 # Systems of Linear Equations

## Presentation on theme: "Systems of Linear Equations"— Presentation transcript:

Systems of Linear Equations
Chapter 5

5-1 Systems of Linear Equations in Two Variables
Deciding whether an ordered pair is a solution of a linear system. The solution set of a linear system of equations contains all ordered pairs that satisfy all the equations at the same time. Example 1: Is the ordered pair a solution of the given system? 2x + y = -6 Substitute the ordered pair into each equation. x + 3y = 2 Both equations must be satisfied. A) (-4, 2) B) (3, -12) 2(-4) + 2 = (3) + (-12) = -6 (-4) + 3(2) = (3) + 3(-12) = 2 -6 = = =  -6  Yes  No

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing. One way to find the solution set of a linear system of equations is to graph each equation and find the point where the graphs intersect. Example 1: Solve the system of equations by graphing. A) x + y = 5 B) 2x + y = -5 2x - y = x + 3y = 6 Solution: {(3,2)} Solution: {(-3,1)}

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing. There are three possible solutions to a system of linear equations in two variables that have been graphed: 1) The two graphs intersect at a single point. The coordinates give the solution of the system. In this case, the solution is “consistent” and the equations are “independent”. 2) The graphs are parallel lines. (Slopes are equal) In this case the system is “inconsistent” and the solution set is 0 or null. 3) The graphs are the same line. (Slopes and y-intercepts are the same) In this case, the equations are “dependent” and the solution set is an infinite set of ordered pairs.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Elimination. Remember: If a=b and c=d, then a + c = b + d. Step 1: Write both equations in standard form Step 2: Make the coefficients of one pair of variable terms opposite (Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either x or y will be zero.) Step 3: Add the new equations to eliminate a variable Step 4: Solve the equation formed in step 3 Step 5: Substitute the result of Step 4 into either of the original equations and solve for the other value. Step 6: Check the solution and write the solution set.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Elimination. Example 2: Solve the system : 2x + 3y = 19 Step 1: Both equations are in standard form 3x - 7y = -6 Step 2: Choose the variable x to eliminate: Multiply the top equation by 3, the bottom equation by -2 3[2x + 3y = 19] x + 9y = 57 -2[3x - 7y = -6] x +14y = 12 Step 3: Add the new equations to eliminate a variable 0x + 23y = 69 Step 4: Solve the equation formed in step y = 3 Step 5: Substitute the result of Step 4 into either of the original equations and solve for the other value. 2x + 3(3) = 19 2x = 10 x = 5 Solution Set: {(5,3)} Step 6: Check the solution and write the solution set.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Elimination. Example 3: Solve the system : 2[2x - 3y = 1] x - 6y = 2 -3[3x - 2y = 9] x + 6y = -27 -5x + 0y = -25 x = 5 3(5) - 2y = 9 -2y = -6 Solution Set: {(5,3)} y = 3

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Elimination. Example 4: Solve the system : x + y = 6 -8x - 4y = -24 4[2x + y = 6] x + 4y = 24 -8x -4y = x - 4y = -24 0 = 0 True Solution Set: {(x,y)| 2x + y = 6} Note: When a system has dependent equations and an infinite number of solutions, either equation can be used to produce the solution set. Answer is given in set-builder notation.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Elimination. Example 5: Solve the system : 4x - 3y = 8 8x - 6y = 14 -2[4x - 3y = 8] x + 6y = -16 8x - 6y = x - 6y = 24 0 = 8 False Solution Set: no solution or null Note: There are no ordered pairs that satisfy both equations. The lines are parallel. There is no solution.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Substitution. Step 1: Solve one of the equations for either variable Step 2: Substitute for that variable in the other equation (The result should be an equation with just one variable) Step 3: Solve the equation from step 2 Step 4: Substitute the result of Step 3 into either of the original equations and solve for the other value. Step 6: Check the solution and write the solution set.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Substitution. Example 6: Solve the system : 4x + y = 5 2x - 3y =13 Step 1: Choose the variable y to solve for in the top equation: y = -4x + 5 Step 2: Substitute this variable into the bottom equation 2x - 3(-4x + 5) = x + 12x - 15 = 13 Step 3: Solve the equation formed in step 2 14x = 28 x = 2 Step 4: Substitute the result of Step 3 into either of the original equations and solve for the other value. 4(2) + y = 5 y = -3 Solution Set: {(2,-3)} Step 5: Check the solution and write the solution set.

4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by Method of Substitution. Example 7: Solve the system : y = -2x + 2 -2x + 5(-2x + 2) = x - 10x + 10 = 22 -12x = 12 x = -1 2(-1) + y = 2 y = 4 Solution Set: {(-1,4)}