Question

solve for x!

Answer 1

solve what

Answer 2

\[\frac{ 1 }{ 2 } + \frac{ 4 }{ 2x }= \frac{ x+4 }{ 10}\]

Answer 3

@Loser66 @amistre64 @myininaya

Answer 4

id suggest multiplying the whole thing by 2x ... that should get rid of the fractions

Answer 5

well, id still have a 5 underneath ... 10x it then

Answer 6

what do you mean?

Answer 7

how do i multiply fractions by 2x

Answer 8

same way you multiply anything else by ... in this case lets make it 10x

Answer 9

\[\frac{ 1(10x) }{ 2 } + \frac{ 4(10x) }{ 2x }= \frac{ 10x(x+4) }{ 10}\]
and cancel what can be canceled

i choose 10x so taht we would be able to work with integers

Answer 10

oh ok i see

Answer 11

i got \[\frac{ 10x }{ 2 } + \frac{ 40x }{ 2x } = \frac{ 10x^2 + 4 }{ 10 }\]

Answer 12

while that is proper, it doesnt really use the notion of WHY we would multiply by 10x in the first place.

\[\frac{ 1(10x) }{ 2 } + \frac{ 4(10x) }{ 2x }= \frac{ 10x(x+4) }{ 10}\]
\[\frac{ 1(\cancel{10}^5x) }{ \cancel2 } + \frac{ 4(\cancel{10x}^5) }{ \cancel{2x} }= \frac{ \cancel{10}x(x+4) }{ \cancel{10}}\]
\[5x+20=x(x+4)\]

Answer 13

hmm i don't understand how that gets us x tho

Answer 14

it just reworks it into a familar looking quadratic expression, and I assume you have been playing with quadratics?

Answer 15

do i solve for x in the equation?

Answer 16

\[5x+20=x(x+4)\]
\[20=x(x+4)-5x\]
\[20=x(x-1)\]
just off hand, id say x is simple to SEE at this moment

Answer 17

but yes, you would work it out so that you could solve for x, which may have 2 solutions

\[5x+20=x(x+4)\]
\[5x+20=x^2+4x\]
\[0=x^2-x-20\]

Answer 18

so x can equal -4 and 5 right

Answer 19

@amistre64