3.2 Solving Linear Systems Algebraically

Honors

2 Methods for Solving Algebraically2 Methods for Solving Algebraically

1. Substitution Method

(used mostly when one of the equations has a variable with a coefficient of 1 or -1)

2. Elimination Method

Substitution MethodSubstitution Method1. Solve one of the given equations for one of

the variables. (whichever is the easiest to solve for)

2. Substitute the expression from step 1 into the other equation and solve for the remaining variable.

3. Substitute the value from step 2 into the revised equation from step 1 and solve for the 2nd variable.

4. Write the solution as an ordered pair (x,y).

ExEx: Solve using substitution method: Solve using substitution method3x – y = 13

2x + 2y = -10

1. Solve the 1st equation for y.

3x - y= 13 -y = -3x +13 y = 3x - 13

2. Now substitute 3x -13 in for the y in the 2nd equation.

2x + 2(3x -13) = -10

Now, solve for x.2x + 6x - 26 = -10 8x = 16

x = 23. Now substitute the 2 in

for x in for the equation from step 1.

y = 3(2) - 13y = 6 - 13

y = -74. Solution: (2,-7)Plug in to check solution

EXAMPLEEXAMPLESolve the linear system using the

substitution method

3x + 4y = -4

x + 2y = 2

Elimination MethodElimination Method1. Multiply one or both equations by a

real number so that when the equations are added together one variable will cancel out.

2. Add the 2 equations together. Solve for the remaining variable.

3. Substitute the value from step 2 into one of the original equations and solve for the other variable.

4. Write the solution as an ordered pair (x,y).

ExEx: Solve using the elimination : Solve using the elimination methodmethod

2x - 6y = 19-3x + 2y = 10

1. Multiply the entire 2nd equation by 3 so that the y’s will cancel.

2x - 6y = 19 -9x + 6y = 30

2. Now add the 2 equations.

-7x = 49and solve for the variable.

x = -7

3. Substitute the -7 in for x in one of the original equations.

2(-7) - 6y = 19 -14 - 6y = 19 -6y = 33

y = -11/24. Now write as an

ordered pair.(-7, -11/2)

Plug into both equations to check.

Solve using elimination methodSolve using elimination method

9x – 5y = -7

-6x + 4y = 2

So far we have had one solution for each of our systems.

Some linear systems can have no solution or infinitely many solutions.

Today we are going to find out how to determine how many solutions a

system has and why.

Exactly one solutiony

x

Infinitely many solutionsy

x

NO solutiony

x

Same number equals same number…. Infinitely many solutions so dependent system

A number equals a different number…. no solution so it is an inconsistent system

ExEx: Solve using either method: Solve using either method9x - 3y = 15

-3x + y = -5

Which method?

Substitution!

Solve 2nd equation for y.

y = 3x - 5

9x - 3(3x - 5) = 15

9x - 9x + 15 =15

15 = 15

OK, so?

What does this mean?

Both equations are for the same line!

Infinitely many solutions

Means any point on the line is a solution.

ExEx: Solve using either method: Solve using either method6x - 4y = 14

-3x + 2y = 7

Which method?

Linear combination!

Multiply 2nd equation by 2.

6x - 4y = 14

-6x + 4y = 14

Add together.

0=28

Huh?

What does this mean?

It means the 2 lines are parallel.

No solution

Since the lines do not intersect, they have

no points in common.

Solution What it means What it means Types graphically algebraically1 solution 2 lines intersect when solving at one point equations you

get a value for x and y

No solution Parallel lines when variables Inconsistent system cancel out & the

statement is false

Infinitely Many Same line when variables

Solutions cancel out & the Dependent System statement is true

Beyonce loves to do her own shopping. Last Saturday she bought a new outfit for her video. Each shirt cost $125 and each pair of pants cost $225. She came home with 26 items and spent exactly $4950. How many pants and shirts did Beyonce buy?

1. DEFINE the variables.2. Write the system of equations.3. Solve.4. State solution.

Beyonce bought 9 shirts and 17 pants

You are in charge of decorating the gym for Prom this year. You purchased 6 bags of balloons and 5 bags of large sparkling hanging stars all for $19.20. You soon realized that this was not enough to decorate the entire gym. On your second trip to the store, you bought 8 bags of balloons and 2 bags of large sparkling hanging stars all for $15.80. What was the price for each item?

1. DEFINE the variables.2. Write the system of equations.3. Solve.4. State solution.

The Ballons cost $1.45 and the Stars cost $2.10