+103 $${\mathtt{12}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{x}}}{{\mathtt{6}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{1}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{x}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{1}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{6}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{x}}}{{\mathtt{1}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{35}}}{{\mathtt{1}}}}\right)$$
through calculater the answer is 60 yet no way of solution please help
+3454 I see you've answered Alan, but I had to post this because I put like 10 minutes into this LaTeX. :)
$$\begin{array}{l}
12(\frac{x}{6}-\frac{5}{1})+5(\frac{x}{5}+\frac{4}{1})+10=6(\frac{x}{1}-\frac{35}{1})\\\\\mbox{Distribute the 12, 5, and 6 into their parentheses}\\\\
(12\times\frac{x}{6})-(12\times \frac{5}{1})+ (5\times \frac{x}{5})+(5\times \frac{4}{1})+10=(6\times \frac{x}{1})-(6\times\frac{35}{1})\\\\
(\frac{12x}{6}) - (\frac{60}{1}) + (\frac{5x}{5}) + (\frac{20}{1}) + 10 =\frac{6x}{1} - \frac{210}{1}\\\\
(\frac{2x}{1}) - (\frac{60}{1}) + (\frac{x}{1}) + (\frac{20}{1}) + 10 =\frac{6x}{1} - \frac{210}{1}\\\\
\mathbf{2x} -60 + \mathbf{x} +20 + 10 =6x -210\\\\
3x \mathbf{- 60} \mathbf{+ 20} \mathbf{+ 10} = 6x - 210\\\\
3x-30 = 6x - 210\\\\
-30= 3x - 210\\\\
180 = 3x\\\\
60 = x\\\\
x = 60
\end{array}$$
+33661 Multiplying through to get rid of the brackets this becomes:
2x - 60 + x + 20 + 10 = 6x - 210
or: 3x - 30 = 6x - 210
Add 210 to both sides:
3x + 180 = 6x
Subtract 3x from both sides:
180 = 3x
Divide both sides by 3:
x = 60
+3454 I see you've answered Alan, but I had to post this because I put like 10 minutes into this LaTeX. :)
$$\begin{array}{l}
12(\frac{x}{6}-\frac{5}{1})+5(\frac{x}{5}+\frac{4}{1})+10=6(\frac{x}{1}-\frac{35}{1})\\\\\mbox{Distribute the 12, 5, and 6 into their parentheses}\\\\
(12\times\frac{x}{6})-(12\times \frac{5}{1})+ (5\times \frac{x}{5})+(5\times \frac{4}{1})+10=(6\times \frac{x}{1})-(6\times\frac{35}{1})\\\\
(\frac{12x}{6}) - (\frac{60}{1}) + (\frac{5x}{5}) + (\frac{20}{1}) + 10 =\frac{6x}{1} - \frac{210}{1}\\\\
(\frac{2x}{1}) - (\frac{60}{1}) + (\frac{x}{1}) + (\frac{20}{1}) + 10 =\frac{6x}{1} - \frac{210}{1}\\\\
\mathbf{2x} -60 + \mathbf{x} +20 + 10 =6x -210\\\\
3x \mathbf{- 60} \mathbf{+ 20} \mathbf{+ 10} = 6x - 210\\\\
3x-30 = 6x - 210\\\\
-30= 3x - 210\\\\
180 = 3x\\\\
60 = x\\\\
x = 60
\end{array}$$
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