Download SYSTEMS OF EQUATIONS System of Equations: A set of two or

SYSTEMS OF EQUATIONS
System of Equations: A set of two or more equations with the same variables. The equations act as constraints on the
variables.
Solution to a System of Equations: The set of all points common to all equations in the system.
We consider the solution of the system to be the intersection point because the values in that point are usually
incredibly important to the problem, like the break-even point.
Systems can be incredibly complex with multiple equations, multiple variables. They don’t have to be linear, but since this
is only Algebra I, we are going to look at solving very basic systems consisting of two linear equations.
There are 3 possible ways 2 lines can interact on the same coordinate grid.
(1) They can intersect (1 solution that is the intersection point)
(2) They can be parallel (No Solution)
(3) They can be the same line (infinitely many points will solve the system)
***Note: The third case is not “All Reals”, it is the set of the infinitely many points that lie on that line. We will state
that there are “Infinitely Many solutions” in that last case. I will show you what set notation looks like in those cases,
later, with the examples.
Solving Systems of Equations.
There are 4 ways to solve a system.
(1) Graphing  Graph each equation on the same set of axes and look for the point of intersection…. Calculators make
life really easy here. This method is not ideal (in the case of “ugly” answers that are not integers.
(2) Substitution (Algebraic Method)  always yields an exact answer and can be used for systems that contain
functions that are not linear
(3) Elimination (Algebraic Method)  always yields an exact answer but can only be used for things that are linear (can
have more than 2 variables)
(4) Matrices (best method for linear systems with more than 2 variables)  computer programmers have to know how
to use these. There are actually several ways to use a matrix to solve a system of equations.
Solving Systems by Graphing.
Example 1: Determine if the following point is a solution to the given system.
(-1, 5)
(a)
x  y  4

 x  1
(b)
 y  x  4

1

 y   5 x
or (a) [x +y =4, x = -1] (b) [y=-x+4,y=-1/5x]
Solution: Just plug in the points and see if it is a solution. (a) -1 + 5 = 4 check and –1 = -1 check. So, yes this point is
common to both graphs and is therefore a solution to the system.
(b) 5 = -(-1) + 4 check and 5 ≠ -1/5(-1). So, no this point is not common to both. It isn’t a point on the second line, so the
point is not a solution to the system of equations.
Example 2: Solve these systems by graphing. If a system has infinite solutions, give one point that will satisfy the
system.
(a)
1

 y  3 x  3

y  1 x 3

3
(b)
y  x

 y  5x
(c)
y 1

y  x
(d)
2 x  y  3

x  2 y  4
(e)
3x  y  7

 y  3x  7
**If you graph with the calculator, the equation has to be in function notation (either slope-intercept or point-slope). IF
you do use a calculator, you must sketch a graph to show your work. If you do not use a calculator, you have to make sure
to graph very precisely, using graph paper. If I ask you to solve with graphing, your solution will consist of integers.
(a)
(b)
(c)
(d)
S = Null Set (lines are parallel)
S = {(0,0)}
S = {(1,1)}
S = {(2, -1)}
SUBSTITUTION METHOD
(Use this method when you have an equation solved given to you in terms of x or y, or if it could easily be solved
to isolate x or y.)
A method for solving a system of equations that involves solving one of the equations for one variable and substituting
the resulting expression into the other equation.
Example 1 : solve
 y  2x  4

 y  3x
Solution Remember what it means to solve a system of equations… you are looking for a shared point and that means that
x and y are equal!
Ex: y = 2x – 4 and y = 3x
So, you know that you are looking for the point where the y’s are equal. So, I am going to take one equation and
replace the y with what I know it is equal to from the other equation. In essence, I am going to set the
equations equal to each other .
Start with y = 2x – 4
Replace y with what it is equal to in the second equation, “3x”. So 3x = 2x – 4
Solve the resulting equation with variables on both sides x = -4
This gives you one number in the ordered pair. Then you substitute this value into one of the equations to solve
for y.
Start with y = 3x (because there is less work) and replace x with –4…y = 3(-4) = -12
Therefore your solution is S = {(-4, -12)}
Be sure to check your solution by either checking the point with both equations, or, use your graphing calculator
and solve by graphing.
Example 2:
 y  3( x  4)  7

x  y  4
Ex3: 3(x-4)+7=3x–12+7=3x–5  simplify the expression first
x – (3x – 5) = 4 you are SUBSTITUTING (get it substitution?) the 1 st equation that tells you what y= into the second
equation where it says ‘y’. Now you have one equation, one variable!
x – 3x + 5 = 4
-2x = -1
x = 1/2
y = 3(1/2) – 5 = 1.5 – 5 = -7/2
S = {(1/2, -7/2)}
Fiona’s favorite movie theater charges $8 for each adult ticket and $4 for each child ticket. At a showing where 200
tickets were sold, the theater collected $1304. How many of each type of ticket were sold?
2 facts, 2 unknowns…. “mixture word problem”
1) Define Variables A = # adult tickets sold, C = # child tickets sold
2) Write a system to represent the information in the problem. A + C = 200, 8A + 4C = 1304
3) Solve the system to solve the problem.
Transform the first one A = 200 – C
Substitute into the 2nd 8(200-C) + 4C = 1304  Remember never sub into the one you transform or else you
get ∞
Solve
1600 – 8C + 4C = 1304
1600 – 4C = 1304
-4C = - 296
C = 74
Be sure to double check to make sure you actually answered the word problem
A = 200 – 74
A = 126
Therefore, there were 74 adult tickets and 126 child tickets
ELIMINATION METHOD
(USE WHEN BOTH EQUATIONS ARE GIVEN IN STANDARD FORM)
Your goal in the elimination method is to eliminate or cancel out one of the variables. For two things to cancel out they
must equal zero. For this to happen the coefficients (wait for it…..the number in front of the variable) of the variable
you want to eliminate must be the same magnitude (value) but opposite signs. For example, -3x and 3x , 8y and -8y etc.
In a nutt shell to eliminate you…
Process:
0) Make sure equations are in the same form
1) Pick the variable you want to eliminate
2) Multiply one, or both, equations by a value to make sure the coefficients of the chosen variable are the same
(for subtraction) or opposite (for addition)
3) Subtract/Add and Solve the resulting equation.
4) Go back to one of the original equations and solve for the other variable.
5) Check your solutions, write them in set notation.
Ex:1
y=3
2x – y = 1
We can eliminate the ys because 1y and -1y add up to 0.
+
2x – y = 12
y=3
2x
= 15  look the y’s dropped off, or were “eliminated”
x
= 7.5  I’ve halfway solved the system
Ex2: 3x – 4y = 14
2x + 2y = 21
Ex3: 5x – 6y = 12
3x – 5y = 24
Ex2 ) Both are standard form
1) Choosing y because just have to multiply second equation by 2 to get opposite coefficients
2) 2nd by 2  4x + 4y = 42
3)
3x – 4y = 14
+ 4x + 4y = 42
7x
= 56
x
=8
4) 3(8) – 4y = 14  y = 2.5
5) S = {(8, 2.5)}
Ex 3:
0) Both are standard form
1) Doesn’t matter which one… just about the same amount of work for both… so I’ll pick x
2) Need to multiply first by 3 and second by –5 to get opposite coefficients of x
5x – 6y = 12  15x – 18y = 36
3x – 5y = 24  -15x + 25y = -120
3)
15x – 18y = 36
+ -15x + 25y = -120
7y = -84
y = -12
4) 5x –6(-12) = 12  x = -12
5) S = {(-12, -12)}
CLASSIFIYING LINEAR SYSTEMS
You can classify linear systems by the number of solutions and by comparing the slopes and y-intercepts of the equations.
A consistent system is a set of equations or inequalities that has at least one solution, and an inconsistent system will
have no solution. An independent system has equation with different slopes. A dependent system has equations with
equal slopes and y interprets.
Application Application Application
Use the strategy below to help with systems problems
1. Read the problem carefully to determine the unknowns and pertinent facts.
2. Choose variables to represent the unknowns.
3. Write equations using your unknowns and the given facts. Drawing a picture or making a chart might help…
4. Solve the equations.
Check to ensure your answer makes sense
Example
DinoCorp opens health food stores, which sell an Trail Mix of raisins and roasted nuts. Raisins sell for $3.50 per kg, and
roasted nuts sell for $4.75 per kg. How many kg of each should be mixed to make 20 kg of Evil Trail Mix worth $4.00 per
kg?
Every number we needed was given to us directly in the problem. There is information in the problem that is not directly
stated that we will have to infer.
Our unknowns are
kg of raisins and kg of nuts.  very straight forward fact
We also have prices per kg, but nothing about a total cost  this is why we need to approach this problem a
little differently.
We are going to use a table to help write the equations.
Fundamentally for this problem we need to know that weight * unit cost = total cost.
Weight
Unit Cost
Total Cost
Raisins
R
3.5
3.5R
Roasted Nuts
N
4.75
4.75N
20
4
80
Mixture
System of Equations:
R + N = 20
3.5R + 4.75N = 80
Our equations are about
(1) weight R + N = 20
(2) Total Cost (we had to determine the total cost of the mixture, it wasn’t directly given in the problem)
3.5R + 4.75N = 80
Solve the system to find the answers…
Distance/Rate/Time problems
At 5:00 pm Mrs. Lakshmanan left ARS driving 25 miles per hour. At 5:30, Ms. DiCuffa left ARS and followed the same
route, driving 40 miles per hour. At what time did Ms. DiCuffa pass slow poke Mrs. Lakshmanan?
It will be good to draw a simple picture to make sure you understand the problems set up!
Rate
Time
Distance
Mrs. Lakshmanan
25
x
25x
Ms. DiCuffa
40
x – 1/2
40(x – 1/2)
System of Equations:
D = 25x
D = 40 (x – 1/2)
Solve using substitution or elimination.
Example; A boat travels 60 km upstream (against the current) in 5 hours. The boat travels the same distance
downstream in 3 hours. What is the rate of the boat in still water? What is the rate of the current?
This sounds like the worst possible word problem an, but they are actually quite easy to solve.
Your two variables will be the two rates (speeds): the boats speed in still water (x), and the speed of the current (y)
Rate
Upstream
Downstream
System of Equations:
Time
Distance
x-y
5
60
x+y
3
60
5(x - y) = 60