CHAPTER 5: SYSTEMS OF LINEAR EQUATIONS
System of Linear Equations: a set (2 or more) of linear equations that has one or
more variables
Unique Solution: the single pair of variables that satisfies BOTH
The solution to a system of equations is a point on BOTH lines; the point of
intersection
Parallel lines have the same slope.
Parallel lines do not intersect….therefore there is NO SOLUTION!
SOLVING EQUATIONS
Draw a line between the two sides of the equation. First remove the constant (the term
that is added or subtracted to the side with the variable by doing the opposite) .
Then remove the coefficient (the number that is multiplied or dividing the variable) by
doing the opposite
STEP 4: Now you need to find the other variable. Pick one equation and plug in the
value you found
STEP 5: Check your work by plugging x and y into the other equation
FINDING THE OTHER VARIABLE AND CHECKING YOUR WORK!
Once you have figured out the answer for one of the variables:
Go back to your original equations.
Substitute the known variable in and solve.
Then…..take the OTHER equation.
Substitute both variables in and see if the answer comes out correctly. If not…..go
back and check your work….something is wrong somewhere!
5.2 SOLVING SYSTEMS OF LINEAR EQUATIONS USING ALGEBRAIC METHODS
Elimination Method:
Example:
-9x - y = 15
5x + y = -7
Step 1: Line them up so that like terms are in the same row:
-9x
-y
5x
+y
=
15
Step 2:
Look to see which terms you could add or subtract to get zero. If, when you line up the equations, you
=
-7
cannot automatically add to eliminate one term.....you must do something to one of the equations (multiply it by an
integer) to make it so that you can.
In the example above, you can add -y and +y to cancel each other out. Do that, along with adding the other terms:
-9x
-y
+5x
=
+y
=
1
-7
____________________________
-4x
+0
=
8
Step 3: This turns into an equation you can solve! See below!
SUBSTITUTION METHOD:
Remember: you can only substitute things of equal value!
Use the substitution method to solve for a system of equations:
Example:
y = x+4
y = 3x +8
STEP 1:
Make a new equation by substituting the value of y from the first equation into the second equation:
x + 4 =
3x + 8
STEP 2:
SIMPLIFY! If you now have an equation with variables on both sides, move the va riable term from one side to the
other.
If you need to combine like terms on each side of the equation, do that!
STEP 3: This turns into an equation you can solve!
5.3 Solving Real World Problems Using Systems of Linear Equations
Example:
The school that Abhasra goes to is selling tickets to the annual dance competition. On the first day
of ticket sales, the school sold 3 senior tickets and 8 child tickets for a total of $74. The school took
in $116 on the second day by selling 6 senior tickets and 8 child tickets. What is the price of one
senior ticket and one child ticket?
STEP 1:
Assign variables. Determine what you need to know and assign a letter to stand for each. You can
usually find this in the question at the end. In the example, you are asked to find the cost of a senior
ticket and the cost of a child ticket. So, I decided to use X and Y as my variables and said:
Let X = the price of one senior ticket
Let Y = the price of one child ticket
STEP 2:
Write one equation that uses both variables. Find the info in the problem. For my equation, I will use
the second sentence: On the first day of ticket sales, the school sold 3 senior tickets and 8 child
tickets for a total of $74.
So that means 3X + 8 Y = 74
STEP 3:
Write a second (different) equation. For this equation I will use the third sentence in the
problem: The school took in $116 on the second day by selling 6 senior tickets and 8 child tickets.
So that means
6X + 8Y = 116
STEP 4:
Solve the system of equations using substitution or elimination.
3X + 8 Y = 74
6X + 8Y = 116
,
To eliminate, I must either subtract or multiply the second equation by ( -1).
My result:
-3X = -42
X = 14
Substitute 14 for X back into the first equation:
3 (14 ) + 8Y = 74
42 + 8Y = 74
8Y = 32
Y=4
So.....A senior ticket costs $14 and a child ticket costs $4.
5.4 Solving a System of Equations by Graphing
FIRST...... A REVIEW OF GRAPHING EQUATIONS!
Given the slope-intercept formula, y = mx + b graph the equation
Step 1: Graph the y -intercept ( the +b part of the equation)
Step 2: Find the slope (m), and use slope to find another point. It's good to find at least 3 points to
make sure you are counting correctly. Remember, slope = RISE/RUN
Step 3. Use a ruler and draw a line through the points, putting arrowheads on the ends
EXAMPLE: y = x + 3
I looked at the equation. The y intercept is 3, so I placed a dot on the 3 on
the y-axis. Then I looked at the slope....no number before the x (where the slope is supposed to be)
so it must be 1, or 1/1. So, from my 3 on the y, I went up 1 (rise) and right 1 (run) and put a new
point. I did a few more to be sure, then drew my line, remembering to add arrowheads on the ends.
Now do this same thing with the system of equations.
Step 1: Graph the first equation
Step 2: Graph the second equation on the same grid
Step 3: Identify the point of intersection as x and y coordinates
Step 4: Plug your x and y into both equations to see if they are true.
EXAMPLE:
I graphed these 2 equations in the same grid (sorry, no picture....you will jus t have to take my word
for it)
Y = 1/4 X - 1
Y = 5/4 X + 3
The two lines met at point ( -4 , 2 ). That means my X is -4 and my Y is 2.
Plug into first equation:
Y = 1/4 X - 1
2 = 1/4 (-4) - 1
2 = -1 - 1
2 = 2
Plug into second equation
Y = 5/4 X + 3
2 = 5/4 (-4) + 3
2 = -5 + 3
2 = 2