Question

For each of these three equations, determine whether there are any solution(s). If there are solutions, say what they are and show how you found them. If there aren't solutions, explain why not.

(a) x/3+x/4+x/5 =2x/3 + 2x/4 + 2x/5

(b) y/3+ y/4+y/5 = y/3+ y/4+ y/5+1

(c) z/3+ z/4+\z/5 = z/4+ z/5+ z/6+1

Answer 1

Solve for x:
\[\frac{47 x}{60} = \frac{47 x}{30}\]Divide both sides by a constant to simplify the equation.
Divide both sides by 47:
\[\frac{x}{60} = \frac{x}{30}\]Make \(\frac{x}{60} = \frac{x}{30}\) simpler by multiplying both sides by a constant.
Multiply both sides by 60:
\[\frac{60 x}{60} = \frac{60 x}{30}\]Cancel common terms in the numerator and denominator of \(\frac{60 x}{60}.
\[\frac{60 x}{60} = \frac{60}{60}×x = x:\]\[x = \frac{60 x}{30}\]In \(\frac{60 x}{30}\), divide 60 in the numerator by 30 in the denominator.
\[\frac{60}{30} = \frac{30×2}{30} = 2:\]\[x = 2 x \]Move terms with x to the left hand side.
Subtract 2 x from both sides:
\[x-2 x = 2 x-2 x \]Combine like terms in x-2 x.
\[x-2 x = -x: \]\[-x = 2 x-2 x \]Look for two terms that sum to zero.
\[2 x-2 x = 0: \]\[-x = 0 \]Multiply both sides by a constant to simplify the equation.
Multiply both sides of \(-x = 0\) by -1:
\[-\frac{x}{-1} = 0\]Any nonzero number divided by itself is one.
\[\frac{-1}{-1} = 1:\]

Answer 2

Sorry for the extra gunk. What does this leave you with for equation a?

Answer 3

Here's a hint.. Only one number can possibly fit \(2x=x\)

Answer 4

oh I got the answer
by the way I got it right

Answer 5

Awesome, nice job :)