Answer:
[tex](\frac{1}{6}, \frac{1}{2})[/tex] or [tex](0.167,0.5)[/tex]
Step-by-step explanation:
Given
[tex]3x + y =1[/tex]
[tex]15x + y = 3[/tex]
Required
Determine the solution
We'll solve using elimination method.
Subtract the first equation from the second:
[tex]15x + y = 3[/tex]
[tex]3x + y =1[/tex]
[tex]--------[/tex]
[tex]15x - 3x + y - y = 3 - 1[/tex]
[tex]15x - 3x + y - y = 3 - 1[/tex]
[tex]15x - 3x = 3 - 1[/tex]
[tex]12x = 2[/tex]
Make x the subject:
[tex]x = \frac{2}{12}[/tex]
[tex]x = \frac{1}{6}[/tex]
[tex]x =0.167[/tex]
Substitute [tex]\frac{1}{6}[/tex] for x in the first equation
[tex]3x + y =1[/tex]
[tex]3(\frac{1}{6}) + y = 1[/tex]
[tex]\frac{1}{2} + y = 1[/tex]
Make y the subject
[tex]y = 1 -\frac{1}{2}[/tex]
[tex]y = \frac{2-1}{2}[/tex]
[tex]y = \frac{1}{2}[/tex]
[tex]y = 0.5[/tex]
Hence, the solution is:
[tex](\frac{1}{6}, \frac{1}{2})[/tex] or [tex](0.167,0.5)[/tex]
