Answer:
[tex](1\frac{1}{2},1)[/tex] - False
[tex](4,6)[/tex] - True
[tex](18,12)[/tex] -- False
[tex](0,0)[/tex] -- True
[tex](2\frac{1}{2},3\frac{3}{4})[/tex] -- True
Step-by-step explanation:
The points are
[tex](1\frac{1}{2},1)[/tex] , [tex](4,6)[/tex], [tex](18,12)[/tex], [tex](0,0)[/tex] and [tex](2\frac{1}{2},3\frac{3}{4})[/tex] ---- missing from the question
Given
[tex]y = 1\frac{1}{2}x[/tex]
Required
Determine if each of the points would be on [tex]y = 1\frac{1}{2}x[/tex]
To do this, we simply substitute the value of x and of each point in [tex]y = 1\frac{1}{2}x[/tex].
(a) [tex](1\frac{1}{2},1)[/tex]
In this case;
[tex]x = 1\frac{1}{2}[/tex] and [tex]y = 1[/tex]
[tex]y = 1\frac{1}{2}x[/tex] becomes
[tex]y = 1\frac{1}{2} * 1\frac{1}{2}[/tex]
[tex]y = \frac{3}{2} * \frac{3}{2}[/tex]
[tex]y = \frac{9}{4}[/tex]
[tex]y = 2\frac{1}{4}[/tex]
The point [tex](1\frac{1}{2},1)[/tex] won't be on the graph because the corresponding value of y for [tex]x = 1\frac{1}{2}[/tex] is [tex]y = 2\frac{1}{4}[/tex]
(b) [tex](4,6)[/tex]
In this case;
[tex]x = 4[/tex]
[tex]y = 6[/tex]
[tex]y = 1\frac{1}{2}x[/tex] becomes
[tex]y = 1\frac{1}{2} * 4[/tex]
[tex]y = \frac{3}{2} * 4[/tex]
[tex]y = \frac{3* 4}{2}[/tex]
[tex]y = \frac{12}{2}[/tex]
[tex]y = 6[/tex]
The point [tex](4,6)[/tex] would be on the graph because the corresponding value of y for [tex]x = 4[/tex] is [tex]y = 6[/tex]
(c) [tex](18,12)[/tex]
In this case:
[tex]x = 18;y = 12[/tex]
[tex]y = 1\frac{1}{2}x[/tex] becomes
[tex]y = 1\frac{1}{2} * 18[/tex]
[tex]y = \frac{3}{2} * 18[/tex]
[tex]y = \frac{3* 18}{2}[/tex]
[tex]y = \frac{54}{2}[/tex]
[tex]y = 27[/tex]
The point [tex](18,12)[/tex] wouldn't be on the graph because the corresponding value of y for [tex]x = 18[/tex] is [tex]y = 12[/tex]
(d) [tex](0,0)[/tex]
In this case;
[tex]x =0; y = 0[/tex]
[tex]y = 1\frac{1}{2}x[/tex] becomes
[tex]y = 1\frac{1}{2} * 0[/tex]
[tex]y = 0[/tex]
The point [tex](0,0)[/tex] would be on the graph because the corresponding value of y for [tex]x = 0[/tex] is [tex]y = 0[/tex]
(e) [tex](2\frac{1}{2},3\frac{3}{4})[/tex]
In this case:
[tex]x = 2\frac{1}{2}[/tex]; [tex]y = 3\frac{3}{4}[/tex]
[tex]y = 1\frac{1}{2}x[/tex] becomes
[tex]y = 1\frac{1}{2} * 2\frac{1}{2}[/tex]
[tex]y = \frac{3}{2} * \frac{5}{2}[/tex]
[tex]y = \frac{15}{4}[/tex]
[tex]y = 3\frac{3}{4}[/tex]
The point [tex](2\frac{1}{2},3\frac{3}{4})[/tex] would be on the graph because the corresponding value of y for [tex]x = 2\frac{1}{2}[/tex] is [tex]y = 3\frac{3}{4}[/tex]
