Algebra Honors ( Period 6)

Using Substitution to Solve a System  6-2
Using Addition to Solve a System  6-3
Using Addition w/ Multiplication to Solve a System 6-4

There are ALGEBRAIC ways (not graphing…solving equations) to find the intersection of 2 (or more) linear equations.

The two Algebraic ways:

1. Substitution

2. Addition or Elimination

Today we’ll look at substitution.

This method works especially well if both equations are solved for the SAME variable (x OR y)

OR

One equation is solved for a SINGLE variable (x or y)

You’ll plug one equation into the other…meaning you’ll substitute it in.

If you’ve ever been on the bench in a game, think of how you hope you’ll be substituted into the game for another player so you can play.

(or if you’re the understudy in a play or if you can substitute one book for another and get the same number of AR points)

Let’s look at some examples and you’ll see how it works.

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’re equal to each other!
 (transitive property of equality)

x + 7 = y = 2x + 1

So just get rid of the “middle man” y and get:

x + 7 = 2x + 1

Solve for x:

x = 6

Now plug into whichever original equation seems easier to you to find the y coordinate:

y = x + 7

y = 6 + 7 = 13

The intersection is (6, 13)

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

13 = 2(6) + 1

13 = 13

A system where one equation is solved for one variable:

y = 2x

5x + 3y = 22

2x is the same value as y.

Since that is true, anywhere you see y, you may 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

WHAT IF YOU HAVE 2 EQUATIONS AND NEITHER ONE IS SOLVE FOR A SINGLE VARIABLE?

You can just solve for one variable in whichever equation is easier.

Example:

x – 3y = 15  and 4x -2y = 20

You would need to pick which variable (x or y) would be easier to solve for in which equation.

Generally, look for a variable with no coefficient (really a coefficient of 1).

So for the above system, I’d pick to 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 + 60 = 20

10y = -40

y = -4

Substitution is often used to solve WORD PROBLEMS.

Example:

The perimeter of a rectangle is 40 in.

The length is 10 less than twice its width.

Find the dimensions of the rectangle.

2l + 2w = 40

l = 2w – 10

Substitute (2w – 10) for l:

2(2w – 10) + 2w = 40

4w – 20 + 2w = 40

6w – 20 = 40

6w = 60

w = 10 in.

l = 2w – 10 = 2(10) – 10 = 20 – 10 = 10 in.

IT’S A SQUARE! ;) 

Using Addition to Solve a System  6-3

The second Algebraic method to solve a system is known as ELIMINATION.

You’ll be eliminating one variable by using the ADDITIVE INVERSE of it in the other equation.

Example where you add the two equations together:

4x + 6y = 32

3x – 6y =  3

--------------------

7x + 0 = 35

x = 5

Plug into either equation to find y:

4(5) + 6y = 32

20 + 6y = 32

6y = 12

y = 2

The solution is (5, 2)

Sometimes you’ll ALMOST have additive inverses, but you’ll need to multiply one equation by -1 first:

5x + 2y =  6

9x + 2y = 22

--------------------

Multiply either the top or bottom by -1 (your choice):

5x + 2y =    6

-9x - 2y = -22

--------------------

-4x + 0 = -16

x = 4

Plug into either ORIGINAL equation:

5(4) + 2y = 6

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) = 22

36 – 14 = 22

22 = 22

Using Addition w/ Multiplication to Solve a System 6-4

This is the same method as Chapter 6-3, but to get ADDITIVE INVERSES of one variable you’ll need to multiply one or both equations by a factor.

MULTIPLYING JUST ONE EQUATION:

5x + 6y = -8

2x + 3y = -5

--------------------

Multiply either the bottom by -2 to eliminate y:

 5x + 6y = -8

-4x - 6y = 10

--------------------

 x +  0  =  2

x = 2

MULTIPLYING BOTH EQUATIONS:

4x + 2y = 8

3x + 3y = 9

--------------------

To eliminate x, you’d need to multiply the top by 3 and the bottom by -4 so that you’d get 12x and -12x

OR

Multiply the top by 3 and the bottom by -2 so that you’d get 6y and -6y.

It’s your choice!

I think keeping the numbers as small as possible is usually easier, so I’ll choose eliminating y.

3(4x + 2y) =  3(8)

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

--------------------

12x + 6y = 24

-6x - 6y = -18

--------------------

 6x +  0  =  6

x = 1