Question

1/10(3x-7) = 3/10x + 5 just need to check my work

Answer 1

is the the final result that your checking?

Answer 2

yes

Answer 3

i got no solution

Answer 4

I'm assuming the problem is:
\[\frac{1}{10(3x-7)}=\frac{3}{10x+5}\]
Cross multiply:
\[10x+5=30(3x-7)\]
\[10x+5=90x-210\]
\[80x=215\]
\[x=\frac{215}{80}\]
\[x=\frac{43}{16}\]

Answer 5

well, lets step thru it, first, is this equational stuff?

\[\frac{1}{10(3x-7)} = \frac{3}{10x} + 5\]

Answer 6

no i really need to learn to put it down the way you do the 1is over the 10

Answer 7

and (3x-7)

Answer 8

like this?
\[\frac{1}{10}(3x-7) = \frac{3}{10x} + 5\]

Answer 9

The method used is \frac{numerator}{denominator}.

Answer 10

the equation editor button at the bottom can be useful, but its might be difficult to use

Answer 11

yes that is right

Answer 12

is that x on the right side under or beside the 3/10?
\[\frac{1}{10}(3x-7) = \frac{3}{10}x + 5\]

Answer 13

that is right it is beside the \[3/10\]

Answer 14

ok

\(\cfrac{1}{10}(3x-7) = \cfrac{3}{10}x + 5\) ; lets multiply it all by 10, do you see why?

\(\cfrac{1(10)}{10}(3x-7) = \cfrac{3(10)}{10}x + 5(10)\)

\(3x-7 = 3x + 50\)

Answer 15

and yes, when we have the same value of x on both sides it dissapears and we get no solution

Answer 16

say 3x = 4 for example...

4-7 = 4+50 ?? no

Answer 17

the only time it would give a solution is when you have the same thing on both sides; then the solution is "all numbers"

Answer 18

right i see

Answer 19

the reason we used ten is that it is LCD and then we divided it by the LCD and do we do that every time?

Answer 20

dont worry about LCM, i never do.

see what you need to do to clear the fractions; by that i mean turn them into "1"

3/5 would be cleared if that 5 on the bottom was gone right?

3(5)/5 = 3 * 5/5 = 3*1 = 3

to clear the fractions, do them one at a time if there are different denominators

Answer 21

3/4 + 7/3 = x ; clear the fractions

3(4)/4 + 7(4)/3 = (4)x

3 + 28/3 = 4x ; and again

3(3) + 28(3)/3 = 4(3)x

9 + 28 = 12x

37 = 12x ; divide off the 12

37/12 = x reduce as wanted