246 
… 5.3‹Graphing Linear Equations
„‹Please complete the given ordered pairs by using the given
ˆˆequation.
ƒ
ˆˆˆ 2x + 3y = 6‰(0, ),‹( ,0)

if x = 0,‹2(0) + 3y = 6Š(0,2)ˆ if y = 0,‹2x + 3(0) = 6Š(3,0)
ˆˆŠ 3y = 6ˆˆˆˆ‰ 2x = 6
ˆˆ‰y = 2ˆˆˆˆˆx = 3
‚S
To complete the ordered pair (0, ), using the equation, 2x + 3y = 6,
substitute '0' in for 'x' and solve the remaining equation for y.

ˆŠ2x + 3y = 6ˆ This value of 'y' is the missing second
ˆ 2(0) + 3y = 6ˆ coordinate in the given ordered pair, (0, ).
ˆˆ 3y = 6ˆ The completed ordered pair is (0,2).
ˆˆ‹y = 2

Similarly to complete the ordered pair, ( ,0), substitute the '0' for 'y'
and solve the remaining equation for 'x'.

ˆˆ‰ 2x + 3y = 6‰The completed ordered pair is (3,0).
ˆˆŠ 2x + 3(0) = 6
ˆˆˆŠ 2x = 6
ˆˆˆ‰x = 3
 1‹Complete the ordered pairs (0, ), ( ,0), and (2, ), using the
ˆŠ equation 3x + 4y = 24.

ˆˆ‰9
 A) (0,6) (8,0) (2,Δ)‰B) (6,0) (0,8) (2,5)Š C) (0,2) (3,0) (2,12)
ˆˆ‰2
Š3x + 4y = 24ˆ‹3x + 4y = 24ˆ‰3x + 4y = 24
ˆ3(0) + 4y = 24ˆ3x + 4(0) = 24ˆŠ3(2) + 4y = 24
ˆˆ4y = 24ˆˆ3x = 24ˆ‰ 6 + 4y = 24
ˆˆ y = 6ˆˆ‹x = 8ˆˆŠ 4y = 18
ˆˆ (0,6)ˆˆ‹(8,0)ˆˆŠ 18Š9‰ 9
ˆˆˆˆˆˆˆ‰ y = ΔΔ = Δ‹(2, Δ)
ˆˆˆˆˆˆˆˆŠ 4Š2‰ 2
€ A
 2‹Complete the ordered pairs, (0, ), ( ,0) and ( ,3), using the
ˆŠ equation, 5x - 3y = 15.

ˆˆˆˆˆˆˆˆˆŠ 24
A) (-5,0) (0,3) (7,3)Š B) (-5,0) (0,3) (6,3)Š C) (0,-5) (3,0) (ΔΔ,3)
ˆˆˆˆˆˆˆˆˆ‰5
Š5x - 3y = 15ˆ‹5x - 3y = 15ˆ‰5x - 3y = 15
ˆ5(0) - 3y = 15ˆ5x - 3(0) = 15ˆŠ5x - 3(3) = 15
ˆ‰ -3y = 15ˆˆ5x = 15ˆ‰5x - 9‹= 15
ˆˆ y = -5ˆˆ x = 3ˆˆŠ 5x = 24
ˆˆ (0,-5)ˆˆ (3,0)ˆˆŠ 24‰ 24
ˆˆˆˆˆˆˆ‰ y = ΔΔ‰(ΔΔ, 3)
ˆˆˆˆˆˆˆˆŠ 5ˆ5
€ C
„‹Please find the intercepts for each equation.
ƒˆˆ‰ -2x + 4y = 16

Š if x=0, -2ω0 + 4y = 16ˆˆif y=0,‹-2x + 4ω0 = 16
ˆˆ‰4y = 16ˆˆˆˆ -2x = 16
ˆˆ‰ y = 4ˆˆˆˆŠ x = -8

Šthe y - intercept is (0,4)ˆ‹the x - intercept is (-8,0)
‚S
To find the intercepts for the equation -2x + 4y = 16, substitute '0'
in for 'x' and solve for 'y'.

ˆˆˆ if x=0, -2ω0 + 4y = 16
ˆˆˆˆˆ‹4y = 16
ˆˆˆˆˆŠy = 4

The y - intercept is the point, (0,4), on the y - axis. This is the
point where the graph crosses the y - axis.

To find the x - intercept, substitute '0' in for 'y' and solve for 'x'.

ˆˆˆ if y=0,‹-2x + 4ω0 = 16
ˆˆˆˆˆ‹-2x = 16
ˆˆˆˆˆŠ x = -8

The x - intercept is the point, (-8,0), on the x - axis.‹This is the
point where the graph crosses the x - axis.
 3
ˆ‰Find the intercepts of the equation, 3x - 9y = 18.
ˆˆˆˆˆˆˆˆˆ †
A)‹(0,-2) (6,0)‰B)‹(6,0) (-2,0)Š C)‹(5,0) (0,-3)Š D)‹of
ˆˆˆˆˆˆˆˆˆ ‡
ˆŠ3x - 9y = 18ˆˆ3x - 9y = 18
ˆˆ 3ω0 - 9y = 18ˆ‰ 3x - 9ω0 = 18
ˆˆ‰ -9y = 18ˆˆ‰3x = 18
ˆˆˆ y = -2ˆˆ‰ x = 6
ˆˆˆ (0,-2)ˆˆ‰ (6,0)
€ A
 4
ˆ‰Find the intercepts of the equation, 2x - 3y = 7.

ˆˆˆˆ 7Š7ˆˆˆ‹2ˆ†
A)‹(3,0) (0,2)‰B)‹(0,Δ Δ) (Δ ,0)Š C)‹(4,0) (0,- Δ)Š D)‹of
ˆˆˆˆ 3Š2ˆˆˆ‹3ˆ‡
ˆŠ2x - 3y = 7ˆˆ 2x - 3y = 7
ˆˆ 2ω0 - 3y = 7ˆˆ2x - 3ω0 = 7
ˆˆ‰ -3y = 7ˆˆ‰ 2x = 7
ˆˆˆ‰7ˆ‹7ˆˆ 7‰ 7
ˆˆ‰ y = - Ċ (0,- Δ)ˆŠx = Δ‰(Δ , 0)
ˆˆˆ‰3ˆ‹3ˆˆ 2‰ 2
€ B
„‹Please graph the following lines by finding the intercepts
ƒˆˆˆŠ2x + 3y = 6

Gƒ [y= -.666x + 2] 6,6,530,151,3,14
 x - intercept‹if y = 0,‹2x = 6
ˆˆˆˆx = 3Š(3,0)

 y - intercept‹if x = 0,‹3y = 6
ˆˆˆˆy = 2Š(0,2)
‚S
The equation, 2x + 3y = 6, is known to have a graph that is a straight
line.‹Since any two points determine a line, it is sufficient to find
the two points where the line crosses the two axes.‹First letting x = 0
and solving for y, and then letting y = 0 and solving for x.

if x = 0, thenŠ2x + 3y = 6ˆŠ if y = 0, then 2x + 3y = 6
ˆˆŠ2ω0‹3y = 6ˆˆˆŠ 2x + 3ω0 = 6
ˆˆŠ 0 + 3y = 6ˆˆˆ‰ 2x + 0 = 6
ˆˆˆ 3y = 6ˆˆˆˆŠ2x = 6
ˆˆˆ‹y = 2ˆˆˆˆŠ x = 3

The ordered pair (0,2) is the y - intercept and the ordered pair (3,0)
is the x - intercept.‹These two intercepts are then plotted on the
coordinate axes and a line is drawn through them.
G 5ˆ Graph 3x + 4y = 12 by finding the intercepts.
A) y = .75x + 3
B) y = -.75x + 3
C) y = .25x + 4
D) y = .5x - 3


If x = 0, thenˆˆˆˆˆŠif y = 0, then
Š3ω0 + 4y = 12ˆˆˆˆˆ‰3x + 4ω0 = 12
ˆŠy = 3ˆˆˆˆˆˆ‰ x = 4
ˆŠ(0,3)ˆˆˆˆˆˆ‰ (4,0)
€ B
G 6ˆ Graph 2x - 5y = 10 by finding the intercepts.
A) y = .5x + 2
B) y = -.5x +2
C) y = .4x - 2
D) y = -.4x + 2


If x = 0, thenˆˆˆˆˆŠif y = 0, then
Š2ω0 - 5y = 10ˆˆˆˆˆ‰2x - 5ω0 = 10
ˆŠy = -2ˆˆˆˆˆˆ‰x = 5
ˆŠ(0,-2)ˆˆˆˆˆˆ‰ (5,0)
€ C
G 7ˆ Graph y = 4x - 5‹by finding the intercepts.
A) y = -3.33x - 5
B) y = 3.33x + 5
C) y = -3.33x + 5
D) y = 3.33x - 5


‹If x = 0, thenˆˆˆˆˆŠif y = 0, then
‰y = 4ω0 - 5ˆˆˆˆˆˆ0 = 4x - 5
‰y = -5ˆˆˆˆˆˆŠ 4x = 5
‰(0,-5)ˆˆˆˆˆˆˆ‹5Š 5
ˆˆˆˆˆˆˆˆ‹x = Ċ(Δ,0)
ˆˆˆˆˆˆˆˆ‰ 4Š 4
€ D
„‹Please graph the following horizontal and vertical lines.
ƒˆˆˆ‹Graph y = 3
Gƒ [y = 0x + 3] 6,6,315,151,3,14
‚S
All equations of the form, y = b, have graphs that are horizontal lines.
They have only one intercept at (0,b).‹All equations of the form, x = a
have graphs that are vertical lines.‹They have only the x - intercept
at (a,0).
G 8ˆˆ‰Graph 2y = 8
A) y = -999x + 4
B) y = -999x - 4
C) y = 0x + 4
D) y = 0x - 4
ˆThis is a horizontal line with y - intercept, (0,4)

ˆŠ 2y = 8
ˆ‰y = 4
€ C
G 9ˆˆŠGraph 2x - 3 = 7
A) y = 0x + 5
B) y = -999x + 5
C) y = 0x - 5
D) y = -999x - 5
ˆThis is a vertical line with x - intercept (7,0)

‹2x - 3 = 7
‰ 2x = 7 + 3
‰ 2x = 10
ˆx = 5

€ B
G 10ˆˆŠ Graph y = -4
A) y = -999x + 4
B) y = x - 4
C) y = -999x - 4
D) y = x + 4
ˆThis is a horizontal line with y - intercept (0,-4)

€ B
G 11ˆˆŠ Graph x = -2
A) y = -999x + 2
B) y = x + 2
C) y = -999x - 2
D) y = x - 2
ˆThis is a vertical line with x - intercept (-2,0)

€ C
„‹Please graph the following lines which pass through the
origin.
ƒˆˆˆ Graph y = 3x

If x = 0,‹y = 3ω0ˆŠ If x = 1,‹y = 3ω1
ˆŠ y = 0ˆˆˆŠy = 3
ˆŠ (0,0)ˆˆˆŠ(1,3)
Gƒ [y = 3x + 0]6,6,530,151,3,14
‚S‹To graph, y = 3x, substitute '0' for 'x' and solve for 'y'.
ˆˆˆˆ‰y = 3x
ˆˆˆˆ‰y = 3ω0
ˆˆˆˆ‰y = 0
The ordered pair, (0,0), is the origin and is both the x - intercept and
the y - intercept.‹The line passes through the origin.‹One additional
point is needed to get a graph.‹Select an x - value, substitute this
value into the equation and solve for y.


If x is '1', y = 3xŠ The additional ordered pair
ˆ‰ y = 3ω1Šis (1,3).‹These two points
ˆ‰ y = 3‰determine the graph.
G‚S [y = 3x + 0]6,6,530,170,3,14
G 12ˆˆŠ Graph y = 2x
A) y = 2x + 0
B) y = -2x + 0
C) y = -.5x + 0
D) y = -2x + 4
Šy = 2xˆˆˆˆˆ‹y = 2x
ˆ‹y = 2ω0ˆˆˆˆˆ y = 2ω2
ˆ‹y = 0ˆˆˆˆˆŠy = 4
ˆ‹(0,0)ˆˆˆˆˆŠ(2,4)
€ A