Answer :
D. y = 6x...subbing in (2,12)...x = 2 and y = 12
12 = 6(2)
12 = 12 (correct)
so (2,12) is a solution to : y = 6x
12 = 6(2)
12 = 12 (correct)
so (2,12) is a solution to : y = 6x
Answer:
D. y = 6x
Step-by-step explanation:
In order to conclude which pair is a real solution to one of these equations, we need to evaluate the point in each equation.
Let:
[tex](x,y)=(2,12)[/tex]
So, this pair will be a solution if when evaluating the value of x=2 in the function the result is y=12.
For A.
[tex]y(x)=\frac{1}{5}x \\\\y(2)=\frac{2}{5} =0.4\neq12[/tex]
This pair is not a solution to this equation.
For B.
[tex]y(x)=\frac{1}{2}x\\\\y(2)=\frac{2}{2} =1 \neq 12[/tex]
This pair is not a solution to this equation.
For C.
[tex]y(x)=2x\\\\y(2)=2*2=4 \neq 12[/tex]
This pair is not a solution to this equation.
For D.
[tex]y(x)=6x\\\\y(2)=6*2=12[/tex]
This pair is a solution to this equation.