systems of equations & inequalities
DESCRIPTION
Systems of Equations & Inequalities. Algebra I . Systems of Equations. Definition . Ways to solve. means two or more linear equations If these two linear equations intersect, that point of intersection is called the solution. Graphing By hand On calculator Substitution Elimination. - PowerPoint PPT PresentationTRANSCRIPT
Systems of Equations & Inequalities
Algebra I
Systems of Equations
Definition Ways to solve means two or more
linear equations
If these two linear equations intersect, that point of intersection is called the solution
Graphing By hand On calculator
Substitution
Elimination
Graphing Method
This method of solving equations is by graphing each equation on a coordinate graph.
The coordinates of the intersection will be the solution to the system.
Graphing System of linear equations with calculator:the following two lines is a system:
y=x+1y=2x
What is the solution?The point
(1,2) is where the
two lines intersect.
Practice Problem:Use Calculator:The following two lines is a system:
y = 2x+1 y = 4x - 1What is the solution?
The solution of this system is the point of
intersection : (1,3)
X = 4 + y x – 3y = 4
~ Find three values for x and y that satisfy each equation.x = 4 + y x – 3y = 4
~ Graph these points and draw straight lines
The point where the two lines cross (4, 0) is the solution of the system.
Graphing System of linear equations by hand ~ Using Table:
2y = 4x + 2 2y = -x + 7
~ Get the equations in slope intercept form , so you have y-intercept and slope to graph line:2y = 4x + 2
y = (4x + 2) / 2y = 2x +1
2y = 8x - 2y = (8x - 2) / 2y = 4x – 1
The point where the two lines cross (1, 3) is the solution of the system.
Solve the following system of linear equations graphing the system by hand using slope intercept form:
Graphing and getting Parallel Lines If the lines are parallel
they do not intersect there is no solution to that system.
Substitution Method
Sometimes a system is more easily solved by the substitution method. This method involves substituting
one equation into another.
x = y + 8 x + 3y = 48If an equation is not already solved for one variable then you need to solve for either x or y in order to substitute.
From the first equation, substitute ( y + 8 ) for x in the second equation.
(y + 8) + 3y = 48
Now solve for y. Simplify by combining y's
y + 8 + 3y = 48 4y + 8 = 48 4y = 48 – 8 4y = 40 y = 40/4 y = 10
Now insert y = 10 in one of the original equations.
x = y + 8x = 10 + 8x = 18
Solution:x = 18 , y = 10(18, 10)
y = 2x + 12y = 3x - 2If an equation is not already solved for one variable then you need to solve for either x or y in order to substitute.
From the first equation, substitute ( 2x + 1 ) for y in the second equation.
2(2x + 1) = 3x - 2
Now solve for x:
4x + 2 = 3x - 2 4x – 3x + 2 = - 2 x + 2 = - 2 x = - 2 – 2 x = - 4
Now insert x = - 4 in one of the original equations.
y = 2x + 1y = 2(-4) + 1y = -8 + 1y = -7
Solution:x = -4 , y = -7(-4, -7)
x + y = 11 3x - y = 5Solve the first equation for either x or yNeed one variable equaling an expression so that you can substitute
From the first equation, for y: x + y = 11 y = 11 – x
Substitute 11 - x for y in the second equation:
3x – (11 – x) = 5
Now solve for x. Simplify by combining x‘s:
3x - 11 + x = 5 4x = 5 + 11 4x = 16 x = 16 / 4 x = 4
Substitute 4 for x in either equation and solve for y:
4 + y = 11y = 11 - 4 y = 7
Solution:x = 4 , y = 7(4, 7)
2x – 3y = 6x + y = -12Solve the first equation for either x or yNeed one variable equaling an expression so that you can substitute
From the second equation, for y:
x + y = -12 y = -12 – x
Substitute -12 - x for y in the second equation:
2x – 3(-12 – x) = 6
Now solve for x: 2x + 36 + 3x = 6 5x + 36 = 6 5x = 6 - 36 5x = - 30 x = - 30 / 5 x = - 6
Substitute – 6 for x in either equation and solve for y:
- 6 + y = -12y = -12 + 6 y = - 6
Solution:x = - 6 , y = - 6(-6, -6)
Individual Practice Problems: Substitution Method Problem 1:
y= x + 1 2y= 3x
Problem 2: y – 5x = - 1 2y= 3x + 12
Problem 3: y = 3x + 1 4y = 12x + 4
Problem 4: y – 3x = 1 4y = 12x + 3
Answers to Practice Problems Problem 1:
y= x + 1 2y= 3x
Answers to Practice Problems Problem 2: y – 5x = - 1 2y= 3x + 12
Solve 1st equation for y: y = 5x - 1
Answers to Practice Problems Problem 3:
y = 3x + 1 4y = 12x + 4
Answers to Practice Problems Problem 4: y – 3x = 1 4y = 12x + 3
Solve 1st equation for y: y = 3x + 1
Elimination MethodHelps eliminate one variable so that you can solve
for the remaining variable.
Steps for Elimination Method Multiply one or both equations by some
number to make the number in front of one of the letters (unknowns) the same in each equation.
Add or subtract the two equations to eliminate one letter.
Solve for the other unknown. Insert the value of the first unknown in
one of the original equations to solve for the second unknown.
x + y = 7x – y = 3~ You have opposites between the two equations a positive and negative y
~ With opposites you can skip step one and add equations together
x + y = 7 + x – y = 3 2x = 10
solve for x: 2x = 10 x = 10 / 2 x = 5
replace x value and find y: 5 + y = 7 y = 7 – 5 y = 2
Solution:x = 5 and y = 2(5, 2)
3x + 3y = 242x + y = 13~ You need to have the same variable in both equations…
~ With the same variable you can subtract equations together
Need to get the same y’s in both:
change 2nd equation to have 3y
3(2x + y) = 3(13)
6x + 3y = 39 3x + 3y =
24 - 6x + 2y =
39 -3x = -
15
solve for x:
-3x = -15 x = -15 / -3 x = 5
replace x value and find y: 2x + y = 13 2(5) + y = 13 10 + y = 13 y = 13 - 10 y = 3
Solution:x = 5 and y = 2(5, 2)
5x + 3y = 7 3x - 5y = -23~ You need to have the same variable in both equations…
~ With the same variable you can subtract equations together
Multiply the second equation by 5 to make the x-coefficient a multiple of 5:
5(3x - 5y) = 5(-23) 15x - 25y = -115
multiply the first equation by 3, to get the same x-coefficient:
3(5x + 3y) = 3(7) 15x + 9y = 21
15x - 25y = -115 -15x + 9y = 21
-34y = -136
solve for y: -34y = -136 y = -136 / -34 y = 4
replace 4 for y in one of the original equations and find x: 5x + 3y = 7 5x + 3(4)= 7 5x + 12 = 7 5x = 7 - 12 5x = -5 x = -5 / 5 x = -1
Solution:x = -1 & y = 4(-1, 4)
Individual Practice Problems: Elimination Method Problem 1:
y = x + 1 y = –x
Problem 2: 4x – 2y = 14 x + 2y = 6
Problem 3: 2x + 3y = 17 4x – 3y = 1
Problem 4: y = 2x + 1 y = -4x + 1
Answers to Practice Problems Problem 1:
y = x + 1 y = –x
Answers to Practice Problems Problem 2:
4x – 2y = 14 x + 2y = 6
Answers to Practice Problems Problem 3:
2x + 3y = 17 4x – 3y = 1
Answers to Practice Problems Problem 4:
y = 2x + 1 y = -4x + 1
Need to get same coefficient fro the x variables
Story Problemswith Systems of Equations
Steps to Solve Define the variables Set up the equations Solve the system
Graphing Substitution Elimination
Put answer back into form of the problem
The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? Define Variables: number of adults: a
number of children: c Write Equations: total number:
a + c = 2200 total income: 4a + 1.5c = 5050
Solve for variables: Solve 1st equation for a: a = 2200 – c Use (2200 – c) for a in
2nd to find c:
4(2200 – c) + 1.5c = 5050
8800 – 4c + 1.5c =
5050 8800 – 2.5c = 5050 -2.5c = 5050 - 8800 –2.5c = –3750 c = -3750 / -2.5 c = 1500 Use 1500 for c to find a: a = 2200 – (1500) =
700 Write out answer There were 1500 children
and 700 adults.
A landscaping company placed two orders with a nursery. The first order was for 13bushes and 4 trees, and totaled $487. The second order was for 6 bushes and 2 trees, and totaled $232. The bills do not list the per-item price. What were the costs of one bush and of one tree?
Define Variables: b = the number of bushes t = the number of trees Write Equations: first order: 13b + 4t = 487
second order: 6b + 2t = 232 Solve for variables: use elimination
Multiplying the second by –2 -2 (6b + 2t) = -2(232) -12b – 4t = -464
13b + 4t = 487 + –12b – 4t = –464
b = 23
Use 23 for b to find t: 13b + 4t = 487 13(23) + 4t = 487 299 + 4t = 487 4t = 487 – 299 4t = 188 t = 188 / 4 t = 47 Write out answer Bushes
cost $23 each; trees cost $47 each.
The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number. Define Variables: t = tens digit of the original number u = units (or "ones") digit Write Equations: 1st Sentence gives you:
t + u = 7 Think about numbers to get original numbers:
26 is 10 times 2, plus 6 times 1 two-digit number will be ten times the (tens digit), plus one
times the (units digit) original number: 10t + 1u New number has the digits reversed:(switch the place of t & u)
new number: 10u + 1t (new number) is (old number) increased by (twenty-seven) 10u + 1t = 10t + 1u + 27
The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number. ~ Continued
System to solve: t + u = 7 10u + t = 10t + u + 27
Simplify the second equation: 10u + t = 10t + u + 27 10u – u + t – 10t = 27 9u – 9t = 27 9(u – t) = 27 u – t = 27 / 9 u – t = 3
Write to look like other: -t + u = 3
Solve for variables: Use elimination! t + u = 7 + -t + u = 3 2u = 10 u = 10 / 2 u = 5 Use 5 for u to find t: t + u = 7 t + 5 = 7 t = 7 – 5 t = 2 Write out answer The number is 25.
A passenger jet took three hours to fly 1800 miles in the direction of the jetstream. The return trip against the jetstream took four hours. What was the jet's speed in still air and the jetstream's speed?
Define Variables: p = the plane's speedometer reading w = the windspeed Write Equations:
the plane is going "with" the wind the two speeds will be added together when the plane is going "against" the wind the windspeed will be subtracted from
the plane's speedometer reading the distance equation will be:
(the combined speed) times (the time at that speed) equals (the total distance travelled)
with the jetstream: against the jetstream: (p + w)(3) = 1800 (p – w)(4) = 1800 p + w = 1800 / 3 p – w = 1800 / 4 p + w = 600 p – w = 450
A passenger jet took three hours to fly 1800 miles in the direction of the jetstream. The return trip against the jetstream took four hours. What was the jet's speed in still air and the jetstream's speed?
Solve for variables: use elimination Method p + w = 600
+ p – w = 450 2p = 1050 p = 1050 / 2 p = 525
Use 525 for p to find w: p + w = 600 525 + w = 600
w = 600 – 525 w = 75
Write out answer The jet's speed
was 525 mph and the jetstream windspeed was 75 mph.
Systems of Inequalities
Solving Systems of Linear Inequalities
The solution to a system of linear inequalities is shown by graphing them.
Need to put the inequalities into Slope-Intercept Form, y = mx + b.
Solving Systems of Linear Inequalities
Lines on the graph
o If the inequality is < or >, make the lines dotted.
o If the inequality is < or >, make the lines solid.
Solving Systems of Linear Inequalities
The solution also includes points not on the line, so you need to shade the region of the graph:
above the line for ‘y >’ or ‘y ’. below the line for ‘y <’ or ‘y ≤’.
Solving Systems of Linear Inequalities
Example: a: 3x + 4y > - 4
b: x + 2y < 2
Put in Slope-Intercept Form:
) 3 4 44 3 4
3 14
a x yy x
y x
) 2 22 2
1 12
b x yy x
y x
Solving Systems of Linear Inequalities
a: dotted
shade above
b:dottedshade below
Graph each line, make dotted or solid and shade the correct area.
Example, continued:
3: 14
a y x 1: 12
b y x
Solving Systems of Linear Inequalities
a: 3x + 4y > - 4
3: 14
a y x
Solving Systems of Linear Inequalities
a: 3x + 4y > - 4
b: x + 2y < 2
3: 14
a y x
1: 12
b y x
Solving Systems of Linear Inequalities
a: 3x + 4y > - 4 b: x + 2y < 2
The area between the green arrows is the region of overlap and thus the solution.