Question

I NEED HELP! Last Question! Giving Medals!!! w/PictureAttached

Answer 1

Note that because K is the midpoint of AB then AK=KB which implies:\[\bf AK=KB \implies 3x+1 = \frac{ x }{ 3 }+9\]Now solve for \(\bf x\).

@Hello_Im_Nick

Answer 2

How? @genius12

Answer 3

Bring the x's on one side and the constants on the other:\[\bf 3x+1=\frac{x}{3}+9 \implies 3x-\frac{x}{3}=8\]Now make the denominator the same, make it a single fraction, and then you can solve.

Answer 4

I don't understand.. :/

Answer 5

Ok you have to put 3x and x/3 over the same denominator so that they're written as a single fraction.

Answer 6

Currently the denominator of 3x is 1, to combine 3x and x/3 in to a single fraction, we must then make the denominator of 3x 3 and so we multiply the top and bottom by 3 to give us 9x/3 which will allow us to write the two as a single fraction.

Answer 7

Can I just get the answer and you show me the steps afterwards? I just need it for an exam. And that's the only question I am missing :/

Answer 8

\[\bf 3x-\frac{ x }{ 3 }=8 \implies \frac{ 3x(3) }{ (3) }-\frac{ x }{ 3 }=8 \implies \frac{ 9x }{3 }-\frac{ x }{ 3 }=8 \implies \frac{ 8x }{ 3 }=8 \]\[\bf \implies 8x=24 \implies x = 3\]

Answer 9

Try to understand those steps. @Hello_Im_Nick

Answer 10

So the length of AK is 3?

Answer 11

@genius12