Unformatted text preview: —"‘W~. EXERCISES 1.4 1 1 28y=57 2:} y=—2§. When we substitute this into the second equation, 24. 25. 7'. 57 285 5 5 The point of intersection is (5/112, 57/28). The lines and point of intersection are shown in
the right ﬁgure above. To find the point of intersection, we solve the ﬁrst equation for :1: in terms of y, getting
3 = 1 + 3y. When we substitute this into the second equation, we obtain —2(1+3y)+6y=2 ==> —2—(iy+6y =2 ==> —2=2.
This ineans that the lines do not intersect; they must be parallel (left ﬁgure below). (a) The plot of the ﬁve points in the right ﬁgure above indicates that the ﬁrst three and the
ﬁfth data points are collinear. This is proved in part (b). (b) Theslope of the line through the ﬁrst two points is m =
of the line through these points is ' 12— 15 3 .
12 _ 10 a —§. The equation y—l5—z—§(a:—10) ==> 2y~30=—3x+30 =2} 3x+2y=60. If we substitute the third point (16, 6) into the left side of this equation we obtain 3(16) +
2(6) = 60. Hence, this point is on the line. If we substitute the ﬁfth point into the left side,
we ﬁnd 3(30) + 2(~15) = 60. This point is also on the line. On the other hand, when we
substitute the fourth point, we get 3(22) + 2(0) = 66 7E 60. This point is not on the line.
(c) We can change either the :1:coordinate or the y—coordinate of the fourth point. If we
wish the mcoordinate to stay as 22, then the ycoordinate is given by 3(22} +2y==60 => 2y=60—66= —6 ==> y: 3.
If we wish the y—cbordinate to stay as 0, the the :1:coordinate is given by
3x+2(0) =60 ==> :1:=20. EXERCISES 1.4
. The inequality 23 + y g 1 includes the points on the line 2:1: + y = 1, whereas 2:1: + y < 1 does not include the points on the line.
(a) For points to be above the line 3; = 2:1: + 3, their y~coordinates must be greater than ' those on the line. Hence, an inequality is y > 29: + 3. (b) For points to he on and below the line y = :1: — 4, their y—coordinates must be less than
or equal to those on_ the line. Hence, an inequality is y g :1: — 4.  (c) For points to be cn and to the right of the line 4:1: 2 3 —~ 3;, their :1:coordinates must be greater than or equal to those on the line. Hence an inequality is 43 2 3 — y.
(d) For points to be to the left of the line :1: = —4  23;, their :1:coordinates must be less
than those on the line. Hence, an inequality is :1: < —4 — 23;. This material has been reproduwd in accordance with copyright law by the Unimily ofManiloha Bookstore. Fm'ther mums“ 1'... ma n. :.. rm} :.. ..._.... cum. :1 — s ...
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 Fall '11
 Dr.Cornell
 Math

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