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Mathematics 9 Online
OpenStudy (alexh107):

Solve the system algebraically: y = sqrt(x) and y=2x

OpenStudy (3mar):

Could I help you?

OpenStudy (alexh107):

Yes please

OpenStudy (3mar):

Thank you. With my pleasure! Do you like to do so by substitution or elimination?

OpenStudy (alexh107):

I started doing substation on it and got 2x = sqrt x but then I blanked out on what to do next.

OpenStudy (alexh107):

substitution*

OpenStudy (steve816):

Both of them are y terms. Basically, set the two equations equal to each other. sqrt(x) = 2x

OpenStudy (alexh107):

I have that so far, but after that how would I solve it? Sorry, I'm blanking out on how to do the basics.

OpenStudy (3mar):

We can use either substitution where we plug one equation into the other, or elimination where we combine the equations.

OpenStudy (steve816):

\[\large \sqrt x = 2x\]you can square both sides of the equation to get rid of the square root\[\large (\sqrt x)^2=(2x)^2\]\[\large x = 4x^2\]\[\large 4x^2-x = 0\] Now do you think you can solve for x now?

OpenStudy (alexh107):

Would I need to do the quadratic formula?

OpenStudy (steve816):

Sure, but you can always factor it since it's easier that way.\[\large x(4x-1) = 0\]\[\large x=0, ~~ x=\frac{ 1 }{ 4 }\]

OpenStudy (alexh107):

Oh yeah, you're completely right. And then after than I plug in the x's to find the y's right?

OpenStudy (steve816):

Yup, nice work!

OpenStudy (alexh107):

Thanks for the help!

OpenStudy (steve816):

You're welcome!

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