Question

0 = 5/6x + 2/7x + 1

My teacher says to multiply by 42 to get rid of the fractions. ...It's not exactly working for me. How should I do it?

Answer 1

is that \( \frac{5}{6x} \) or \( \frac{5}{6}x \)?

Answer 2

I can't seem to be able to type it out using the Equation system, but it's the second one. Same with 5/7x, they're both fractions.

Answer 3

just use brackets

Answer 4

So you have
\[ 0 = \frac{5}{6}x + \frac{2}{7}x + 1 \]
Now multiply both sides by 42 and you have
\[ 42(0) = 0 = 42\frac{5}{6}x + 42\frac{2}{7}x + 1 \]

Now simplify that expression.

Answer 5

*correction:
\[ 0 = 42\frac{5}{6}x + 42\frac{2}{7}x + 42 \]

Answer 6

Okay, I now have:
\[0 = \frac{257}{6}x +\frac{296}{7}x + 42\]
\[0 =\frac{3575}{42}x + 42\] or \[0 = 85\frac{5}{42}x + 42\]

Do I divide by 42 to get rid of that fraction?

Answer 7

If I were you, I would cancel first.
for example:
\[ \frac{42\cdot 5}{6}= 7\cdot5= 35\]
do that for both fractions before taking the next step.

Answer 8

That is why you multiply by 42, to get rid of the 6 and the 7 in the denominators

Answer 9

I think I got it!

0 = 5/6x + 2/7x + 1
(42)0 = 42(5/6x) + 42(2/7x) + 42(1)
0 = 35x + 12x + 42
0 = 47x + 42
-42 = 47x
-42/47 = x

Answer 10

looks good. It is easier working with smaller numbers, isn't it?

Answer 11

Definitely! Thanks to the both of you. :D