# Algebra-I really need help, please?

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I have to solve sqrt(2x-7) >=5
x>=7/2
7/2<=x<=16
x<=16

I don't get this concept at all and I have 20 of these kinds of problems for homeowrk-can someone direct me how to start or give me an example or just help please? I really need to figure out this concept
Thank you

• Algebra-I really need help, please? - ,

√(2x-7) > 5

Since both left and right side are positive we don't have to worry about any changes in direction of the inequality sign.
So square both sides

2x-7 ≥ 25
2x ≥ 32
x ≥ 16

and of course for √(2x-7) to be a real number, the inside must be ≥ 0
that is, 2x-7 ≥ 0
2x ≥ 7
x ≥ 7/2

so now you have x ≥ 7/2 AND x ≥ 16

which set of numbers satisfies BOTH conditions?
It would be x ≥ 16

(take x = 5 for an example
it satisfies the first condition, in that it is ≥ 7/2
but does not work for the original inequation. )

• Algebra-I really need help, please? - ,

Thank you for the detailed example-I can really use those to figure out the rest of my homeowrk-I should be good now

Thanks again

• Algebra-I really need help, please? - ,

welcome, come back if you run into more problems.

• Algebra-I really need help, please? - ,

sqrt(2x-7) >= 5

What don't you understand?
to solve, square both sides to get rid of the radical

2x - 7 >= 25
2x >= 32
x >= 16