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) $\textcolor{white}.~~\frac x3+\frac x4+\frac x5 ~=~ \frac{2x}3 + \frac{2x}4 + \frac{2x}5$

(b) $\textcolor{white}.~~\frac y3+\frac y4+\frac y5 ~=~ \frac y3+\frac y4+\frac y5+1$

(c) $\textcolor{white}.~~\frac z3+\frac z4+\frac z5 ~=~ \frac z4+\frac z5+\frac z6+1$

Answer 1

Answer:

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Answer 2

(a) We can rewrite the right side of the equation by factoring:

$$\frac x3+\frac x4+\frac x5 = 2\cdot\left(\frac x3+\frac x4+\frac x5\right).$$Then we can subtract $\frac x3+\frac x4+\frac x5$ from both sides of the equation to get

$$0 = 2\left(\frac x3+\frac x4+\frac x5\right)-\left(\frac x3+\frac x4+\frac x5\right)=\frac x3+\frac x4+\frac x5,$$so

$$0 = \left(\frac 13+\frac 14 + \frac 15\right)x.$$Dividing both sides by $\left(\frac 13+\frac 14 + \frac 15\right)$, we have one solution, $x=0$.

(b) The two sides of this equation always differ by $1$, no matter what is substituted for $y$. So, there cannot be any solutions.

(c) We can subtract $\frac z4+\frac z5$ from both sides of the equation to get

$$\frac z3=\frac z6+1,$$then we can carefully multiply both sides by $6$ to get

$$6\cdot\frac z3 = 6\cdot\left(\frac z6 + 1\right).$$Distributing on the right side gives

$$6\cdot\frac z3 = 6\cdot\frac z6 + 6\cdot 1,$$which simplifies to

$$2z = z+6.$$Finally, subtracting $z$ from each side gives $z=6$. So there is one solution: $z=6$.