Solve for x (x^2)/(x^2+1)=(7x)/(x^2+1)

$\frac{x^{2}}{x^{2} + 1} = \frac{7 x}{x^{2} + 1}$

Step 1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x^{2} = 7 x$

Step 2

Subtract $7 x$ from both sides of the equation.

$x^{2} - 7 x = 0$

Step 3

Factor $x$ out of $x^{2} - 7 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 7 x = 0$

Factor $x$ out of $- 7 x$ .

$x \cdot x + x \cdot - 7 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$x (x - 7) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 7 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 7

The final solution is all the values that make $x (x - 7) = 0$ true.

$x = 0, 7$