Unformatted text preview: 1o _ _ EXERCISES 1.3 21 . To ﬁnd the point of intersection, we multiplyr the ﬁrst equation by 3 and the second equatiou ' byz. 9x — 6y 2 12,
103 + 6y =14. When we add these equations together, we get 26
When we substitute this into the ﬁrst equation,
25 7'8 2 l
3(E) —2y—4 ls» ~2y24—ﬁ=~ﬁ ——‘* 9—K) The point of intersection is (26/ 19, 1 / 19). The lines and point of intersection are shown in
the right ﬁgure above. 22. To ﬁnd the point of interSection, we multiply the ﬁrst equation by 4 and the second equation by 5.
83: + 20y = —12,
15.1: — 20y = 15.
When we add these equations together, we get
' 3
'When we substitute this into the ﬁrst equation, '
3 . 6 7'5 __ 15 The point of intersection is (3/23, —15/23). The lines and point of intersection are shown
in the left ﬁgure below. 23. Toﬁnd the point of intersection, we multiply the second equation by 3, 123 +133; = 2?, '
12x — 15;; = ~30. When we subtract these equations, we get This material has been reproduced in accordance with cogyright law by the Univclsity of Manitoba Bookstore. Further repmdudim in oar: m in ﬁm ,‘g glﬁnﬂu was”: /"‘\. ...
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 Fall '11
 Dr.Cornell
 Math

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