Topic: Solving Systems of Linear Equations Algebraically.

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Presentation transcript:

Topic: Solving Systems of Linear Equations Algebraically

Substitution Method Isolate a variable in one of the equations (doesn’t matter which equation or which variable). Substitute the result into the other equation & solve for the variable in question. Use that solution to substitute and solve for the other variable. Elimination (Linear Combination) Method Combine the two equations by adding or subtracting them, creating a single equation in which one variable has been eliminated, making it easy to solve for the other variable. Like terms & “=“ signs must be aligned. Each equation should have the same coefficient in front of one of the variables. Use that solution to substitute and solve for the other variable. Solving Linear Systems Algebraically

The variable y is already isolated in the 2 nd equation. Substitute the right side of that equation for y in the 1 st equation; always protect your substitution with ( ). Now solve this equation for x. Substitute this value into one of the original equations and solve for y; it doesn’t matter which equation, but to keep things simple we’ll use the 2 nd one. The solution to the system is the point (5, -2). { Solving Linear Systems: Substitution

The variable x is already isolated in the 2 nd equation. Substitute the right side of that equation for x in the 1 st equation; remember to protect your substitution with ( ). Now solve this equation for y. Our variable disappeared and left us with a true statement, meaning this equation is an identity and is always true. Therefore, our system is dependent and has infinite solutions. { Solving Linear Systems: Substitution

Each y term has the same coefficient. Since one is positive and one is negative, we can add the two equations together to eliminate y. Now solve this equation for x. Substitute this value into one of the original equations and solve for y. Since y is positive in the 1 st equation, that’s probably the easier one to simplify. Solving Linear Systems: Elimination { The solution to the system is the point (-9, 1).

Neither equation has the same coefficient in front of one of the variables. However, we can simplify the 2 nd equation by a factor of 2, which would give both x terms the same coefficient. Solving Linear Systems: Elimination {

Now we can combine the equations to eliminate x. Since both coefficients are the same sign, we must subtract (WATCH YOUR SIGNS!) That was easy! Now we substitute this into one of the original equations and solve for x. Solving Linear Systems: Elimination { The solution to the system is the point (-4, 7).

How do I know what method to use? Graphing Both equations are already in slope-intercept form, or can easily be rewritten in slope-intercept. Substitution One of the equations already has a variable isolated, or a variable can be easily isolated. Elimination One of the variables has the same coefficient in both equations, or one equation can be easily divided or multiplied to get the same coefficient. REMEMBER: Each method is equally valid. The goal is to select the most efficient method depending on the system given.

JOURNAL ENTRY TITLE: Checking My Understanding: Solving Systems of Linear Equations Algebraically Review your notes from this presentation & create and complete the following subheadings in your journal: “Things I already knew:” Identify any information with which you were already familiar. “New things I learned:” Identify any new information that you now understand. “Questions I still have:” What do you still want to know or do not fully understand?

Homework Quest: Solving Systems of Equations Algebraically Due 1/23 (A-day) or 1/24 (B-day)