y = 3

No longer Challenging! Now, in this problem we need to subtract to eliminate the "y" variable. 16x + 12y = -4 -15x - 12y = - 3 x = - 7 Substitute "x = -7" into either of the ORIGINAL equations to get the value for "y". 5x + 4y = 1 5(-7) + 4y = 1 -35 + 4y = 1 4y = 36 y = 9

Elimination Method Check:  substitute x = -7 and y = 9 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE! 4x + 3y = -1 4(-7) +3(9) = -1   -28 + 27 = -1   -1 = -1  (check!) 5x + 4y = 1 5(-7) + 4(9) = 1   -35 + 36 = 1    1 = 1  (check!)

Practice 4x + 8y = 20 −4x + 2y = −30

Practice 8x + y = −16 −3x + y = −5

Practice −3x + 7y = −16 −9x + 5y = 16

Practice 5x + 4y = −30 3x − 9y = −18