Plug the x and y-values into the given equation. If it makes a true statement, then it is a solution.
y = x + [tex] \frac{2}{5} [/tex]
(0,0) → 0 = 0 + [tex] \frac{2}{5} [/tex] FALSE
(1, [tex] \frac{2}{5} [/tex]) → [tex] \frac{2}{5} [/tex] = 1 + [tex] \frac{2}{5} [/tex] FALSE
(2, 2 [tex] \frac{2}{5} [/tex]) → 2 [tex] \frac{2}{5} [/tex] = 2 + [tex] \frac{2}{5} [/tex] TRUE
Answer: (2, 2 [tex] \frac{2}{5} [/tex])
The ordered pair which is a solution to the given equation is:
[tex](2,2\dfrac{2}{5})[/tex]
We are given a equation in terms of x as follows:
[tex]y=x+\dfrac{2}{5}[/tex]
Now, we are asked to find which ordered pair is a solution to the given equation.
i.e. we will put each of the given points in the equation and see which holds true.
a)
(0,0)
we put x=0 and y=0 in the equation.
[tex]0=0+\dfrac{2}{5}\\\\i.e.\\\\0=\dfrac{2}{5}[/tex]
which is not a true statement.
Hence, (0,0) is not a solution to the equation.
b)
(1,2/5)
we put x=1 and y=2/5
[tex]\dfrac{2}{5}=1+\dfrac{2}{5}\\\\i.e.\\\\\dfrac{2}{5}=\dfrac{5+2}{5}\\\\\dfrac{2}{5}=\dfrac{7}{5}[/tex]
which is again incorrect.
Hence, (1,2/5) is not a solution to the equation.
c)
(2, 2 2/5)
Now, we put [tex]x=2\ and\ y=2\dfrac{2}{5}[/tex]
i.e.
[tex]x=2\ and\ y=\dfrac{12}{5}[/tex]
Hence,
[tex]\dfrac{12}{5}=2+\dfrac{2}{5}\\\\\dfrac{12}{5}=\dfrac{2\times 5+2}{5}\\\\\dfrac{12}{5}=\dfrac{12}{5}[/tex]
which is correct.
Hence,
[tex](2,2\dfrac{2}{5})\ \text{is a solution to the given equation}[/tex]
