Question

solve for x!

Answer 1

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

Answer 2

multiply the 1st fraction (top and bottom) by x, add everything on the left side and cross multiply.

Answer 3

how exactly do you do that?

Answer 4

\[\frac{1}{2}+\frac{4}{2x}=\frac{x+4}{10}\]\[\frac{1 \times x}{2 \times x}+\frac{4}{2x}=\frac{x+4}{10}\]\[\frac{x}{2x}+\frac{4}{2x}=\frac{x+4}{10}\]\[\frac{x+4}{2x}=\frac{x+4}{10}\]\[2x=10\]\[x=5\]

Answer 5

This should be helpful.

Answer 6

thanks. how about this one?

solve for x :\[\frac{ 2 }{ 5}+ \frac{ 3 }{ 5x}= \frac{ x+5 }{ 10 }\]

Answer 7

@SolomonZelman

Answer 8

\[\frac{2}{5}+\frac{3}{5x}=\frac{x+5}{10}\]

Answer 9

\[\frac{2 \times x}{5 \times x}+\frac{3}{5x}=\frac{x+5}{10}\]

Answer 10

\[\frac{2x}{5x}+\frac{3}{5x}=\frac{x+5}{10}\]

Answer 11

\[\frac{2x+3}{5x}=\frac{x+5}{10}\]

Answer 12

\[\frac{2x+3}{x}=\frac{x+5}{2}\]

Answer 13

\[4x+6=x^2+5x\]

Answer 14

\[x^2+x-6=0\]

Answer 15

\[(x+3)(x-2)=0\]

Answer 16

\[x=-3,~~~2\]

Answer 17

so you basically factor it?

Answer 18

when it is \[x^2+x-6=0\]after cross multiplying, yes!

Answer 19

ha! i think i get it! ok ill just post and let you check my answers so you don't have to do all that again

Answer 20

\[\frac{ 5 }{x^2-4 } + \frac{ 2 }{ x }+\frac{ 2 }{ x-2}\]

Answer 21

i got x = 8. is that right?

Answer 22

@SolomonZelman

Answer 23

retype the question please, (with the equal sign)

Answer 24

oh whoops. just replace the last plus with equal

Answer 25

\[\frac{5}{x^2-4}+\frac{2}{x}=\frac{2}{x-2}\]

Answer 26

\[\frac{2}{x}=\frac{2}{x-2}-\frac{5}{x^2-4}\]

Answer 27

\[\frac{2}{x}=\frac{2}{x-2}-\frac{5}{(x+2)(x-2)}\]\[\frac{2}{x}=\frac{2(x+2)}{(x-2)(x+2)}-\frac{5}{(x+2)(x-2)}\]

Answer 28

\[\frac{2}{x}=\frac{2(x+2)+5}{(x+2)(x-2)}\]\[\frac{2}{x}=\frac{2x+7}{(x+2)(x-2)}\]

Answer 29

\[2x^2+7x=2x^2-8\]\[x=-7/8\]

Answer 30

2 more?

Answer 31

\[\frac{ 2 }{ x +2 }+\frac{ 1 }{ 5 } = \frac{ 6 }{ x + 5}\]