Solving inequality $\frac{7x+12}{x} \geq 3$

Question

I was sitting with my friend in a park, when he thought of and asked for solving this inequality:

$$ \frac{7x+12}{x} \geq 3 $$

I was confident enough that I could solve this example as continued:

$$ 7x + 12 \geq 3x $$

$$ 4x \geq -12 $$

$$ x \geq -3 $$

He simply told me to put $x = -4$, and in the equation, I applied:

$$ \frac{-28+12}{-4} $$

It got $\frac{-16}{-4}$ which is 4, so it is false that $x \geq -3$. Where I am doing a mistake, and what would be the correct solution.

Answer 1

The problem is just when you send the variable "$x$" to the right hand side of the inequality because you are imposing (implicitly) that $x>0$ so as not to alter the direction of the inequality given by $\geqslant$.

If we want to avoid this, we can rewrite from the beginning $$\frac{7x+12}{x}\geqslant 3\iff \frac{7x+12}{x}-3\geqslant 0\iff \frac{4(x+3)}{x}\geqslant 0.$$

Now use the following fact:

$$\forall (x,y)\in \mathbf{R}\times\mathbf{R}^*: \frac{x}{y}\geqslant 0\iff [(x\geqslant 0)\wedge (y>0)]\vee [(x\leqslant 0)\wedge(y<0)]$$ Translation is:

• Let $x$ be a real and $y$ be a real different of zero, then $\frac{x}{y}\geqslant 0$ if and only if either $x\geqslant 0$ and $y>0$, either $x\leqslant 0$ and $y<0$.

Then $\frac{4(x+3)}{x}\geqslant 0$ iff either:

• $4(x+3)\geqslant 0$ and $x>0$: it has the interval solution given bythe intersection of intervals of solution as $]0,+\infty[$.

• $4(x+3)\leqslant 0$ and $x<0$: it has interval solution given by the intersection of intervals of solution as $]-\infty,-3]$.

Therefore, the set solution for $\frac{7x+12}{x}\geqslant 3$ is the union of intervals $$]-\infty,-3]\cup ]0,+\infty[.$$

Answer 2

The inequality $\,\dfrac{7x+12}x\ge3\,$ is equivalent to $7x+12\ge3x$ if and only if $x>0$. If $x<0$, it is equivalent to $7x+12\le3x$.

Answer 3

As suggested by Another User, you can split the inequlity in two cases. Alternatively, you can study the sign of your expression: $$\frac{7x+12}{x} \geq 3 \iff \frac{7x+12}{x}-3 \ge 0 \iff\frac{x+3}{x}\ge0$$ The last expression is true if $x+3 \ge 0$ and $x>0$ or if $x+3 <0$ and $x<0$.

Answer 4

You made a mistake when you multiplied both sides of the inequality by $x$. Since multiplying an inequality by a negative number reverses the direction of the inequality, you have to consider cases when you multiply both sides of the inequality by $x$.

Method 1

Case 1: $x > 0$ \begin{align*} \frac{7x + 12}{x} & \geq 3\\ 7x + 12 & \geq 3x\\ 4x + 12 & \geq 0\\ 4x & \geq -12\\ x & \geq -3 \end{align*} Since $x > 0$ and $x \geq -3$, $x > 0$.

Case 2: $x < 0$ \begin{align*} \frac{7x + 12}{x} & \geq 3\\ 7x + 12 & \leq 3x\\ 4x + 12 & \leq 0\\ 4x & \leq -12\\ x & \leq -3 \end{align*} Since $x < 0$ and $x \leq -3$, $x \leq -3$.

Since the two cases are mutually exclusive and exhaustive, $x > 0$ or $x \leq -3$. Therefore, the solution set is $$S = (-\infty, -3] \cup (0, \infty) = ]-\infty, -3] \cup ]0, \infty[$$

Method 2:

We can avoid cases if we first subtract $3$ from each side of the inequality. \begin{align*} \frac{7x + 12}{x} & \geq 3\\ \frac{7x + 12}{x} - 3 & \geq 0\\ \frac{7x + 12}{x} - \frac{3x}{x} & \geq 0\\ \frac{4x + 12}{x} & \geq 0\\ \frac{x + 3}{x} & \geq 0 \end{align*} Equality holds when $x = -3$. The strict inequality holds when $x + 3$ and $x$ have the same sign. $x + 3 > 0$ and $x > 0 \implies x > 0$. $x + 3 < 0$ and $x < 0 \implies x < -3$. Thus, the solution set is $$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$

Method 3

Since the direction of the inequality is preserved if we multiply both sides of the inequality by a positive number, we multiply both sides of the inequality by $x^2 > 0$. \begin{align*} \frac{7x + 12}{x} & \geq 3\\ x(7x + 12) & \geq 3x^2\\ 7x^2 + 12x & \geq 3x^2\\ 4x^2 + 12x & \geq 0\\ x^2 + 3x & \geq 0\\ x(x + 3) & \geq 0 \end{align*} Equality holds if $x = 0$ or $x = -3$. However, the original expression is not defined when $x = 0$, so equality holds if and only if $x = -3$. The strict inequality holds when $x + 3$ and $x$ have the same sign. $x + 3 > 0$ and $x > 0 \implies x > 0$. $x + 3 < 0$ and $x < 0 \implies x < -3$. Thus, the solution set is $$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$