Unformatted text preview: Chapter 4 Solving Systems of
Linear Equations Chapter Sections
4.1 – Solving Systems of Linear Equations by Graphing
4.2 – Solving Systems of Linear Equations by Substitution
4.3  Solving Systems of Linear Equations by Addition
4.5 – Systems of Linear Equations and Problem Solving MartinGay, Beginning and Intermediate Algebra, 4ed 2 § 4.1 Solving Systems of Linear
Equations by Graphing Systems of Linear Equations
A system of linear equations consists of two
or more linear equations.
This section focuses on only two equations at
a time.
The solution of a system of linear equations
in two variables is any ordered pair that solves
both of the linear equations. MartinGay, Beginning and Intermediate Algebra, 4ed 4 Solution of a System
Example:
Determine whether the given point is a solution of the following
system.
point: (– 3, 1)
system: x – y = – 4 and 2x + 10y = 4
•Plug the values into the equations.
First equation: – 3 – 1 = – 4 true
Second equation: 2(– 3) + 10(1) = – 6 + 10 = 4 true
the point (– 3, 1) produces a true statement in both
equations, it is a solution.
•Since MartinGay, Beginning and Intermediate Algebra, 4ed 5 Solution of a System
Example:
Determine whether the given point is a solution of the following
system
point: (4, 2)
system: 2x – 5y = – 2 and 3x + 4y = 4
Plug the values into the equations.
First equation: 2(4) – 5(2) = 8 – 10 = – 2
true
Second equation: 3(4) + 4(2) = 12 + 8 = 20 ≠ 4 false
Since the point (4, 2) produces a true statement in only one
equation, it is NOT a solution. MartinGay, Beginning and Intermediate Algebra, 4ed 6 Finding a Solution by Graphing
• Since a solution of a system of equations is a
solution common to both equations, it would also
be a point common to the graphs of both equations.
• To find the solution of a system of 2 linear
equations, graph the equations and see where the
lines intersect.
• (Note that since our chances of guessing the right
coordinates to try for a solution are not that high,
we are usually more successful if we try a different
technique.)
MartinGay, Beginning and Intermediate Algebra, 4ed 7 Finding a Solution by Graphing
y Example: (5, 5) Solve the following
system of equations
by graphing.
2x – y = 6 and
x + 3y = 10 (2, 4) (6, 6) (1, 3) (4, 2)
x (3, 0) First, graph 2x – y = 6.
Second, graph x + 3y = 10.
The lines APPEAR to intersect at (4, 2).
MartinGay, Beginning and Intermediate Algebra, 4ed (0, 6)
Continued.
8 Finding a Solution by Graphing
Example continued:
Although the solution to the system of equations appears
to be (4, 2), you still need to check the answer by
substituting x = 4 and y = 2 into the two equations.
First equation,
2(4) – 2 = 8 – 2 = 6 true
Second equation,
4 + 3(2) = 4 + 6 = 10 true
The point (4, 2) checks, so it is the solution of the
system.
MartinGay, Beginning and Intermediate Algebra, 4ed 9 Finding a Solution by Graphing
y Example:
Solve the following
system of equations
by graphing.
– x + 3y = 6 and
3x – 9y = 9 (6, 4)
(0, 2)
(3, 0)
(6, 0) (6, 1)
x (0, 1) First, graph – x + 3y = 6.
Second, graph 3x – 9y = 9.
The lines APPEAR to be parallel.
MartinGay, Beginning and Intermediate Algebra, 4ed Continued.
10 Finding a Solution by Graphing
Example continued:
Although the lines appear to be parallel, you still need to check
that they have the same slope. You can do this by solving for y.
First equation,
– x + 3y = 6
3y = x + 6
Add x to both sides.
y= 1
3 x+2 Divide both sides by 3. Second equation,
3 x – 9y = 9 –9y = –3x + 9
y= 1
3 x–1 Subtract 3x from both sides.
Divide both sides by –9. 1
3 Both lines have a slope of , so they are parallel and do not
intersect. Hence, there is no solution to the system .
MartinGay, Beginning and Intermediate Algebra, 4ed 11 Finding a Solution by Graphing
y Example:
Solve the following
system of equations
by graphing.
x = 3y – 1 and
2x – 6y = –2 (5, 2)
(1, 0) (2, 1) x (4, 1)
(7, 2) First, graph x = 3y – 1.
Second, graph 2x – 6y = –2.
The lines APPEAR to be identical.
MartinGay, Beginning and Intermediate Algebra, 4ed Continued.
12 Finding a Solution by Graphing
Example continued: Although the lines appear to be identical, you still need to check
that they are identical equations. You can do this by solving for y.
First equation,
x = 3y – 1
3y = x + 1
Add 1 to both sides.
y= 1
x
3 + 1
3 Divide both sides by 3. Second equation,
2x – 6y = – 2
–6y = – 2x – 2
y= 1
x
3 + 1
3 Subtract 2x from both sides.
Divide both sides by 6. The two equations are identical, so the graphs must be identical.
There are an infinite number of solutions to the system (all the
points on the line).
MartinGay, Beginning and Intermediate Algebra, 4ed 13 Types of Systems
• There are three possible outcomes when
graphing two linear equations in a plane.
One point of intersection, so one solution
• Parallel lines, so no solution
• Coincident lines, so infinite # of solutions
• •
• If there is at least one solution, the system is
considered to be consistent.
If the system defines distinct lines, the
equations are independent.
MartinGay, Beginning and Intermediate Algebra, 4ed 14 Types of Systems
Since there are only three possible outcomes
with two lines in a plane, we can determine
how many solutions of the system there will
be without graphing the lines.
Change both linear equations into slopeintercept form.
We can then easily determine if the lines
intersect, are parallel, or are the same line.
MartinGay, Beginning and Intermediate Algebra, 4ed 15 Types of Systems
Example:
How many solutions does the following system have?
3x + y = 1 and 3x + 2y = 6
Write each equation in slopeintercept form.
First equation,
3x + y = 1
y = –3x + 1
Subtract 3x from both sides.
Second equation,
3x + 2y = 6
2y = –3x + 6
Subtract 3x from both sides.
3
y = − x+3
2 Divide both sides by 2. The lines are intersecting lines (since they have different slopes), so
there is one solution.
MartinGay, Beginning and Intermediate Algebra, 4ed 16 Types of Systems
Example:
How many solutions does the following system have?
3x + y = 0 and 2y = –6x
Write each equation in slopeintercept form.
First equation,
3x + y = 0
y = – 3x
Subtract 3x from both sides.
Second equation,
2y = –6x
y = – 3x
Divide both sides by 2.
The two lines are identical, so there are infinitely many solutions.
MartinGay, Beginning and Intermediate Algebra, 4ed 17 Types of Systems
Example:
How many solutions does the following system have?
2x + y = 0 and y = –2x + 1
Write each equation in slopeintercept form.
First equation,
2x + y = 0
y = –2x
Subtract 2x from both sides.
Second equation,
y = –2x + 1
This is in slopeintercept form.
The two lines are parallel lines (same slope, but different yintercepts), so there are no solutions.
MartinGay, Beginning and Intermediate Algebra, 4ed 18 § 4.2 Solving Systems of Linear
Equations by Substitution The Substitution Method
Another method (beside getting lucky with
trial and error or graphing the equations) that
can be used to solve systems of equations is
called the substitution method.
You solve one equation for one of the
variables, then substitute the new form of the
equation into the other equation for the solved
variable. MartinGay, Beginning and Intermediate Algebra, 4ed 20 The Substitution Method
Example:
Solve the following system using the substitution method.
3x – y = 6 and – 4x + 2y = –8 Solving the first equation for y,
3x – y = 6
–y = –3x + 6
y = 3x – 6 Subtract 3x from both sides.
Multiply both sides by – 1. Substitute this value for y in the second equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8
–4x + 6x – 12 = –8
2x – 12 = –8
2x = 4
x=2 Replace y with result from first equation.
Use the distributive property.
Simplify the left side.
Add 12 to both sides.
Divide both sides by 2. MartinGay, Beginning and Intermediate Algebra, 4ed Continued.
21 The Substitution Method
Example continued: Substitute x = 2 into the first equation solved for y.
y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0
Our computations have produced the point (2, 0).
Check the point in the original equations.
First equation,
3x – y = 6
3(2) – 0 = 6
true
Second equation,
–4x + 2y = –8
–4(2) + 2(0) = –8
true
The solution of the system is (2, 0).
MartinGay, Beginning and Intermediate Algebra, 4ed 22 The Substitution Method
Solving a System of Linear Equations by the
Substitution Method
1)
2)
3)
4)
5) Solve one of the equations for a variable.
Substitute the expression from step 1 into the
other equation.
Solve the new equation.
Substitute the value found in step 3 into
either equation containing both variables.
Check the proposed solution in the original
equations.
MartinGay, Beginning and Intermediate Algebra, 4ed 23 The Substitution Method
Example:
Solve the following system of equations using the
substitution method.
y = 2x – 5 and 8x – 4y = 20 Since the first equation is already solved for y, substitute
this value into the second equation.
8x – 4y = 20
8x – 4(2x – 5) = 20
Replace y with result from first equation.
8x – 8x + 20 = 20
Use distributive property.
20 = 20
Simplify left side. Continued. MartinGay, Beginning and Intermediate Algebra, 4ed 24 The Substitution Method
Example continued:
When you get a result, like the one on the previous
slide, that is obviously true for any value of the
replacements for the variables, this indicates that the
two equations actually represent the same line.
There are an infinite number of solutions for this
system. Any solution of one equation would
automatically be a solution of the other equation.
This represents a consistent system and the linear
equations are dependent equations.
MartinGay, Beginning and Intermediate Algebra, 4ed 25 The Substitution Method
Example:
Solve the following system of equations using the substitution
method.
3x – y = 4 and 6x – 2y = 4
Solve the first equation for y.
3x – y = 4
– y = – 3x + 4
Subtract 3x from both sides.
y = 3x – 4
Multiply both sides by –1.
Substitute this value for y into the second equation.
6x – 2y = 4
6x – 2(3x – 4) = 4
Replace y with the result from the first equation.
6x – 6x + 8 = 4
Use distributive property.
8=4
Simplify the left side.
Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 26 The Substitution Method
Example continued:
When you get a result, like the one on the previous
slide, that is never true for any value of the
replacements for the variables, this indicates that the
two equations actually are parallel and never
intersect.
There is no solution to this system.
This represents an inconsistent system, even though
the linear equations are independent.
MartinGay, Beginning and Intermediate Algebra, 4ed 27 § 4.3 Solving Systems of Linear
Equations by Addition The Addition Method
Another method that can be used to solve
systems of equations is called the addition or
elimination method.
You multiply both equations by numbers that
will allow you to combine the two equations
and eliminate one of the variables. MartinGay, Beginning and Intermediate Algebra, 4ed 29 The Addition Method
Example:
Solve the following system of equations using the addition method.
6x – 3y = –3 and 4x + 5y = –9
Multiply both sides of the first equation by 5 and the second
equation by 3.
First equation,
5(6x – 3y) = 5(–3)
30x – 15y = –15
Use the distributive property.
Second equation,
3(4x + 5y) = 3(–9)
12x + 15y = –27
Use the distributive property.
Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 30 The Addition Method
Example continued:
Combine the two resulting equations (eliminating the
variable y).
30x – 15y = –15
12x + 15y = –27
42x
= –42
x = –1 Divide both sides by 42. Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 31 The Addition Method
Example continued:
Substitute the value for x into one of the original
equations.
6x – 3y = –3
6(–1) – 3y = –3
– 6 – 3y = – 3
–3y = –3 + 6 = 3
y = –1 Replace the x value.
Simplify the left side.
Add 6 to both sides and simplify.
Divide both sides by –3. Our computations have produced the point (–1, –1).
Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 32 The Addition Method
Example continued:
Check the point in the original equations.
First equation,
6x – 3y = –3
6(–1) – 3(–1) = –3
Second equation,
4x + 5y = –9
4(–1) + 5(–1) = –9 true true The solution of the system is (–1, –1). MartinGay, Beginning and Intermediate Algebra, 4ed 33 The Addition Method
Solving a System of Linear Equations by the Addition or
Elimination Method
1)
2)
3)
4)
5)
6) Rewrite each equation in standard form, eliminating
fractional coefficients.
If necessary, multiply one or both equations by a number
so that the coefficients of a chosen variable are opposites.
Add the equations.
Find the value of one variable by solving the equation from
step 3.
Find the value of the second variable by substituting the
value found in step 4 into either original equation.
Check the proposed solution in the original equations.
MartinGay, Beginning and Intermediate Algebra, 4ed 34 The Addition Method
Example:
Solve the following system of equations using the
addition method.
2
1
3
x+ y =−
3
4
2
1
1
x − y = −2
2
4 First multiply both sides of the equations by a number
that will clear the fractions out of the equations.
Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 35 The Addition Method
Example continued:
Multiply both sides of each equation by 12. (Note: you don’t
have to multiply each equation by the same number, but in this
case it will be convenient to do so.)
First equation,
2
1
3
x+ y =−
3
4
2
1
2
12 x +
4
3 3
y = 12 − 2 8 x + 3 y = −18 Multiply both sides by 12.
Simplify both sides.
Continued. MartinGay, Beginning and Intermediate Algebra, 4ed 36 The Addition Method
Example continued:
Second equation, 1
1
x − y = −2
2
4
1
1
12 x − y = 12( − 2 )
4
2
6 x − 3 y = −24 Combine the two equations.
8x + 3y = – 18
6x – 3y = – 24
14x
= – 42
x = –3 Multiply both sides by 12.
Simplify both sides. Divide both sides by 14. MartinGay, Beginning and Intermediate Algebra, 4ed Continued.
37 The Addition Method
Example continued:
Substitute the value for x into one of the original
equations.
8x + 3y = –18
8(–3) + 3y = –18
–24 + 3y = –18
3y = –18 + 24 = 6
y=2 Our computations have produced the point (–3, 2).
Continued.
MartinGay, Beginning and Intermediate Algebra, 4ed 38 The Addition Method
Example continued:
Check the point in the original equations. (Note: Here you should
use the original equations before any modifications, to detect any
computational errors that you might have made.)
First equation,
Second equation,
2
1
3
x+ y =−
3
4
2 1
1
x − y = −2
2
4 2
1
3
(−3) + (2) = −
3
4
2 1
1
(−3) − (2) = −2
2
4 1
3
−2+ = −
2
2 31
− − = −2
22 true true The solution is the point (–3, 2).
MartinGay, Beginning and Intermediate Algebra, 4ed 39 Special Cases
In a similar fashion to what you found in the
last section, use of the addition method to
combine two equations might lead you to
results like . . .
5 = 5 (which is always true, thus indicating that
there are infinitely many solutions, since the two
equations represent the same line), or
0 = 6 (which is never true, thus indicating that
there are no solutions, since the two equations
represent parallel lines).
MartinGay, Beginning and Intermediate Algebra, 4ed 40 § 4.5 Systems of Linear
Equations and Problem
Solving Problem Solving Steps
Steps in Solving Problems
1) 1)
2)
3) Understand the problem.
•
Read and reread the problem.
•
Choose a variable to represent the unknown.
•
Construct a drawing, whenever possible.
•
Propose a solution and check.
Translate the problem into two equations.
Solve the system of equations.
Interpret the results.
•
Check proposed solution in the problem.
•
State your conclusion.
MartinGay, Beginning and Intermediate Algebra, 4ed 42 Finding an Unknown Number
Example:
One number is 4 more than twice the second number. Their
total is 25. Find the numbers.
1.) Understand
Read and reread the problem. Suppose that the second number
is 5. Then the first number, which is 4 more than twice the
second number, would have to be 14 (4 + 2•5).
Is their total 25? No: 14 + 5 = 19. Our proposed solution is
incorrect, but we now have a better understanding of the
problem.
Since we are looking for two numbers, we let
x = first number
y = second number
Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 43 Finding an Unknown Number
Example continued:
2.) Translate x = 4 + 2y x + y = 25 Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 44 Finding an Unknown Number
Example continued:
3.) Solve
We are solving the system x = 4 + 2y and x + y = 25 Using the substitution method, we substitute the solution for x
from the first equation into the second equation.
x + y = 25
(4 + 2y) + y = 25
4 + 3y = 25
3y = 21
y=7 Replace x with result from first equation.
Simplify left side.
Subtract 4 from both sides and simplify.
Divide both sides by 3. Now we substitute the value for y into the first equation.
Continued
x = 4 + 2y = 4 + 2(7) = 4 + 14 = 18
MartinGay, Beginning and Intermediate Algebra, 4ed 45 Finding an Unknown Number
Example continued:
4.) Interpret
Check: Substitute x = 18 and y = 7 into both of the equations.
First equation,
x = 4 + 2y
18 = 4 + 2(7) true Second equation,
x + y = 25
18 + 7 = 25 true State: The two numbers are 18 and 7. MartinGay, Beginning and Intermediate Algebra, 4ed 46 Solving a Problem about Prices
Example:
Hilton University Drama club sold 311 tickets for a play.
Student tickets cost 50 cents each; nonstudent tickets cost
$1.50. If total receipts were $385.50, find how many tickets
of each type were sold.
1.) Understand
Read and reread the problem. Suppose the number of students
tickets was 200. Since the total number of tickets sold was
311, the number of nonstudent tickets would have to be 111
(311 – 200).
Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 47 Solving a Problem about Prices
Example continued:
1.) Understand (continued)
Are the total receipts $385.50? Admission for the 200 students
will be 200($0.50), or $100. Admission for the 111 nonstudents will be 111($1.50) = $166.50. This gives total receipts
of $100 + $166.50 = $266.50. Our proposed solution is
incorrect, but we now have a better understanding of the
problem.
Since we are looking for two numbers, we let
s = the number of student tickets
n = the number of nonstudent tickets
Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 48 Solving a Problem about Prices
Example continued:
2.) Translate
Hilton University Drama club sold 311 tickets for a play.
s + n = 311
total receipts were $385.50 0.50s Total
receipts Admission for
non students Admission for
students
+ 1.50n = 385.50
Continued MartinGay, Beginning and Intermediate Algebra, 4ed 49 Solving a Problem about Prices
Example continued:
3.) Solve
We are solving the system s + n = 311 and 0.50s + 1.50n = 385.50
Since the equations are written in standard form (and we might like
to get rid of the decimals anyway), we’ll solve by the addition
method. Multiply the second equation by –2.
s + n = 311
−2(0.50s + 1.50n) = −2(385.50) simplifies to s + n = 311
−s – 3n = −771
−2n = −460
n = 230 Now we substitute the value for n into the first equation.
s + n = 311
s + 230 = 311
s = 81
⇒
⇒
MartinGay, Beginning and Intermediate Algebra, 4ed Continued
50 Solving a Problem about Prices
Example continued:
4.) Interpret
Check: Substitute s = 81 and n = 230 into both of the equations.
First equation,
s + n = 311
81 + 230 = 311 true Second equation,
0.50s + 1.50n = 385.50
0.50(81) + 1.50(230) = 385.50
40.50 + 345 = 385.50 true State: There were 81 student tickets and 230 nonstudent
tickets sold.
MartinGay, Beginning and Intermediate Algebra, 4ed 51 Finding Rates
Example:
Terry Watkins can row about 10.6 kilometers in 1 hour
downstream and 6.8 kilometers upstream in 1 hour. Find how fast
he can row in still water, and find the speed of the current.
1.) Understand
Read and reread the problem. We are going to propose a
solution, but first we need to understand the formulas we will be
using. Although the basic formula is d = r • t (or r • t = d), we
have the effect of the water current in this problem. The rate
when traveling downstream would actually be r + w and the rate
upstream would be r – w, where r is the speed of the rower in
still water, and w is the speed of the water current.
Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 52 Finding Rates
Example continued:
1.) Understand (continued)
Suppose Terry can row 9 km/hr in still water, and the water
current is 2 km/hr. Since he rows for 1 hour in each direction,
downstream would be (r + w)t = d or (9 + 2)1 = 11 km
Upstream would be (r – w)t = d or (9 – 2)1 = 7 km
Our proposed solution is incorrect (hey, we were pretty close
for a guess out of the blue), but we now have a better
understanding of the problem.
Since we are looking for two rates, we let
r = the rate of the rower in still water
w = the rate of the water current
Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 53 Finding Rates
Example continued:
2.) Translate
rate
downstream ( r + w) time
downstream • rate
upstream ( r – w) 1 distance
downstream = time
upstream • 1 10.6
distance
upstream = 6.8
Continued MartinGay, Beginning and Intermediate Algebra, 4ed 54 Finding Rates
Example continued:
3.) Solve
We are solving the system r + w = 10.6 and r – w = 6.8 Since the equations are written in standard form, we’ll solve by the
addition method. Simply combine the two equations together.
r + w = 10.6
r – w = 6.8
2r = 17.4
r = 8.7 Now we substitute the value for r into the first equation.
r + w = 10.6 ⇒ 8.7 + w = 10.6 ⇒ w = 1.9 MartinGay, Beginning and Intermediate Algebra, 4ed Continued
55 Finding Rates
Example continued:
4.) Interpret
Check: Substitute r = 8.7 and w = 1.9 into both of the
equations.
First equation,
(r + w)1 = 10.6
(8.7 + 1.9)1 = 10.6 true Second equation,
(r – w)1 = 1.9
(8.7 – 1.9)1 = 6.8 true State: Terry’s rate in still water is 8.7 km/hr and the rate of
the water current is 1.9 km/hr.
MartinGay, Beginning and Intermediate Algebra, 4ed 56 Solving a Mixture Problem
Example:
A Candy Barrel shop manager mixes M&M’s worth $2.00 per
pound with trail mix worth $1.50 per pound. Find how many
pounds of each she should use to get 50 pounds of a party mix
worth $1.80 per pound.
1.) Understand
Read and reread the problem. We are going to propose a
solution, but first we need to understand the formulas we will be
using. To find out the cost of any quantity of items we use the
formula
price per unit • number of units = price of all units Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 57 Solving a Mixture Problem
Example continued:
1.) Understand (continued)
Suppose the manage decides to mix 20 pounds of M&M’s.
Since the total mixture will be 50 pounds, we need 50 – 20 = 30
pounds of the trail mix. Substituting each portion of the mix
into the formula,
M&M’s
$2.00 per lb • 20 lbs = $40.00
trail mix $1.50 per lb • 30 lbs = $45.00 Mixture $1.80 per lb • 50 lbs = $90.00 Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 58 Solving a Mixture Problem
Example continued:
1.) Understand (continued)
Since $40.00 + $45.00 ≠ $90.00, our proposed solution is
incorrect (hey, we were pretty close again), but we now have a
better understanding of the problem.
Since we are looking for two quantities, we let
x = the amount of M&M’s
y = the amount of trail mix Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 59 Solving a Mixture Problem
Example continued:
2.) Translate
Fifty pounds of party mix
x + y = 50 Using price per unit • number of units = price of all units
Price of
M&M’s
2x Price of
trail mix
+ 1.5y Price of
mixture
= 1.8(50) = 90 Continued
MartinGay, Beginning and Intermediate Algebra, 4ed 60 Solving a Mixture Problem
Example continued:
3.) Solve
We are solving the system x + y = 50 and 2x + 1.50y = 90
Since the equations are written in standard form (and we might like
to get rid of the decimals anyway), we’ll solve by the addition
method. Multiply the first equation by 3 and the second equation
by –2.
3(x + y) = 3(50) 3x + 3y = 150 simplifies to – 4x – 3y = – 180 –2(2x + 1.50y) = –2(90) –x = – 30
x = 30 Now we substitute the value for x into the first equation.
x + y = 50 ⇒ 30 + y = 50 ⇒ y = 20 MartinGay, Beginning and Intermediate Algebra, 4ed Continued
61 Solving a Mixture Problem
Example continued:
4.) Interpret
Check: Substitute x = 30 and y = 20 into both of the equations.
First equation,
x + y = 50
30 + 20 = 50 true Second equation,
2x + 1.50y = 90
2(30) + 1.50(20) = 90
60 + 30 = 90 true State: The store manager needs to mix 30 pounds of M&M’s
and 20 pounds of trail mix to get the mixture at $1.80 a pound.
MartinGay, Beginning and Intermediate Algebra, 4ed 62 ...
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 Winter '11
 J.King
 Linear Equations, Addition, Equations, Systems Of Linear Equations, Elementary algebra

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