Transcript Introduction To Logic

Solving Quadratic/Linear Systems
Algebraically
Integrated A2/trig
Come on! Solve it!
• Solve for “x”
 x2 = 3x – 18
• Solve for “x” and “y” using substitution method
x+y=5
2x = 4y – 8
Solving a Quadratic/Linear System
• A quadratic/linear system is when there are
two equation but one equation has an “x2” term
and the other just has an “x”.
 If there are NO “x2” terms then the system is
called a linear system.
• To solve a quadratic/linear system you need
to use the “substitution method” and factoring
method of a quadratic equation.
How to solve a Quadratic/Linear
System
Step 1:
 Start by taking both
equations and set the
both equal to “y”.
 Use algebraic skills to
manipulate the
equation and move
terms from one side of
the equal sign to the
other
Ex:
y = x2 + 2x + 1
-x + y = 3
y = x2 + 2x + 1 (OK)
How to solve a Quadratic/Linear
System
Step 1:
 Start by taking both
equations and set the
both equal to “y”.
 Use algebraic skills to
manipulate the
equation and move
terms from one side of
the equal sign to the
other
Ex:
y = x2 + 2x + 1
-x + y = 3
y = x2 + 2x + 1 (OK)
-x + y = 3
+x
+x
y = x + 3 (OK)
How to solve a Quadratic/Linear
System
Step 2:
 Now that both
equations are equal to
“y”, set them equal to
each other eliminating
the “y” and having one
equation containing
only “x’s”
 Use algebraic skills to
manipulate the
equation so that one
side is equal to “0”.
Make sure the “x2”
term is positive.
y = x2 + 2x + 1
y=x+3
x2 + 2x + 1 = x + 3
-x
-x
x2 + x + 1 = 3
-3
x2 + x – 2 = 0
-3
How to solve a Quadratic/Linear
System
Step 3:
 Factor the
quadratic
equation and
solve for both
values of “x”
Ex:
x2 + x – 2 = 0
(
)(
)=0
(x
)(x
)=0
(x + 2)(x – 1) = 0
x+2=0
x = -2
x–1=0
x=1
Solve a Quadratic/Linear System
 Since there are 2
values of “x” you must
solve for 2 values of
“y” – one for each “x”.
 It is probably best to
use the linear equation
substitute in to.
 After you solve for “y”
there are two solutions
and express your
answer as a
coordinate.
x = -2
x=1
-x + y = 3
-x + y = 3
-(-2) + y = 3 -(1) + y = 3
2+y=3
-1 + y = 3
-2
+1
-2
+1
y=1
y=4
(-2, 1)
(1, 4)
YOU CAN DO IT!!!!
1)
y = x2 – 4x + 3
y=x–1
2)
x2 – y = 5
y = 3x - 1
solutions
1) (4,3), (1,0)
y = x2 – 4x + 3
y=x–1
2)(4,11), (-1,-4)
x2 – y = 5
y = 3x - 1
A circle and a line
• Try this one!
(x  2)  (y  3)  16
xy5
2
2
A circle and a line
• Solution:
(x  2)  (y  3)  16
xy5
2
(-2, 7) and (2,3)
2