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Unformatted text preview: fore, another point that sa+sﬁes the equa+on is (6, 3). Let us plug in x =
6 into the equa+on. y = (3/4)(
6) – 3/2 =
9/2 – 3/2 =
12/2 =
6 Graph of
3x + 4y =
6 Now we must ﬁnd at least two addi+onal points. Let us plug in x = 6 into the equa+on. y = (3/4)(6) – 3/2 = 9/2 – 3/2 = 6/2 = 3 Therefore, another point that sa+sﬁes the equa+on is (6, 3). Let us plug in x =
6 into the equa+on. y = (3/4)(
6) – 3/2 =
9/2 – 3/2 =
12/2 =
6 Therefore, another point that sa+sﬁes the equa+on is (
6,
6). Graph of
3x + 4y =
6 Now we must ﬁnd at least two addi+onal points. Let us plug in x = 6 into the equa+on. y = (3/4)(6) – 3/2 = 9/2 – 3/2 = 6/2 = 3 Therefore, another point that sa+sﬁes the equa+on is (6, 3). Let us plug in x =
6 into the equa+on. y = (3/4)(
6) – 3/2 =
9/2 – 3/2 =
12/2 =
6 Therefore, another point that sa+sﬁes the equa+on is (
6,
6). We will plot the following points in the table, and the connect the points together with a straight line: X y
6
6 0
3/2 2 0 6 3 Graph of
3x + 4y =
6 It is easy to see by looking at the graph that this equation is a
function and that its domain and range are all real numbers. Linear Equa+ons, y = a Example. Graph y = 3. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what the value of x is, y is always 3. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what the value of x is, y is always 3. We will ﬁnd the x
intercept by plugging in y = 0 into the equa+on and solving for x. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what...
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 Spring '11
 johnson
 Algebra

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