Solve (x+4)div3=3div(x-4)

$$(x+4) \div 3=3 \div (x-4)$$

Solve for x

$$x=-5$$

$$x=5$$

Steps by Finding Square Root

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-4\right), the least common multiple of 3,x-4.

$$\left(x-4\right)\left(x+4\right)=3\times 3$$

Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$x^{2}-16=3\times 3$$

Multiply 3 and 3 to get 9.

$$x^{2}-16=9$$

Add 16 to both sides.

$$x^{2}=9+16$$

Add 9 and 16 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-4\right), the least common multiple of 3,x-4.

$$\left(x-4\right)\left(x+4\right)=3\times 3$$

Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$x^{2}-16=3\times 3$$

Multiply 3 and 3 to get 9.

$$x^{2}-16=9$$

Subtract 9 from both sides.

$$x^{2}-16-9=0$$

Subtract 9 from -16 to get -25.

$$x^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{0±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{0±10}{2}$$

Now solve the equation x=\frac{0±10}{2} when ± is plus. Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{0±10}{2} when ± is minus. Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$