Algebra III
Lesson 19
Nonlinear Systems – Factoring Exponentials
– Sum and Difference of Two Cubes
Nonlinear Systems
- More than one equation with at least one nonlinear equation
- Nonlinear equations involve equations where powers greater
than one on the variables, the variable is the power, or mixed
variables in one term.
-- Examples:
x2 + y = 7
y4 – x2 = 12
y = 4x
xy = 3
Example 19.1
Solve: (a)
(b)
x2 + y2 = 9
y –x =1
(circle)
(line)
Solving systems of equations is done by elimination or
substitution.
When there are mismatched powers or mixed terms between the
various equations in the system then substitution is the main
method of solution.
Before solving it can be helpful to consider the shapes of the
equations in the system by graphing the shapes (not necessarily
the exact equations). In this case a circle and a line.
In this case there are three possible ways to put a circle and a
line on the same graph.
(1)
(2)
(3)
Every time the graphs touch is a real solution to the system of equations.
This situation can have from 0 to 2 real solutions. The missing real
solutions might end up being imaginary. Or like in (2) you get the same
answer more than once.
Solve: (a)
(b)
x2 + y2 = 9
y –x =1
(circle)
(line)
Since there are mixed powers substitution is the best method of solution.
Solve (b) for y
y=x+1
(c)
Sub this into (a)
x2 + (x + 1)2 = 9
Solve
x2 + x2 + 2x + 1 = 9
2x2 + 2x – 8 = 0
2x2 + 2x – 8 = 0
Reduce by a 2. (Optional, but strongly recommended.)
x2 + x – 4 = 0
Use the quadratic formula to find x.
a=1
b=1
− b ± b 2 − 4ac
x=
2a
x=
x=
− (1) ±
(1)2 − 4(1)(− 4)
2(1)
− 1± 17
2
c = -4
x=
− 1± 17
2
1
17
1
17
x=− +
or − −
2
2
2
2
Now find y using (c)
y=x+1
(c)
Since there are two possible values for x, both have to be used to
find a y to make an ordered pair.
1
17
x=− +
2
2
1
17
x=− −
2
2
1
17
y=− +
+1
2
2
1
17
y=− −
+1
2
2
y=
1
17
+
2
2
 1
17 1
17 
− +

,
+
 2
2 2
2 

y=
1
17
−
2
2
 1
17 1
17 
− −

,
−
 2
2 2
2 

This makes version (3) of the graphs correct for this situation.
(3)
Example 19.2
Solve: (a) x2 + y2 = 9 (circle)
(b) 2x2 – y2 = -6 (hyperbola)
Do the pictures first.
0 real solutions
1 real solutions
2 real solutions
2 real solutions
3 real solutions
4 real solutions
Doing the graphs only helps to realize how many possible answers
there could be. Carefully graphing the actual equations will give a very
good estimate of the real answers.
Solve: (a) x2 + y2 = 9 (circle)
(b) 2x2 – y2 = -6 (hyperbola)
Elimination can be used since the powers are all the same.
(a) + (b)
x2 + y2 = 9
2x2 – y2 = -6
3x2 = 3
x2 = 1
x = ±1
Now use these two answers to find y’s
x = +1
x = -1
(+1)2 + y2 = 9
(-1)2 + y2 = 9
y2 = 8
y2 = 8
y=± 8
y=± 8
y = ±2 2
y = ±2 2
Note: this x value resulted
in two y values.
(1,2 2 )
(1,−2 2 )
Note: this x value also
resulted in two y values.
(− 1,2 2 )
(− 1,−2 2 )
Ended up with four real solutions, so the last graph is the
correct picture.
Example 19.3
Solve: (a) xy = 9
(hyperbola)
(b) y = -x – 2 (line)
Potential pictures.
0 real solutions
1 real solution
2 real solutions
Solve: (a) xy = -4
(hyperbola)
(b) y = -x – 2 (line)
Because of the mixed term, xy, in (a) substitution is the best method to
use. Sub (b) into (a).
x(−x − 2) = −4
− x 2 − 2 x = −4
− x2 − 2x + 4 = 0
Use the quadratic formula.
a = -1
b = -2
− b ± b 2 − 4ac
x=
2a
c=4
a = -1
b = -2
− b ± b 2 − 4ac
x=
2a
x=
− (− 2 ) ±
x=
(− 2)2 − 4(− 1)(4)
2(− 1)
2 ± 4 + 16
−2
x=
x=
2 ± 20
−2
−2±2 5
2
x = −1± 5
c=4
Sub these two answers into (b) to find y
y = -x – 2
x = −1+ 5
(
x = −1− 5
)
(
)
y = − −1+ 5 − 2
y = − −1− 5 − 2
y = 1− 5 − 2
y = 1+ 5 − 2
y = −1− 5
y = −1+ 5
(− 1 +
5 , −1 − 5
)
(− 1 −
5 ,−1 + 5
)
Factoring Exponentials
Two basic ways:
Æ(
)(
)
Æ Pull out common terms (undistribute)
Example 19.4
Factor: 3x2n+2 + 12x3n+3
Rewrite by separating the powers
3x2nx2 + 12x3nx3
Pull out a 3, a x2n, and a x2
3x2nx2(1 + 4xnx)
Rewrite consolidating the powers
3x2n+2(1 + 4xn+1)
Example 19.5
x 2a − y 2b
Factor : a
x + yb
Rewrite top such that it is the difference of squares. Things
can be regrouped as squares if there are even powers.
(x ) − (y )
=
a 2
b 2
xa + yb
Factor the top like it was c2 – d2 = (c + d) ( c – d)
( x a − y b )( x a + y b )
=
xa + yb
= xa - yb
Sum & Difference of Cubes
There is a pattern to the solution. Just learn it.
same
a3 + b3 = (a + b) (a2 _ ab +b2)
always opposite
a3 - b3 = (a - b) (a2 + ab + b2)
same
For factoring looks at powers as multiples of 3.
always
positive
Example 19.6
Factor: x3y3 – p3
Rewrite as cubes.
= (xy)3 – p3
Use the correct pattern: a3 - b3 = (a - b) (a2 + ab + b2)
= ((xy) – p) ((xy)2 + xyp + p2)
=(xy – p) (x2y2 + pxy + p2)
Example 19.7
Factor: 8m3y6 + x3
Regroup
= (2my2)3 + x3
Use correct pattern:
a3 + b3 = (a + b) (a2 _ ab +b2)
= ((2my2) + x) ((2my2)2 – 2my2x + x2)
= ( 2my2 + x) (4m2y4 – 2mxy2 + x2)
Practice
(a) Factor: 4x3n+2 – 6x4n+1
Rewrite
= 4x3nx2 – 6x4nx
Pull out 2, x3n, and x
= 2x3nx(2x – 3xn)
= 2x3n+1 (2x – 3xn)
x 4 a − y 4b
(b) Simplify by factoring the numerator:
x 2 a + y 2b
The difference of even powers on top leads to using the
difference of squares.
(x ) − (y )
=
2a 2
2b 2
x 2 a + y 2b
(x
=
2a
)(
− y 2b x 2 a + y 2b
x 2 a + y 2b
= x2a – y2b
)
Å not done, this can be factored
= (xa)2 – (yb)2
= (xa – yb) (xa + yb)
(c) Factor: 27x12y6 – z9
Regroup
= (3x4y2)3 – (z3)3
Use correct pattern:
a3 - b3 = (a - b) (a2 + ab + b2)
=((3x4y2) – (z3)) ((3x4y2)2 + 3x4y2z3 + (z3)2)
= (3x4y2 – z3) (9x8y4 + 3x4y2z3 + z6)
V
(d) Givens: UV ≅ WV
VX ⊥ UW
Prove:
∆UVX ≅ ∆WVX
U
Statements
1) UV ≅ WV
Reasons
W
X
Notes
1) Given
Reflexive
VX ⊥ UW
2) VX ≅ VX
2) Reflexive
3) ∆UVX ≅ ∆WVX
3) HL
HL