Solve the system algebraically: y = sqrt(x) and y=2x
Could I help you?
Yes please
Thank you. With my pleasure! Do you like to do so by substitution or elimination?
I started doing substation on it and got 2x = sqrt x but then I blanked out on what to do next.
substitution*
Both of them are y terms. Basically, set the two equations equal to each other. sqrt(x) = 2x
I have that so far, but after that how would I solve it? Sorry, I'm blanking out on how to do the basics.
We can use either substitution where we plug one equation into the other, or elimination where we combine the equations.
\[\large \sqrt x = 2x\]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?
Would I need to do the quadratic formula?
Sure, but you can always factor it since it's easier that way.\[\large x(4x-1) = 0\]\[\large x=0, ~~ x=\frac{ 1 }{ 4 }\]
Oh yeah, you're completely right. And then after than I plug in the x's to find the y's right?
Yup, nice work!
Thanks for the help!
You're welcome!
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