Transcript No Slide Title

LINEAR AND QUADRATIC EQUATION
SYSTEMS
In this chapter, you will learn how
to:
 solve two simultaneous where at
least one is a non-linear equation
 solve three simultaneous linear
equations
Linear and quadratic equation systems
Linear equation
systems in two
variable
Linear equation
systems in three
variables
Simultaneous
equations: one linear
and one quadratic
Simultaneous
Equations: Two
Quadratics
Applications
Exercises
Linear equations system in two
variables
We have seen in Junior High school
Mathematics that a set of simultaneous
linear equations can be solved by either
the method of:



Elimination
Substitution
Combination both elimination and
substitution
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
In this lesson you will study two algebraic methods for solving
linear systems. The first method is called substitution.
THE SUBSTITUTION METHOD
1
Solve one of the equations for one of its variables.
2
Substitute expression from Step 1 into other equation
and solve for other variable.
3
Substitute value from Step 2 into revised equation
from Step 1. Solve.
The Substitution Method
Solve the linear system using
the substitution method.
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
SOLUTION
Solve Equation 2 for x.
x + 2y  2
Write Equation 2.
x  – 2y + 2
Revised Equation 2.
Substitute the expression for x into Equation 1 and solve for y.
3x + 4y  – 4
3(– 2y + 2) + 4y  – 4
y5
Write Equation 1.
Substitute – 2y + 2 for x.
Simplify.
The Substitution Method
Solve the linear system using
the substitution method.
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
Substitute the expression for x into Equation 1 and solve for y.
3x + 4y  – 4
3(– 2y + 2) + 4y  – 4
y5
Write Equation 1.
Substitute – 2y + 2 for x.
Simplify.
Substitute the value of y into revised Equation 2 and solve for x.
x  – 2y + 2
Write revised Equation 2.
x  – 2(5) + 2
Substitute 5 for y.
x  –8
Simplify.
The solution is (– 8, 5).
The Substitution Method
Solve the linear system using
the substitution method.
CHECK
3 x + 4y  – 4
x + 2y  2
Equation 1
Equation 2
Check the solution by substituting back into the original equation.
3x + 4y  – 4
?
3 (– 8) + 4 (5)  – 4
–4  –4
Write original equations.
Substitute x and y.
Solution checks.
x + 2y  2
?
– 8 + 2 (5)  2
22
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
In the first step of the previous
example, you could have solved for either x or y in either
Equation 1 or Equation 2. It was easiest to solve for x in
Equation 2 because the x-coefficient was 1. In general you
should solve for a variable whose coefficient is 1 or –1.
CHOOSING A METHOD
If neither variable has a coefficient of 1 or –1, you can still use
substitution. In such cases, however, the linear combination
method may be better. The goal of this method is to add the
equations to obtain an equation in one variable.
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
THE LINEAR COMBINATION METHOD
1
Multiply one or both equations by a constant to
obtain coefficients that differ only in sign for one
of the variables.
2
Add revised equations from Step 1. Combine like
terms to eliminate one of the variables. Solve for
remaining variable.
3
Substitute value obtained in Step 2 into either
original equation and solve for other variable.
The Linear Combination Method: Multiplying One Equation
Solve the linear system using the
linear combination method.
2 x – 4y  13
4 x – 5y  8
Equation 1
Equation 2
SOLUTION
Multiply the first equation by – 2 so that x-coefficients differ only in
sign.
2 x – 4y  13
•
–2
4 x – 5y  8
Add the revised equations
and solve for y.
– 4x + 8y  – 26
4 x – 5y  8
3y  –18
y  –6
The Linear Combination Method: Multiplying One Equation
2 x – 4y  13
4 x – 5y  8
Solve the linear system using the
linear combination method.
Add the revised equations and solve for y.
Equation 1
Equation 2
y  –6
Substitute the value of y into one of the original equations.
CHECK
Write Equation 1.
2
x
–
4y

13
You can check the solution algebraically using
the method shown in the previous example.
Substitute – 6 for y.
2 x – 4(– 6)  13
2 x + 24  13
x–
The solution is
(
–
Simplify.
11
2
11
, –6
2
Solve for x.
).
The Linear Combination Method: Multiplying Both Equations
Solve the linear system using the
linear combination method.
7 x – 12 y  – 22
– 5 x + 8 y  14
Equation 1
Equation 2
SOLUTION
Multiply the first equation by 2 and the second equation by 3
so that the coefficients of y differ only in sign.
7 x – 12 y  – 22 •
2
14 x – 24y  – 44
– 5 x + 8 y  14
3
– 15 x + 24y  42
•
Add the revised equations and
solve for x.
–x  –2
x 2
The Linear Combination Method: Multiplying Both Equations
Solve the linear system using the
linear combination method.
7 x – 12 y  – 22
– 5 x + 8 y  14
Add the revised equations and solve for x.
Equation 1
Equation 2
x2
Substitute the value of x into one of the original equations.
Solve for y.
– 5 x + 8 y  14
– 5 (2) + 8y  14
y=3
Write Equation 2.
Substitute 2 for x.
Solve for y.
The solution is (2, 3). Check the solution algebraically or graphically.
Linear Systems with Many or No Solutions
Solve the linear system
x – 2y  3
2x – 4y  7
SOLUTION
Since the coefficient of x in the first equation is 1, use
substitution.
x – 2y  3
x  2y + 3
Solve the first equation for x.
Linear Systems with Many or No Solutions
Solve the linear system
x  2y + 3
x – 2y  3
2x – 4y  7
Solve the first equation for x.
Substitute the expression for x into the second equation.
2x – 4y  7
Write second equation.
2(2 y + 3) – 4 y  7
Substitute 2 y + 3 for x.
67
Simplify.
Because the statement 6 = 7 is never true, there is no solution.
Linear Systems with Many or No Solutions
6 x – 10 y  12
– 15 x + 25y  – 30
Solve the linear system
SOLUTION
Since no coefficient is 1 or –1, use the linear combination
method.
6 x – 10 y  12
•
5
30 x – 50 y  60
– 15 x + 25 y  – 30 •
2
– 30 x + 50 y  –60
Add the revised equations.
00
Because the equation 0 = 0 is always true, there are
infinitely many solutions.
Exercise 1.
Work in pair and answer these questions:
 x  y  8
system: 
2 x  y  1
1.
Solve each
2.
Translate to a system of equations and
solve: The sum of two numbers is 82.
One is twelve more than the other.
Find the larger number.
Exercise 1
3. Translate to a system of equations and
solve: Alfa Rent-a-Car rents compact
cars at daily rate of $23.95 plus 20¢ per
kilometer. Giant Rent-a-Car rents
compact cars at daily rate of $22.95 plus
22¢ per kilometer. For what kilometer is
cost will be same?
Exercise 1
4. Abu works for 6 days which 4 days are
overtime to get Rp74.000,00. Budi works
for 5 days which 2 days are overtime to
get Rp55.000,00. Abu, Budi, and Catur
work under the same payment system.
Catur works for 5 days overtime. Find
the payment that he will receive.
Linear equation systems in three
variables
General form of linear equation systems in
three x, y, z
variables is:
 a x  b y  c z  d
1
1
1
1

 a 2 x  b2 y  c 2 z  d 2
a x  b y  c z  d
3
3
3
 3
whereas
a1 , a 2 , a 3 , b1 , b2 , b3 , c1 , c 2 , c 3 , d 1 , d 2 , d 3  R
Answering the linear equation system with
three
variables
is
determining
 x 0 , y 0 , z 0  that is a simultaneous
solution
from
the
equation system, so the solution set
is: ( x , y z )
0
0,
0
Example 1
Find the solution set of:  2 x  3 y  z  15
Solution:

 3 x  2 y  z  20
 x  y  3z  1

2 x  3 y  z  15  z  2 x  3 y  15
Substitute to: 3 x  2 y  z  20 and x  y  3 z  1,
3 x  2 y  ( 2 x  3 y  15 )  20 , and x  y  3 ( 2 x  3 y  15 )  1
 5 x  5 y  35
 -5x - 8y  - 44
 x y 7
 5 x  8 y  44
Then we will find the linear equation system:

x  y  7

 5 x  8 y  44
y
 7  x
And it is substituted into: 5 x + 8 y = 4 4 we will find:
5 x  8 ( 7  x )  44
 5 x  8 x  56  44
  3 x  12
 x  4 substitute
d to y  7  x , get y  7  4  3 , and x  4
And put the result into:
 the solution
z  2 ( 4 )  3 ( 3 )  15  2
set is ( 4 , 3 , 2 ) 
Exercise 2
 5 x  3 y  2 z  28

 7 x  4 y  z  24
 3 x  5 y  4 z  28

1.
Solve this system:
2.
Three lines 3x-y+1=0, 2x-y-3=0, and
x-ay-7=0 are concurrent. Find the value of a
The price of 2 guavas, 2 bananas, and a
mango are Rp1400,00. The price of a guava,
a banana, and 2 mangoes are Rp1300,00.
The price of 3 guavas, a banana, and a
mango are Rp1500,00. Find the price when
you buy a guava, a banana, and a mango.
3.
Exercise 2
4. Do the exercises in: … p. …. no…., p.
…. number
Simultaneous equations, one
linear and one quadratic
General form of this system is:
where
 mx  ny  k

2
y

ax
 bx  c

m , n , k , a , b , c  R and a  0
Solve the following system:
3x  y  1

2
y

4
x
 6x  1

(1)
(2)
Solution:
From the linear equation : y = 3x-1 substitute to (2)
3x  1  4x  6x  1  4x2  9x  2  0
2
(4x-1)(x-2)=0
 x = ¼ or x =2
The solution is either x = ¼ => y = - ¼ or x = 2=> y = 5
Exercise 3
1. Solve the following system: 
x  2y  4
 2
2
4
x

12
xy

9
y
9

2. Find the intersection of line y = 6x-26 and
parabola y  x 2  4 x  5 and sketch.
2
3. Find tangent to the parabola y  x at the
point A(3,9), and sketch
2
2
x

y
 25
4. The line m intersects the circle
at A(3,4) and B. Find coordinate of point B.
Simultaneous Equations:
Two Quadratics

Solution of system:
 y  ax 2  bx  c

2
y

mx
 nx  q

can be found by: ax 2  bx  c  mx 2  nx  q
or ( a  m ) x 2  (b  n ) x  ( c  q )  0
after x has been found from the last quadratic
equation then the value of y can be obtained
by substituting the value of x into one of
equations in the system
Find the intersection of parabolas
2
2
y

x
 3x  5
y  2 x  x  2 and
2 x  x  2  x  3x  5
2
2
 x  4x  3  0
2
 ( x  3 )( x  1)  0
 x = 3 or x = 1
When x = 3, y = 13, and x = 1, y = - 1
The intersection are (3,13) and (1, - 1)
Exercise 4
1.Find the value of m so that the line y = mx + 2
touches the parabola y  mx 2  mx  2 , also
find the coordinate of intersection.
2. If x and y are acute angles that satisfy the
following system:
2
2
 cos x  cos y  1

2
1
1
sin
x

cos
y

2
2

find the angles x and y in degrees.
Exercise
1
Exercise 2
Exercise 3
Exercise 4
Simultaneous equations: one is
linear and the other one is quadratic
General form of this system is:
 mx  ny  k

2
 y  ax  bx  c
whereas
m , n , k , a , b , c  R and a  0