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Algebra/Geometry Blend Name Unit #4: Systems of Linear Equations & Inequalities Period Lesson 3: Solving Systems of Linear Equations Algebraically Date
[DAY #1]
Earlier we talked about how to find the solution to a system of equations by graphing each equation. That is often time a LOT more work and not always the best choice. Who walks around with graph paper or a graphing calculator?
You can also solve equations algebraically. In fact, there are 2 different methods for solving equations algebraically. Today we are going to learn and explore how to solve systems of equations using substitution.
Substitution
To substitute is to a variable with something of equal value.
Examples : Find the solution to the following system of equations algebraically
#1. y = 2x Steps 6x – y = 8 1) Substitute….. plug in the equation that has the variable __________ into the other equation.
Solve the equation for the first ______________
Substitute…… plug in the answer for the variable that you just found into one of the ________________ equations and ____________ for the other variable.
Write out your answer.
#2. 3x – 5y = 11 #3. y = x – 4 x = 3y + 1 2x + y = 5
#4. x + 4y = 6 #5. 2x + y = 1 x = -y + 3 x = 23 + 4y
Try these…
[DAY #3]
In your own words , write what you think the word Elimination means?
Example #1: Solve this system of linear equations using the elimination method.
Let’s solve this system of equations by eliminating the ‘x’ variable. If you look at the ‘x’ terms, we have an ‘x’ and a ‘2x’. What will cancel out the ‘2x’ term?. We need to multiply the ‘x’ term by to turn it into so it will cancel out with the ‘2x’ term. But when we do, we need to multiply EVERYTHING in that equation by the same number.
ORIGINAL SYSTEM NEW SYSTEM SOLUTION
Once you have the NEW SYSTEM , add the two equations together. One of the variables should cancel out. Solve the remaining equation for the variable. Substitute that variable into EITHER of the original equations (YOUR CHOICE!!!) and solve. Write out your answer!
Now let’s solve this system of equations by eliminating the ‘y’ variable. If you look at the ‘y’ terms, we have an ‘y’ and a ‘-3y’. What will cancel out the ‘-3y’ term?. We need to multiply the ‘y’ term by to turn it into so it will cancel out with the ‘-3y’ term. But when we do, we need to multiply EVERYTHING in that equation by the same number.
ORIGINAL SYSTEM NEW SYSTEM SOLUTION
Try these… Solve each system of equations by writing a new system that eliminates one of the variables. 1.) 5x + y = 6 2.) 2x + 5y = - 3x – 4y = 22 7x + 5y = 5
3.) 7x – 3y = 2 4.) 9x + 5y = 5 2x + 6y = - 20 2x + 10y = -
[DAY #4]
Some times when solving a system of equations by elimination, you have to multiply BOTH equations in order for something to cancel out. 1.) 3x – 5y = 29 2.) 8x – 3y = - 13 2x + 3y = -6 3x + 5y = - 11
3.) 3x + 4y = -10 4.) 6x – 4y = 38 2x + 5y = -2 5x – 3y = 31
[DAY #5]
We have explored 3 different ways to solve a system of equations. There will be times you MUST solve a system a particular way. Other times, it is up to you to choose the method. But how do you know when to use which method?
Solve by graphing only when
Solve by substitution when or of the variables are alone.
Solve by elimination when the are “lined up”
1.) Method: 2.) Method: 5x + y = 15 4x + 3y = 5 3x – 2y = -4 6x + 5y = 9
3.) Method: 4.) Method: y = x + 7 5x – 6y = - 4x – 2y = -24 10x + 3y = -
Let’s look at some Regents questions that have appeared on this topic. 5.) A system of equations is given below x + 2y = 5 2x + y = 4
Which system of equations does not have the same solution?
(1) 3x + 6y = 15 (2) x + 2y = 5 2x + y = 4 6x + 3y = 12
(3) 4x + 8y = 20 (4) x + 2y = 5 2x + y = 4 4x + 2y = 12
6.) Which pair of equations could not be used to solve the following equations for x and y? 4x + 2y = 22 -2x + 2y = -
(1) 4x + 2y = 22 (2) 12x + 6y = 66 2x – 2y = 8 6x – 6y = 24
(3) 4x + 2y = 22 (4) 8x + 4y = 44 -4x + 4y = - 16 -8x + 8y = -
