solving systems of equations modeling real-world problems
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Solving Systems of Solving Systems of EquationsEquations
Solving Systems of Solving Systems of EquationsEquations
Modeling real-world problemsModeling real-world problems
In business, the point at which income equals expenses is called the break-even
point.When starting a business, people want to
know the point a which their income equals their expenses, that’s the point where they
start to make a profit. In the example above the values of y on the blue line represent dollars made and the value of y on the dotted red line represent dollars spent.
Profit
Loss
A system of equations is a set of two or more equations that have variables in
common.
The common variables relate to similar quantities. You can think of an equation as a condition imposed on one or more variables, and a system as several conditions imposed simultaneously.
Remember, when solving systems of equations, you are looking for a solution that makes each equation true.
In earlier chapters, you learned to solve an equation for a specified variable, graph the
equation, and find how pairs of lines represented by linear equations in two
variables are related.
When solving a system of equations, you look for a solution that makes each equation true. There are several strategies you can use. To begin with we will be using tables and graphs. Let’s look at the following example and work through the problem step-by-step to find a solution.
Edna leaves the trailhead at dawn to hike 12 miles toward the lake, where her friend Maria is camping. At the same time, Maria starts her hike toward the trailhead. Edna is walking uphill so she averages only 1.5
mi/hr, while Maria averages 2.5 mi/hr walking downhill. When and where will
they meet?
Here’s what we need to complete to solve this example.
• Define variables for time (x) and for distance (y) from the trailhead.
• Write a system of two equations to model this situation.
• Solve this system by creating a table and finding values for the variables that make both equations true. Then locate this solution on a graph.
• Check your solution and explain its real-world meaning.
Let x represent the time in hours. Both
women hike the same amount of
time. Let y represent the
distance in miles from the trailhead.
When Edna and Maria meet they will both be the
same distance from the trailhead,
although they will have hiked different
distances.
The system of equations that models this situation is grouped in a brace.
Edna starts at the trailhead so she increases her distance from it as she hikes 1.5 mi/hr for x hours. Maria starts 12 miles
from the trailhead and reduces her distance from it as she hikes 2.5 mi/hr for x
hours.
xy
xy
5.212
5.1
Create a table from the equations. Fill in the times and calculate each distance. The table shows the x-value that gives equal y-values
for both equations. When x = 3, both y-values are 4.5. So the solution is the ordered pair (3, 4.5). We say these values “satisfy”
both equations.
Hiking Times and Distances
X y = 1.5x y = 12 – 2.5x 0 0 12 1 1.5 9.5 2 3 7 3 4.5 4.5 4 6 2 5 7.5 -0.5
Let’s create the data on your graphing calculator. Enter the equations in y = on
your calculator and create a table.
Do both tables model this situation?What do you notice about y when x
increases in each equation?Why are the values different?
Hiking Times and Distances
X y = 1.5x y = 12 – 2.5x 0 0 12 1 1.5 9.5 2 3 7 3 4.5 4.5 4 6 2 5 7.5 -0.5
On the graph this solution is the point where the two lines intersect. You can use trace function or calculate function on your calculator to approximate the coordinates of the solution point, though sometimes
you’ll get an exact answer as in our example here.
Solving Systems of Solving Systems of EquationsEquations
Solving Systems of Solving Systems of EquationsEquations
Graphing MethodGraphing Method
A system of linear equations is a set of two or more equations with the same
variable. The solution of a system in x and y is any ordered pair (x, y) that
satisfies each of the equations in the system.
The solution of a system of equations is the intersection of the graphs of the
equations.
If you can graph a straight line, you can
solve systems of equations
graphically!
The process is very easy. Simply graph
the two lines and look for the point
where they intersect (cross).
Remember using the graphing method many times only approximates the
solution, so sometimes it can be
unreliable.
Solving Systems of Equations by Graphing.
To solve a system of equations graphically, graph both equations and see where they intersect. The intersection point is the
solution.
4x – 6y = 124x = 6y + 124x – 12 = 6y6y = 4x – 12 6 6 6y = 2/3x – 2slope = 2/3y-intercept = -2
2x + 2y = 62y = -2x + 6 2 2 2y = -x + 3slope = -1/1y-intercept = 3
Graph the equations.
The slope-intercept method of graphing was
used in this example.
The point of intersection of the two lines (3, 0) is
the solution to the system of equations.
This means that (3, 0), when substituted into either equation, will
make them both true.
Use a graph to solve the system of equations below. Graph both equations on
the same coordinate plane.Graph x + y = 5 using the intercepts: (5, 0) and (0, 5)
Graph y = 2x – 1 using the slope-intercept method.
12
5
xy
yx
Locate the point where the lines intersect. From the graph, the solution appears to (2,
3).Check to be sure that (2, 3) is the solution,
substitute 2 for x and 3 for y into each equation.
x + y = 5 y = 2x – 12 + 3 = 5 3 = 2(2) - 1
Use a graph to solve each system of equations. If the system has no solution,
write none.
3
5
1
5
5
1
xy
xy
xy
xy
xy
xy
7
3
22
12
2
4
21
41
xy
yx
yx
yx
xy
xy
Summary of Solutions of Systems of Linear
EquationsThe lines intersect so there is one solution.
The lines are parallel so there are no solutions.
The lines are the same so there are infinitely many solutions.
x + 2y = 7
x = y + 4
y – 2x = 7
Y = 2x + 3
-3x = 5 – y
2y = 6x + 10
Note
Some systems of equations may be very difficult to solve using the graphing method. The exact solution would be
hard to determine from a graph because the coordinates are
not integers. Solving a system algebraically is better than graphing when you need an
accurate solution.Take for example the system of
equations:3x + 2y = 12
x – y = 3x = 3/8 and y = 15/4
Solving Systems of Solving Systems of Equations Equations
AlgebraicallyAlgebraically
Solving Systems of Solving Systems of Equations Equations
AlgebraicallyAlgebraicallySubstitution MethodSubstitution Method
The substitution method is used to
eliminate one of the variables by
replacement when solving a system of
equations.
Think of it as “grabbing” what one variable equals from
one equation and “plugging” it into the
other equation.
Solve this system of equations using the substitution method.
Step 13y – 2x = 11Y + 2x = 9
Solve one of the equations for either “x” or “y”.In this example it is easier to solve the second equation for “y”, since it only involves one step.
Y = 9 – 2x
Step 2
Replace the “y” value in the first equation by what “y” now equals (y = 9 – 2x). Grab
the “y” value and plug it into the other equation.
3y – 2x = 113(9 – 2x) – 2x = 11
Step 3
Solve this new equation for “x”.
3(9 – 2x) – 2x = 1127 – 6x – 2x = 1127 – 6x – 2x = 11
27 – 8x = 11-8x = -16
x = 2
Step 4
Now that we know the “x” value (x = 2), we place it into either of the ORIGINAL
equations in order to solve for “y”. Pick the easier one to work with!
Y + 2x = 9y + 2(2) = 9
y + 4 = 9y = 5
Step 5
Check: substitute x = 2 and y = 5 into BOTH ORIGINAL equations. If these
answers are correct BOTH equations will be true!
3y – 2x = 113(5) – 2(2) = 11
15 – 4 = 11 11 = 11 True?
Y + 2x = 95 + 2(2) = 9
5 + 4 = 99 = 9 True?
The Substitution Method
• Step 1 Solve one equation for x (or y).• Step 2 Substitute the expression from Step 1
into the other equation.• Step 3 Solve for y (or x).• Step 4 Take the value of y (or x) found in
Step 3 and substitute it into one of the original equations. Then solve for the other variable.
• Step 5 The ordered pair of values from Steps 3 and 4 is the solution. If the system has no solution, a contradictory statement will result in either Step 3 or 4.
Use the substitution method to solve each system of equations. Check your answers.
42
52
52
4
4
15
yx
yx
yx
xy
y
xy
53
3
52
632
13
132
51
yx
yx
xy
yx
xy
yx
Solving systems of equations in real-world problems.
April sold 75 tickets to a school play and collected a total of $495. If the adult tickets cost $8 each and child tickets cost $5 each, how many adult tickets and how many child tickets did she sell?
Solution: Let a represent the adult tickets and c represent the child tickets.
Individual tickets sold equaled 75, so a + c = 75All total April sold $495 in tickets, since adult tickets are $8
and child tickets are $5, so 8a + 5c = 495.
System of Equations a + c =
758a + 5c == 495
Solutiona + c = 75
8a + 5c = 495a = 75 – c
8(75 – c) + 5c = 495600 – 8c + 5c = 495
600 - 3c = 495105 = 3c
35 = c
a + c = 75a + 35 = 75
a = 40There were 40 adult tickets and 35 child
tickets sold. 40 + 35 = 758(40) + 5(35) = 495
320 + 175 = 495
Write a system of equations and solve.
At a baseball game, Jose bought five hot
dogs and three sodas for $17. At the same time, Allison bought
two hot dogs and four sodas for $11. Find the cost of one hot dog and one soda.
Solving Systems of Solving Systems of EquationsEquations
Solving Systems of Solving Systems of EquationsEquations
Elimination methodElimination method
You can use the Addition and Subtraction Properties of Equality to solve a system by the
elimination method. You can add or subtract equations to eliminate (getting rid of)
a variable.
Step 15x – 6y = -323x + 6y = 48
Eliminate y because the sum of the coefficients of y is zero
5x – 6y = -323x + 6y = 488x + 0 = 16
x = 2
Addition Property of EqualitySolve for x
If you add the two equations together, the +6y and -6y cancel each other out because of the
Property of Additive Inverse
Step 2
Solve for the eliminated variable y using either of the original equations.
3x + 6y = 483(2) + 6y = 48
6 + 6y = 486y = 42y = 7
Choose the 2nd equationSubstitute 2 for x
Simplify. Then solve for y.
Remember x = 2
Since x = 2 and y = 7, the solution is (2, 7)
Check5x – 6y = -32 3x + 6y = 48
5(2) – 6(7) = -32 3(2) + 6(7) = 4810 – 42 = -32 6 + 42 = 48-32 = -32 48 = 48
Remember, the order pair (2, 7) must make both equations true.
True
True
Suppose your community center sells a total of 292 tickets for a basketball
game. An adult ticket cost $3. A student ticket cost $1. The sponsors collected $470 in ticket sales. Write
and solve a system to find the number of each type of ticket sold.
Let a = number of adult ticketsLet s = number of student tickets
total number of ticket total number of sales
a + s = 292 3a + 1s = 470
Solve by elimination (get rid of s) because the difference of the coefficients of s is
zero.a + s = 2923a + s = 470
-2a + 0 = -178a = 89
That means you must subtract the two equations
so,-3a – a = -470 is what you
must subtract. Next Step
This is the number of adult
tickets sold.
Solve for the eliminated variable using either of the original equations.
a + s = 29289 + s = 292
s = 203
There were 89 adult tickets sold and 203 student tickets sold.
Is the solution reasonable? The total number of tickets is 89 + 203 = 292. The total sales is $3(89) + $1(203)
= $470. The solution is correct.
This is the number of student
tickets sold.
If you have noticed in the last few examples that to eliminate a variable its
coefficients must have a sum or difference of zero.
Sometime you may need to multiply one or both of the equations by a nonzero number first so that you can
then add or subtract the equations to eliminate one of the variables.
2x + 5y = 17 7x + 2y = 10 2x + 5y = -226x – 5y = -19 -7x + y = -16 10x + 3y = 22
If you notice the systems of equations above, two of them have something in common. The third doesn’t.
We can add these two equations together to eliminate the y
variable.
We can add these two equations together to
eliminate the x variable.
What are we going to do with these equations, can’t eliminate a variable the way they are written?
Multiplying One Equation
Solve by Elimination2x + 5y = -2210x + 3y = 22
Step 12x + 5y = -225(2x + 5y = -22) 10x + 25y = -11010x + 3y = 22 10x + 3y = 22 -(10x + 3y = 22) 0 + 22y = -
132 y = -6
Start with the given system.
To prepare for eliminating x, multiply the first equation by
5.
Subtract the equations to eliminate x.
NEXT
Be careful when you subtract. All the
signs in the equation that is being
subtracted change.-10x – 3y = -22
Ask:Is one
coefficient a factor of the
other coefficient for
the same variable?
Step 2
Solve for the eliminated variable using either of the original equations.
2x + 5y = -22 Choose the first equation.
2x +5(-6) = -22 Substitute -6 for y.
2x – 30 = -22 Solve for x.
2x = 8 x = 4
The solution is (4, -6).
Solve by elimination.
-2x + 5y = -327x – 5y = 17
3x – 10y = -254x + 40y = 20
2x – 3y = 612x + y = -7
Ask:Is one
coefficient a factor of the
other coefficient for
the same variable?
Multiplying Both Equations
To eliminate a variable, you may need to multiply both equations in a system by a
nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to
eliminate a variable.
4x + 2y = 147x + 3y = -8
In these two equations you cannot use graphing or
substitution very easily. However ever if we
multiply the first equation by 3 and the second by 2,
we can eliminate the y variable.
Find the least common multiple LCM of the coefficients of one
variable, since working with smaller
numbers tends to reduce the likelihood
of errors.
4 x 7 = 28
2 x 3 = 6
NEXT
4X + 2Y = 14 3(4X + 2Y = 14) 12X + 6Y = 427X – 3Y = - 8 2(7X – 3Y = -8) 14X – 6Y = -16
26X + 0 = 26
26X = 26 X = 1
Solve for the eliminated variable y using either of the original equations.
4x + 2y = 144(1) + 2y = 14 4 + 2y = 14 2y = 10 y = 5 The solution is (1, 5).
Start with the given system. To prepare to eliminate y,
multiply the first equation by 3 and the second equation by
2.
Add the equations to eliminate y.
Practice and Problem Solving
Solve by elimination:1) 2x + 5y = 17 2) 7x + 2y = 10 6x + 5y = -9 -7x + y = -16
3) 2x – 3y = 61 4) 24x + 2y = 52 2x + y = -7 6x – 3y = -36
5) y = 2x 6) 9x + 5y = 34 y = x – 1 8x – 2y = -2
Word Problems
You choose what method you
what to use to solve question 5
thru 10.
7) The sum of two numbers is 20. Their difference is 4. Write and solve a system of
equations.
8) Your school sold 456 tickets for a school play. An adult ticket cost $3.50. A student
ticket cost $1. Total ticket sales equaled $1131. Let a = adult tickets sold, and s = student
tickets sold. How many tickets of each were sold?
9) Suppose the band sells cans of popcorn for $5 each and mixed nuts for $8 each. The band sells a total of 240 cans and makes a total of $1614. Find the number of cans of each sold.
One more problem,my favorite.
10) A farmer raises chicken and cows. He has a total of 34 animals in his barnyard.
His six-year son came in one day all excited saying, “Daddy, daddy, did you
know all your animals have a total of 110 legs.”
Write a system of equations to represent this situation.
How many chickens and how many cows does the farmer have?
When you solve systems using elimination, plan a strategy. The flowchart like the one
below can help you decide how to eliminate a variable.
When you solve systems using elimination, plan a strategy. Here are a few hints to
help you decide how to eliminate a variable.
Answer these questions.Can I eliminate a variable by
adding or subtracting the given equations? YES
OrNO Can I multiply one of
the equations by a number, and then add or subtract the
equations?YESorNO
Multiply both equationsby different numbers.
then add or subtract theequations
Do It.
Do It.
Solving Systems of Equations The Best Time to Use Which Method
Graphing: Used to estimate the solution, since graphing usually does
not give an exact solution. Y = 2x - 3 y = x - 1
Substitution: I f one of the variables in either equation has a
coeffi cient of 1 or –1 3y + 2x = 4 -6x + y = -7
Elimination Using Addition: I f one of the variables has opposite coeffi cients in the
two equations 5x – 6y = -32 3x + 6y = 48
Elimination Using Subtraction: I f one of the variables has the same coeffi cient in the
two equations 2x + 3y = 11 2x + 9y = 1
Elimination Using Multiplication: I f none of the coeffi cients are 1 or –1 and neither of the
variables can be eliminated by simply adding and subtracting the equations. 5(2x + 5y = -22) 10x + 3y = 22 3(4x + 2y = 14) 2(7x = 3y = -8)