Classifying Systems, Solving Systems by Graphing and Substitution

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Presentation transcript:

Classifying Systems, Solving Systems by Graphing and Substitution Systems of Equations Classifying Systems, Solving Systems by Graphing and Substitution

Basics of Systems of Linear Equations A system of equations is a collection or set of 2 or more equations in the same variables The solution to a system is the point at which the lines intersect. There are 3 possibilities for systems in 2 variables: intersecting lines (1 point), coinciding lines (same line, infinite points), and parallel lines (no points, no solution)

Classifying Systems of Equations Classified Two Ways: Has at least one solution – CONSISTENT (intersecting and coinciding) Exactly 1 solution – INDEPENDENT (intersecting) Infinite solutions – DEPENDENT (coinciding) No solution – INCONSISTENT (parallel)

Three Methods of Solving Systems Graphing Method Substitution Method Elimination Method

Solving by Graphing Steps: Put both equations in slope-intercept form. Graph one line. Graph second line, extending the line accurately until the lines intersect. Identify the point of intersection (x, y), that is the solution.

Examples: y = 2x – 1 6x – y = 13 3x – y = 2 -3x + y = 1

Solving Systems by Substitution This method is algebraic; meaning it is done without graphing the lines You will be manipulating the equations and solving for x and y

Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y= ; x= ; a=) 2. Substitute the expression from step one into the other equation. 3. Simplify and solve the equation. 4. Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of the system.

Example #1: y = 4x 3x + y = -21 Step 1: Solve one equation for one variable. y = 4x (This equation is already solved for y.) Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3

y = 4x 3x + y = -21 Step 4: Substitute back into either original equation to find the value of the other variable. 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12).

y = 4x 3x + y = -21 3x + y = -21 y = 4x 3(-3) + (-12) = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). 3x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21 -21= -21 y = 4x -12 = 4(-3) -12 = -12

Example #2: x + y = 10 y = -x +10 5x - y = 2 5x -(-x +10) = 2 Step 1: Solve one equation for one variable. x + y = 10 y = -x +10 Step 2: Substitute the expression from step one into the other equation. 5x - y = 2 5x -(-x +10) = 2

5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x = 2 x + y = 10 Step 3: Simplify and solve the equation. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x = 2

Solution to the system is (2,8). x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8).

5x – y = 2 x + y =10 5(2) - (8) = 2 2 + 8 =10 10 – 8 = 2 10 =10 2 = 2 Step 5: Check the solution in both equations. Solution to the system is (2, 8). 5x – y = 2 5(2) - (8) = 2 10 – 8 = 2 2 = 2 x + y =10 2 + 8 =10 10 =10

Solve by substitution: 1. 2.