`1+sqrt((7x)/3)=8`

As for most equations, we try to isolate the variable by using inverse operations.

Subtract 1 from each side:

`sqrt((7x)/3)=7`

Square both sides -- be aware that squaring both sides of an equation can lead to extraneous solutions.

`(7x)/3=49`

Multiply both sides of the equation by 3/7:

x=21

Check the answer in the original equation. (This must be done in order to check if we have introduced an extraneous solution.)

`1+sqrt((7*21)/3)=1+sqrt(49)=8` as required.

Note that `sqrt((7x)/3)` is only defined in the reals if the argument is nonnegative; so `x>=0` .``

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The solution is x=21. The expression on the left side of the equation is only defined in the reals for x nonnegative.

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The graph of the left and right side expressions: