SME
Grade 9 Math · Algebra 1
Systems by elimination, and systems of linear inequalities
Use this for “Graph y > x − 2 and y ≤ −2x + 4. Is (0, 0) a solution?”
Chart
Where both sides are shaded, both inequalities are true: those points are the solutions.
A dashed line is left out; a solid line is included.
The boundaries cross at (2, 0)
Test (0, 0)
1 × 0 − 2 = −2, and 0 > −2 is true
−2 × 0 + 4 = 4, and 0 ≤ 4 is true
It is a solution.
This is an example. Type your own values to start a new problem.
y
>
m₁
x
+
(
b₁
)
y
≤
m₂
x
+
b₂
x
Corner x
worked out (not typed)
y
Corner y
worked out (not typed)
x₀
Test point x
example
y₀
Test point y
example
d₁
Test point’s height above the first line
worked out (not typed)
d₂
Test point’s height above the second line
worked out (not typed)
Formulas
x
=
b₂
−
b₁
m₁
−
m₂
2
=
4 − (−2)
1 − (−2)
y
=
m₁
×
x
+
b₁
0 = 1 × 2 + (−2)
d₁
=
y₀
−
(
m₁
×
x₀
+
b₁
)
2 = 0 − (1 × 0 + (−2))
d₂
=
y₀
−
(
m₂
×
x₀
+
b₂
)
−4 = 0 − (−2 × 0 + 4)
Assumptions
•
A dashed line (< or >) is not included; a solid one (≤ or ≥) is.
•
The solutions are where the two shadings overlap.
•
Test a point by putting it into both inequalities.
Step-by-step
WE KNOW
x₀ = 0, y₀ = 0, m₁ = 1, b₁ = −2, m₂ = −2, b₂ = 4
FIND
corner x (x), corner y (y), test point’s height above the first line (d₁), test point’s height above the second line (d₂)
Step 1 · Find corner x (x)
Use x = (b₂ − b₁) ÷ (m₁ − m₂)
Where
the
boundaries
cross:
set
the
two
right
sides
equal,
gather
x
on
one
side,
then
divide.
x
=
4 − (−2)
1 − (−2)
x
=
6
3
x
=
2
Step 2 · Find corner y (y)
Use y = m₁ × x + b₁
Put
x
into
the
first
boundary.
y
=
1
×
2
−
2
y
=
0
Step 3 · Find test point’s height above the first line (d₁)
Use d₁ = y₀ − (m₁ × x₀ + b₁)
Put
the
test
point
into
y
>
m₁
x
+
b₁
:
it
is
true
when
y₀
is
above
the
line,
d₁
>
0.
d₁
=
0
−
(1
×
0
−
2)
d₁
=
2
Step 4 · Find test point’s height above the second line (d₂)
Use d₂ = y₀ − (m₂ × x₀ + b₂)
And
into
y
≤
m₂
x
+
b₂
:
true
when
y₀
is
on
or
below
the
line,
d₂
≤
0.
d₂
=
0
−
(−2
×
0
+
4)
d₂
=
0
−
4
d₂
=
−4
→
both
are
true:
the
point
is
a
solution
Check
Put the numbers back in. Both sides should match.
2
=
4 − (−2)
1 − (−2)
✓
0 = 1 × 2 − 2 ✓
2 = 0 − (1 × 0 − 2) ✓
−4 = 0 − (−2 × 0 + 4) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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