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FTCE Middle Grades Math 5-9
Solving Linear Systems of Equations by Graphing
Graph both equations on one set of axes.
The point of intersection is the solution of
the system.
Skills 6.13-6.16 and 6.19
Example: Solve the following linear system of
equations using the elimination method.
4 x  2 y  4
1.
x  y  4
Multiply the second equation by 4 so that
the coefficients of the x terms are opposites.
4 x  2 y  4
2.
4 x  4 y  8
Add the equations together.
4 x  2 y  4
Solving Linear Systems of Equations by
Substitution
1.
2.
3.
Solve one of the equations for one of the
variables.
Substitute what you found for the chosen
variable back into the other equation and
solve for the other variable.
Solve for the first variable.
4 x  4 y  16
6 y  12
3.
Example: Solve the following linear system of
equations using the substitution method.
1.
2.
y2
Solve for x by substituting the value found
for y into one of the original equations.
4 x  2 y  4
4 x  2( 2)  4
4 x  2 y  4
4 x  4  4
x  y  4
4 x  8
Solve for y in the second equation to
get y  x  4 .
Substitute into first equation for y and solve
for x.
x  2
The solution is (-2, 2).
Graphing Quadratic Functions
y  ax 2  bx  c
4 x  2( x  4)  4
4 x  2 x  8  4
6 x  12
x  2
3.
If a is positive, the parabola opens up.
If a is negative, the parabola opens down.
To graph a parabola:
1. Find the x-coordinate of the vertex using the
Solve for y by substituting the value found
for x.
4 x  2 y  4
4( 2)  2 y  4
8  2 y  4
2y  4
formula x 
2.
3.
y2
The solution is (-2, 2).
Solving Linear Systems of Equations by
Elimination
1.
2.
3.
Multiply or divide either or both of the
equations by a number so that the
coefficients of one variable are opposites of
each other.
Add the equations to eliminate one of the
variables.
Substitute what you found for the chosen
variable back into the other equation and
solve for the eliminated variable.
4.
b
.
2a
To find the y-coordinate of the vertex,
substitute the x-coordinate of the vertex into
the function.
To find additional points on the parabola,
pick x-values on both sides of the vertex and
substitute into the function to find the
corresponding y-values.
Plot the points found in step 3 and connect
with a smooth curve.
Graphing Quadratic Inequalities
1.
2.
3.
4.
Graph as if it is a quadratic equation.
Use a dotted line for > or < symbols. Use a
solid curve for ≥ or ≤ symbols.
Select a point inside the parabola and
substitute the x and y values of that point
into the function.
If the point is a solution of the inequality,
shade inside the parabola. If the point is not
a solution, shade outside of the parabola.
FTCE Middle Grades Math 5-9
Factoring
Skills 6.13-6.16 and 6.19
Factoring Trinomials with a Leading Coefficient
other than 1:
Factor out GCF:
1.
2.
3.
4.
Find GCF of coefficients.
Find smallest power of each variable.
Multiply the terms from steps 1 and 2, this is
the GCF.
Divide each term in the polynomial by the
GCF.
1.
2.
3.
4.
Factor out the GCF if possible.
Multiply leading coefficient and 3rd term.
Find the factors that add to middle term.
Split the middle term.
Factor by grouping.
Example: Factor the polynomial:
2 x 2 y 3 z 4  4 xyz 3  10 x 4 y 2 z 2 
2 xyz 2 ( xy 2 z 2  2 z  5x 2 y )
Factor by Grouping
Factoring by grouping is usually done when you have
4 terms and they do not look to have anything in
common.
1.
2.
3.
4.
Remove the GCF if possible.
Group terms with common factors.
Remove the GCF of each group.
Use the distributive law to rewrite the
expression.
Example: Factor
Example: Factor the trinomial: 2 x  x  6
2
No GCF. Multiply 2 x 6 = 12. What two factors of 12 add up to 1? 4 and -3. Split the middle:
2 x 2  4 x  3x  6
(Notice the polynomial has not changed)
Factor by grouping:
2 x 2  4 x  3x  6 
2 x  x  2   3( x  2) 
 2 x  3 x  2 
6x2  9x  4x  6
Special Factoring Patterns:
6x  9x  4x  6 
2
3x( 2 x  3)  2( 2 x  3) 
(3x  2)( 2 x  3)
Factoring Trinomials with a Leading Coefficient
of 1:
1.
2.
Factor out the GCF if possible.
Take factors of 3rd term that add up to the
middle term and factor into 2 binomials.
Example: Factor the trinomial:
x2  x  6 
( x  3)( x  2)
Difference of Squares: a  b
Difference of Cubes (SOAP):
2
2
 (a  b)(a  b)
a3  b3  (a  b)(a 2  ab  b2 )
Sum of Cubes (SOAP):
a3  b3  (a  b)( a 2  ab  b2 )
SOAP: Same, Opposite, Always Positive
FTCE Middle Grades Math 5-9
Skills 6.13-6.16 and 6.19