Question

The number of real numbers x such that $\frac{x}{x+4} = \frac{5[x]-7}{7[x]-5}$ is ______ ([x] denotes greatest integer $\leq x$).

Answer 1

3

Answer 2

Let $k = [x]$. Then $k$ is an integer and $k \leq x < k+1$. The given equation is $\frac{x}{x+4} = \frac{5[x]-7}{7[x]-5}$. Substitute $[x]=k$: $\frac{x}{x+4} = \frac{5k-7}{7k-5}$

For the equation to be defined, we must have $x+4 \neq 0$ and $7k-5 \neq 0$. Since $k$ is an integer, $7k-5 \neq 0$. Cross-multiply: $x(7k-5) = (x+4)(5k-7)$ $7kx - 5x = 5kx - 7x + 20k - 28$ $7kx - 5x - 5kx + 7x = 20k - 28$ $2kx + 2x = 20k - 28$ $2x(k+1) = 20k - 28$ $x(k+1) = 10k - 14$

Case 1: $k+1 \neq 0$, i.e., $k \neq -1$. In this case, we can solve for $x$: $x = \frac{10k - 14}{k+1}$

We use the condition $[x] = k$, which means $k \leq x < k+1$. Substitute the expression for $x$: $k \leq \frac{10k - 14}{k+1} < k+1$

We split this into two inequalities:

1. $k \leq \frac{10k - 14}{k+1}$
2. $\frac{10k - 14}{k+1} < k+1$

To solve these inequalities, we consider the sign of $k+1$.

Subcase 1.1: $k+1 > 0$, i.e., $k > -1$. Since $k$ is an integer, $k \geq 0$.

1. $k(k+1) \leq 10k - 14$ $k^2 + k \leq 10k - 14$ $k^2 - 9k + 14 \leq 0$ The roots of $k^2 - 9k + 14 = 0$ are $k=2$ and $k=7$. The inequality holds for $2 \leq k \leq 7$. Since $k \geq 0$, the integer values satisfying this are $k \in \{2, 3, 4, 5, 6, 7\}$.

2. $\frac{10k - 14}{k+1} < k+1$ $10k - 14 < (k+1)^2$ $10k - 14 < k^2 + 2k + 1$ $0 < k^2 - 8k + 15$ The roots of $k^2 - 8k + 15 = 0$ are $k=3$ and $k=5$. The inequality holds for $k < 3$ or $k > 5$. Since $k \geq 0$, the integer values satisfying this are $k \in \{0, 1, 2, 6, 7, 8, ...\}$.

For $k \geq 0$, we need to find integers $k$ that satisfy both $2 \leq k \leq 7$ and ($k < 3$ or $k > 5$). The integers satisfying both conditions are $k \in \{2, 6, 7\}$. Let's check the value of $x = \frac{10k-14}{k+1}$ for these values of $k$: If $k=2$, $x = \frac{10(2)-14}{2+1} = \frac{6}{3} = 2$. $[x] = [2] = 2$, which matches $k=2$. $x=2 \neq -4$. This is a solution. If $k=6$, $x = \frac{10(6)-14}{6+1} = \frac{46}{7}$. $[x] = [\frac{46}{7}] = [6 + \frac{4}{7}] = 6$, which matches $k=6$. $x=46/7 \neq -4$. This is a solution. If $k=7$, $x = \frac{10(7)-14}{7+1} = \frac{56}{8} = 7$. $[x] = [7] = 7$, which matches $k=7$. $x=7 \neq -4$. This is a solution.

Subcase 1.2: $k+1 < 0$, i.e., $k < -1$. Since $k$ is an integer, $k \leq -2$. When multiplying or dividing by $k+1$, we reverse the inequality sign.

1. $k \leq \frac{10k - 14}{k+1}$ $k(k+1) \geq 10k - 14$ $k^2 - 9k + 14 \geq 0$ This holds for $k \leq 2$ or $k \geq 7$. Since $k \leq -2$, the integers satisfying this are $k \in \{..., -4, -3, -2\}$.

2. $\frac{10k - 14}{k+1} < k+1$ $10k - 14 > (k+1)^2$ $10k - 14 > k^2 + 2k + 1$ $0 > k^2 - 8k + 15$ This holds for $3 < k < 5$. There are no integers $k \leq -2$ that satisfy $3 < k < 5$. So, there are no solutions in this subcase.

Case 2: $k+1 = 0$, i.e., $k = -1$. The equation $x(k+1) = 10k - 14$ becomes $x(0) = 10(-1) - 14$, which is $0 = -24$. This is a contradiction, so there are no solutions when $k = -1$.

Combining all cases, the possible integer values for $[x]$ are $k=2, 6, 7$. For each of these values of $k$, we found a corresponding value of $x$: If $k=2$, $x=2$. $[2]=2$. Valid. If $k=6$, $x=46/7$. $[46/7]=6$. Valid. If $k=7$, $x=7$. $[7]=7$. Valid.

We also need to check the original constraint $x \neq -4$. The solutions are $x=2, x=46/7, x=7$. None of these are equal to $-4$.

Thus, there are exactly 3 real numbers $x$ that satisfy the given equation.