Answer:
[tex]y=\sqrt{\dfrac{3}{7}x}+5[/tex]
Step-by-step explanation:
Suppose [tex]y=3t+5[/tex] and [tex]x=21t^2.[/tex]
Express t from the first equation:
[tex]3t=y-5\\ \\t=\dfrac{y-5}{3}[/tex]
Substitute it into the second equation:
[tex]x=21\left(\dfrac{y-5}{3}\right)^2\\ \\x=21\cdot\dfrac{(y-5)^2}{9}\\ \\(y-5)^2=\dfrac{9}{21}x\\ \\y-5=\pm \sqrt{\dfrac{3}{7}x}[/tex]
Since [tex]t\ge 0,[/tex] then [tex]y-5\ge 0,[/tex] so
[tex]y-5=\sqrt{\dfrac{3}{7}x}\\ \\y=\sqrt{\dfrac{3}{7}x}+5[/tex]
