Download Nonlinear Systems of Equations

Review
Decompose into partial fractions.
9 x + 21
x + 2 x − 15
Nonlinear Systems of Equations
2
Objective: To solve a nonlinear
system of equations.
Review
What?
{
{
{
Review
How?
System – 2 or more equations
together
Solution of system – any ordered
pair that
pa
t at makes
a es ______ equat
equations
o s
true
Possible solutions:
1. One point
2. No solution
3. Infinitely many solutions
{
1. Graphing
2. Substitution
3. Elimination
Review
Solve using Substitution.
{
{
2x – y = 6
y = 5x
What methods have we used to
solve linear systems of equations?
Review
Steps for using SUBSTITUTION
1.
2.
3.
4.
5.
______ one equation for one variable.
(Hint: Look for an equation already solved for
one of the variables or for a variable with a
coefficient of 1.)
_________ into the other equation.
Solve this equation to find a ______ for a
variable.
Substitute again to find the value of the
other variable.
______.
1
Review
Solve using elimination.
What if one of the equations is not already
(or easily) solved for a variable?
{ Use
ELIMINATION!
{
{
3x + 5y = 11
2x + 3y = 7
{ Elimination
method – to use
math operations to eliminate a
variable
Review
STEPS for ELIMINATION
With equations in ______ form, look for a
variable that has OPPOSITE coefficients. If
found, ADD the two equations. Solve,
substitute, solve.
If no opposite coefficients, look for variables
with the SAME coefficients. If found,
SUBTRACT one equation from the other.
Then, solve, substitute, solve.
1.
2.
What if neither variable can be
eliminated by simply adding or
subtracting?
{
{
We may have to MULTIPLY before
adding or subtracting!
An Example…
(Just watch and listen.)
{ Solve
the following system of
equations:
What’s New?
•
{
{
{
A non-linear system is one in which
one or more of the equations has a
graph that is not linear.
With non
non-linear
linear systems, the solution could
be one or more points of intersection or
no point of intersection.
We’ll solve non-linear systems using
substitution or elimination.
A graph of the system will show the points
of intersection.
An Example…
{
We use the substitution method.
First, we solve equation (2) for y.
x2 + y2 = 9
(1) The graph is a circle.
2x − y = 3
2x − y = 3
(2) The graph is a line.
− y = −2 x + 3
y = 2x − 3
2
An Example… …
An Example…
{
Next, we substitute
y = 2x − 3 in equation (1) and solve
for x: x 2 + (2 x − 3) 2 = 9
x + 4 x − 12 x + 9 = 9
2
2
5 x 2 − 12 x = 0
x (5 x − 12) = 0
x = 0 or x =
{
{
Now, we substitute these numbers for x in equation
(2) and solve for y.
x=0
x = 12 / 5
y = 2x − 3
y = 2x − 3
y = 2(0) − 3
y = −3
y = 2( 125 ) − 3
y=
9
5
⎛ 12 9 ⎞
12
5
An Example…
(0, −3) and ⎜ , ⎟
⎝ 5 5⎠
Example
Check: (0, −3)
{
x2 + y2 = 9
2x − y = 3
02 + 32 = 9
2(0) − (−3) = 3
9 =9
3=3
Visualizing the
Solution
Solve by substitution.
{ x2 = y – 1
{ 4x – y = -1
⎛ 12 9 ⎞
⎜ , ⎟
⎝ 5 5⎠
2
2
x + y =9
2x − y = 3
Check:
( 125 ) +( 95)
2
2
=9
9 =9
2(125 ) −(95) = 3
3=3
Recognizing Graphs by Equation
{
{
{
{
no x or y with power > 1 = line
only one variable squared = parabola
x & y both squared = circle, ellipse, or
hyperbola
xy = special hyperbola
Example
{ Solve by substitution.
x + 2y = 0
(x – 1)2 + (y – 1)2 = 5
3
Another example to watch…
{
xy = 4
4
y=
x
Solve xy = 4 for y.
Solve the following system of
equations:
xy = 4
3x + 2y = −10
10
3 x + 2 y = −10
3 x + 2( 4x ) = −10
Substitute into
3x + 2y = −10.
3 x + 8x = −10
x ( 3 x + 8x ) = −10( x )
3x 2 + 8 = −10 x
3x 2 + 10 x + 8 = 0
x=
{
Use the quadratic
formula
to solve:
−10 ± 102 − 4(3 ⋅ 8)
x=
2(3)
−10 ± 100 − 96
6
−10 ± 4 −10 ± 2
=
x=
6
6
−10 + 2
−10 − 2
and
x=
6
6
−4
and − 2
x=
3
x=
3 x 2 + 10 x + 8 = 0
Substitute values of x to find y.
−b ± b 2 − 4ac
2a
3x + 2y = −10
x = −4/3
x = −2
3 ( −34 ) + 2 y = −10
3 ( −2 ) + 2 y = −10
−4 + 2 y = −10
−6 + 2 y = −10
2 y = −6
y = −3
{
Visualizing the
Solution
2 y = −4
y = −2
The solutions are
(−4/3, −3) and (−2, −2).
5 x 2 − 2 y 2 = −13
3 x 2 + 4 y 2 = 39
Need to watch another one?
{
Solve the system of equations:
5 x − 2 y = −13
2
2
3x 2 + 4 y 2 = 39
{
Solve by elimination.
Multiply equation (1) by 2 and
add to eliminate the y2 term.
10 x 2 − 4 y 2 = −26
3 x 2 + 4 y 2 = 39
= 13
13x 2
x =1
x = ±1
2
4
Substituting x = ±1 in equation (2) gives us:
x=1
x = -1
3 x 2 + 4 y 2 = 39
3 x 2 + 4 y 2 = 39
3(1) 2 + 4 y 2 = 39
3( − 1) 2 + 4 y 2 = 39
4 y 2 = 36
4 y 2 = 36
y =9
y2 = 9
2
y = ±3
{
{
Visualizing the Solution
All four pairs check,
so they are the
solutions.
y = ±3
The possible solutions are
(1, 3), (−1, 3), (−1, −3) and (1, −3).
Example
Solve by elimination.
{ 3x2 + 2y2 = 35
{ 4x2 + 3y2 = 48
Example
Solve by elimination.
{ y = x2 + 5
{ x2 + y2 = 25
5