How do you solve this system of equations: #y= \sqrt { 2x + 1} and y = \sqrt { 3- 4x }#?

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Answer 1 · 2018-01-12T03:46:56

#x=1/3# and #y=sqrt(5/3)#

We have that:

#y=sqrt(2x+1)# and #y=sqrt(3-4x)#

Begin by setting #y=y# so that:

#sqrt(2x+1)=sqrt(3-4x)#

Now square:

#2x+1=3-4x#

#->6x=2 therefore x=1/3#

Now #y#; substitute into either equation for #y#:

#y=sqrt(2x+1)=sqrt(2(1/3)+1)=sqrt(2/3+1)=sqrt(5/3)#

We can check if we are correct with the other equation:

#y=sqrt(3-4(1/3))=sqrt(3-4/3)=sqrt(5/3)#

So #x=1/3# and #y=sqrt(5/3)#