Using Table to Graph Lines Exs

# Let us plug in x 6 into the equaon graph of 3x 4y

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Unformatted text preview: 2 = x Therefore, the x ­intercept is (2,0). Now, we will ﬁnd the y ­intercept by plugging in x = 0 into the equa+on and solving for y. y = (3/4)(0) – 3/2 Graph of  ­3x + 4y =  ­6 Example. Graph  ­3x + 4y =  ­6. Solu+on: We must ﬁrst solve for y in the equa+on. 4y = 3x – 6 y = (3/4)x – 3/2 Now, we will ﬁnd the x ­intercept by plugging in y = 0 into the equa+on and solving for x. 0 = (3/4)x – 3/2 3/2 = (3/4)x 2 = x Therefore, the x ­intercept is (2,0). Now, we will ﬁnd the y ­intercept by plugging in x = 0 into the equa+on and solving for y. y = (3/4)(0) – 3/2 y =  ­3/2 Therefore, the y ­intercept is (0, ­3/2). Graph of  ­3x + 4y =  ­6 Now we must ﬁnd at least two addi+onal points. 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. 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 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). 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. 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 There...
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