Algebra ( Period 5)

Using Addition With Multiplication to Solve a System 6-4

This is still  second Algebraic method to solve a system but to get the additive inverses of one variable—you will need to multiply ONE or BOTH equations by a factor.

EXAMPLE 1

Multiplying just ONE equation

5x + 6y = -8
2x + 3y = -5

Multiply the bottom by -2 to eliminate y

5x + 6y = -8
-4x -6y = 10

x = 2

EXAMPLE 2

Multiply BOTH Equations

4x + 2y = 8
3x + 3y = 9

To eliminate x, you would need to multiply the top by 3 and the bottom by -4
so you would get 12x and – 12x
OR

Multiply the top by 3 and the bottom by -2 so that you will get 6y and -6y

It is your choice…
 [and you could actual divide the second equation by 3 to start out with
x + y = 3  Making it even easier… but if you did not see that—use the other two…]

I think keeping the numbers as SMALL as possible is usually easier so I will eliminate y

3(4x + 2y) =  3(8)
-2(3x + 3y) = -2(9)

12x + 6y = 24
-6x -6y = -18

6x = 6
x= 1

Algebra ( Period 5)

Using Addition to Solve a System 6-3

The second Algebraic method to solve a system is known as ELIMINATION
You will be eliminating one variable by using the ADDITIVE INVERSE of it in the other equation

4x + 6y = 32
3x -6y = 3

7x + 0 = 35
7x = 35
x = 5

Plug into EITHER equation to find y
4(5) + 6y = 32                   20 +   6y = 32               
6y = 12                      y = 2
So the solution is ( 5, 2)

Sometimes you almost have additive inverses, but you need to multiply ONE EQUATION by -1 ...FIRST

5x + 2y = 6
9x + 2y = 22

Multiply EITHER the Top or the Bottom by -1 (Your choice)

5x + 2y = 6
-9x -2y = -22

-4x + 0 = -16                  

-4x = -16
x = 4

Plug into either ORIGINAL equation

5(4) + 2y = 6   so 20 + 2y = 6
2y = -14 

y = -7
The solution is ( 4, -7)

Now you can plug this point into the other equation to check that you haven’t made a mistake:  9x + 2y = 22  

9(4) +2(-7) = 2   

36 – 14 = 22  YES!!!

Algebra (Period 5)

Using Substitution to Solve a System 6-2

There are two Algebraic ways to find the intersection of 2 (or more) linear equations:

·        1,   .Substitution

·       2.     Addition ( also called Elimination)

Substitution
This method works especially well if both equations are solved for the SAME variable (either x OR y)  OR
ONE  equation is solved for a SINGLE variable ( x or y)

You plug one equation into the other—meaning you will substitute it in. If you have ever been on the bench in a game think of how you hope you will be substituted into the game for another play so you can play—or if you are the understudy in a play – or if you can substitute one book for another and get the same number of AR points!

EXAMPLE 1
A system where both equations are already solved for one variable
y = x + 7  and y = 2x + 1

Since both equations are equal to y—they are EQUAL to each other
(Remember: Transitive property of equality)         
 x + 7 = y = 2x + 1
just get rid of the “middle man” and get

x + 7 = 2x + 1              
 solve for x
x= 6

Now plug into whichever ORIGINAL equation seems easier to find the y coordinate
y = x + 7                           
y = (6) + 7 = 13
The intersection is ( 6, 13)
Make sure to give the solution as an ordered pair  (Ordered, Pair)

What if we plug in this point to the other equation?
It SHOULD work because both equations have ( 6, 13) as a solution.
y = 2x + 1                      y = 2(6) = 1
y = 12 + 1                       y = 13

EXAMPLE 2
A system where one equation is solved for one of the variables
y = 2x      AND          5x + 3y =22
2x is the same value as y
Since that is true, anywhere you see “y” you my use 2x instead
“Substituting into the game for y is 2x”

5x + 3(2x) = 22          
5x + 6x = 22
11x = 22                    
 x = 2

Now plug into the other equation to find y
y = 2x = 2(2) = 4
The solution (intersection is (2,4)
Check: plug in ( 2, 4) into the other equation 5(2) + 3(4) = 10 + 12 = 22
IT WORKS!!

EXAMPLE 3

What if you have 2 equations and neither one is solved for a single variable?
You can just solved for one of the variables in whichever equation is easier

x – 3y = 15  and 4x – 2y = 20

You would need to pick which variable ( x or y) would be easier to solve in one of the equations.
Generally look for a variable with no coefficient ( or really it has a coefficient of 1)
So for the above system, pick the solve for x in the first equation
x = 3y + 15
So, wherever you see “x” in the other equation, substitute in ( 3y + 15)

4(3y + 15) – 2y = 20                       
12y +60 -2y = 20
10y = -40                                         
y = -4

Substitute is often used to solve word problems.
The perimeter of a rectangle is 40 in The length is 10 less than twice its width. Find the dimension f the rectangle.

let l = the length of the rectangle           Let w = the width of the rectangle

 2l + 2w = 40              l = 2w -10

substitute ( 2w– 10) for l           2( 2w– 10) + 2w = 40

4w -20 + 2w = 40                      6w = 60
w = 10                                       10 inches
Since the width is 10inches substitute that in

l = 2(10) – 10 = 10
The length is also 10 inches.
It is a SQUARE