Transcript Do Now 2/8/07

Do Now 1/20/10
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Take out HW from last night.
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Text p. 462, #1-8 all, #10, #12, #16-30 evens, #36
Copy HW in your planner.
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Benchmark Test #1 evens
Text p. 469, #3-8 all, #10-38 evens
Quiz sections 7.5 – 7.6 Friday
Homework
Text p. 462, #1-8 all, #10, #12, #16-30 evens & 36
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1) inconsistent
 12) infinitely many
solutions
 2) consistent dependent
 3) lines have same slope  16) infinitely many
solutions
but different y-intercepts
 4) the graph would show  18) infinitely many
solutions
only one line
 20) no solution
 5) B; one solution
 22) (3,0)
 6) C; no solution
 24) C
 7) A; infinitely many
solutions
 26) no solution
 8) no solution
 28) one solution
 10) one solution
 30) one solution
 36) No, there are infinitely
many solutions
Objective
 SWBAT
solve systems of linear
inequalities in two variables
Section 6.7 “Graph Linear
Inequalities”
Linear Inequalitiesthe result of replacing the = sign
in a linear equation with an inequality sign.
2x + 3y > 4
y ≤ ½x + 3
y ≥ 4x - 3
7y < 8x - 16
Graphing Linear Inequalities
 Graphing
Boundary Lines:
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Use a dashed line for < or >.
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Use a solid line for ≤ or ≥.
Graph an Inequality
Graph the inequality
STEP 1
Graph the equation
y  4x  3
STEP 2
Test (0,0) in the
original inequality.
y  4x  3
0  4(0)  3
True
y > 4x - 3.
STEP 3
Shade the half-plane that
contains the point (0,0),
because (0,0) is a solution
to the inequality.
Graph an Inequality
Graph the inequality
STEP 1
Graph the equation
x  3 y  1
STEP 2
Test (1,0) in the
original inequality.
x  3 y  1
1  3(0)  1
True
x + 3y ≥ -1.
STEP 3
Shade the half-plane that
contains the point (1,0),
because (1,0) is a solution
to the inequality.
Graph an Inequality
Graph the inequality
STEP 1
Graph the equation
y  3
STEP 2
Test (2,0) in the
original inequality.
Use only the ycoordinate, because
the inequality does
not have a x-variable.
y  3
( 0 )  3
True
y ≥ -3.
STEP 3
Shade the half-plane that
contains the point (2,0),
because (2,0) is a solution
to the inequality.
Graph an Inequality
Graph the inequality
STEP 1
Graph the equation
x  1
STEP 2
Test (3,0) in the
original inequality.
Use only the ycoordinate, because
the inequality does
not have a x-variable.
x  1
( 0 )  1
False
x ≤ -1.
STEP 3
Shade the half-plane that
does not contain the
point (3,0), because (3,0)
is not a solution to the
inequality.
Section 7.6 “Solve Systems of Linear
Inequalities”
SYSTEM OF INEQUALITIESconsists of two or more linear inequalities in the
same variables.
x–y>7
Inequality 1
2x + y < 8
Inequality 2
A solution to a system of inequalities is an
ordered pair (a point) that is a solution to
both linear inequalities.
Solving a System of Inequalities by
Graphing
(1) Graph both inequalities in the same plane.
(2) Find the intersection of the two half-planes. The
graph of the system is this intersection.
(3) Check a coordinate by substituting into EACH
inequality of the system, to see if the point is a
solution for both inequalities.
Graph a System of Inequalities
Inequality 1
y > -x – 2
Inequality 2
y ≤ 3x + 6
Graph both inequalities in the same
coordinate plane. The graph of the
system is the intersection of the two
half-planes, which is shown as the
darker shade of blue.
(0,1)
?
1>0–2
1>–2
?
1>0+6
1>6
Graph a System of Inequalities
Inequality 1
y<x–4
Inequality 2
y ≥ -x + 3
Graph both inequalities in the
same coordinate plane. The graph
of the system is the intersection of
the two half-planes, which is
shown as the darker shade of
blue.
(5,0)
?
0<5–4
0< 1
?
0 ≥ -5 + 3
0 ≥ -2
Graph a System of THREE Inequalities
Inequality 1:
Inequality 2:
Inequality 3:
Check
(0,0)
y ≥ -1
x > -2
x + 2y ≤ 4
?
y ≥ -1
0 ≥ -1
Graph all three inequalities in the
same coordinate plane. The graph
of the system is the triangular
region, which is shown as the
darker shade of blue.
?
x > -2
0 > -2
?
x + 2y ≤ 4
0+0≤4
Graph a System of THREE Inequalities
Inequality 1:
Inequality 2:
Inequality 3:
y ≥ -x + 2
y<4
x<3
Inequality 1:
Inequality 2:
Inequality 3:
y > -x
y≥x–4
y<5
Write a System of Linear Inequalities
Write a system of inequalities for the shaded region.
INEQUALITY 1: One boundary line for
the shaded region is y = 3. Because the
shaded region is above the solid line, the
inequality is y ≥ 3.
INEQUALITY 2: Another boundary line for the shaded region
has a slope of 2 and a y-intercept of 1. So, its equation is y = 2x + 1.
Because the shaded region is above the dashed line, the
inequality is y > 2x + 1.
y≥3
y > 2x + 1
Inequality 1
Inequality 2
NJASK7 Prep
Homework
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Text p. 469, #3-8 all, #10-38 evens