2x + 3y = 6
4x + 6y = 12
GRAPHING:
eg. 1
𝟐𝒙
3y = -
𝟑
𝟐
y= -
𝟔
𝟑
x +2
𝟑
m=-
+
𝟐
𝟑
b=2
eg. 2
𝟔𝒚
𝟔
=-
𝟒𝒙
𝟔
+
𝟏𝟐
𝟔
𝟐
y=- +2
𝟑
m=
−𝟐
𝟑
b=2
Hence: the 2x + 3y = 6 and 4x + 6y = 12 is an identical, consistent and
dependent.
2x + 3y = 6
4x + 6y = 12
SUBSTITUTION:
𝟐𝒙
𝟐
=-
X=-
𝟑𝒚
𝟐
𝟑
+
𝟔
2x + 3 (1) =6
𝟐
𝒚+𝟑
𝟐
2x + 3 = 6
𝟑
4(- 𝒚 + 𝟑) + 𝟔𝒚 = 𝟏𝟐
𝟐
-6 + 12 + 6y + 12
6 + 6y = 12
𝟔𝒚
𝟔
=
𝟏𝟐
𝟔
−
𝟑
𝟐
𝟐𝒙
𝟐
=
X=
𝟔
𝟔
y = 2-1
y=1
The solution is
2x = 6 – 3
,𝟏
CHECK :
2X = 3Y = 6
𝟑
2 ( ) + 3 (1) = 6
𝟐
3+3=6
6=6 √
4x + 6y = 12
𝟑
4 ( ) + 6 (1) = 12
𝟐
6 + 6 = 12
12 = 12 √
𝟑
𝟐
𝟑
𝟐
2x + 3y = 6
4x + 6y = 12
ELIMINATION:
2(2x + 3y = 6)
4x + 6y = 12
4x + 6y = 12
4x + 6y = 12
0=0
The system is infinitely many solution. Its solution set is all
numbers for each variable.
GRAPHICALLY:
x = -2y + 3
y = 2x + 4
x = -2y + 3
Let y = 0 , x = -2 (0) + 3
x= 0+3
x=3
Let x = 0 , 0 = -2y + 3
2y = 3
𝟐𝒚 𝟑
=
𝟐
𝟐
𝟏
y=1
𝟐
y = 2x + 4
Let y = 0 , 0 = 2x + 4
−𝟐𝒙
𝟒
=
−𝟐
𝟐
x = -2
Let x = 0 , y = 2 (0) + 4
y=0+4
y=4
CHECK:
(-1 ,2)
x = -2y + 3
-1 = -2(2) + 3
-1 = -4 + 3
-1 = -1 True
y = 2x + 4
2 = 2(-1) + 4
2 = -2 + 4
2 = 2 True
SUBSTITUTION:
x = -2y + 3
y = 2x + 4
x = -2y + 3
y = 2(-2y + 3) + 4
y = -4y + 6 + 4
y + -4y + 10
y + 4y = 10
𝟓𝒚
𝟓
=
𝟏𝟎
𝟓
y=2
Solve for X
x = -2y + 3
x = -2(2) + 3
x = -4 + 3
x = -1
CHECK: (-1, 2)
x = -2y + 3
-1 = -2(2) + 3
-1 = -4 + 3
-1 = -1 True
y = 2x + 4
2 = 2(-1) + 4
2 = -2 + 4
2 = 2 True
ELIMINATION:
x = -2y + 3
y = 2x + 4
2(x + 2y = 3) 2
-2x + y = 4
2x + 4y = 6
+ -2x + y = 4
𝟓𝒚
=
𝟓
y=2
𝟏𝟎
𝟓
Solve for x:
x = -2y + 3
x = -2(2) + 3
x = -4 + 3
x=1
CHECK: (-1, 2)
x = -2y + 3
-1 = -2(2) + 3
-1 = -4 + 3
-1 = -1 True
y = 2x + 4
2 = -2 + 4
2 = 2 True
A. Sketch the graph of each linear inequality in x-y plane
3.
5x + 2y ≥ 11
𝟐𝒚
𝟐
−
Y=
𝟓𝒙
𝟐
−𝟓𝒙
𝟐
+
+
𝟏𝟏
𝟐
𝟏𝟏
𝟐
B. Sketch the graph of each system in x-y plane
3.
𝒙>𝟔
𝒚 ≤ −𝟐
x=6
y = -2
 If the inequality symbol is < or ≤ the shaded region is below the boundary
line.
 For x interpret if the inequality symbol is > or ≥ the shaded region is to the
right of the boundary line.
 If the inequality symbol is < or ≤ the shaded region is to the left of the
boundary line.
NOTE: for y interpret of the inequality symbol is > or ≥ the shaded region is
above the boundary line.
TEST POINT:
(eg. 1) 𝒙 > 𝟔 , (7,0)
7 > 6 TRUE √
(eg. 2) 𝒚 ≤ −𝟐 , (0, -3)
-3 ≤ -2 TRUE √
The solution is the set of points common to both Inequalities. It is the part of the
graph that contains the two shades, including those [parts of the line that satisfy
both inequalities.