Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

this has to be done algebraically?

yeah, idk how to do it

i'm not sure how to do the sqrt part along with the y^3 part

i have a suggestion

set \[x^3+\sqrt{x+7}=11\] and solve for \(x\)

you can do it using algebra, but i would not
there is a simpler way

since 11 is an integer, \(x+7\) must be a perfect square

7+2=9, so 2?

an easy check now right?

\[2^3+\sqrt{2+7}=?\]

8+3=11.. so its really 2
wait that's it??

lol yeah, that's it

oh okay thanks.. it's like a trial and error method then