Solve 10xleq7x^2

$$10 x \leq 7 x ^ { 2 }$$

Solve for x

$$x\in (-\infty,0]\cup [\frac{10}{7},\infty)$$

Solution Steps

Subtract 7x^{2} from both sides.

$$10x-7x^{2}\leq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 10x-7x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-10x+7x^{2}\geq 0$$

Factor out x.

$$x\left(7x-10\right)\geq 0$$

For the product to be ≥0, x and x-\frac{10}{7} have to be both ≤0 or both ≥0. Consider the case when x and x-\frac{10}{7} are both ≤0.

$$x\leq 0 x-\frac{10}{7}\leq 0$$

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

Consider the case when x and x-\frac{10}{7} are both ≥0.

$$x-\frac{10}{7}\geq 0 x\geq 0$$

The solution satisfying both inequalities is x\geq \frac{10}{7}.

$$x\geq \frac{10}{7}$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq \frac{10}{7}$$