Solve 3+2*3div2-6+kleft(3*3-2*3div2-6right)=0

$$3+2 \times 3 \div 2-6+k \left( 3 \times 3-2 \times 3 \div 2-6 \right) = 0$$

Solve for k

$$k\in \mathrm{R}$$

Solution Steps

Multiply both sides of the equation by 2.

$$6+2\times 3-12+2k\left(3\times 3-\frac{2\times 3}{2}-6\right)=0$$

Multiply 2 and 3 to get 6.

$$6+6-12+2k\left(3\times 3-\frac{2\times 3}{2}-6\right)=0$$

Add 6 and 6 to get 12.

$$12-12+2k\left(3\times 3-\frac{2\times 3}{2}-6\right)=0$$

Subtract 12 from 12 to get 0.

$$2k\left(3\times 3-\frac{2\times 3}{2}-6\right)=0$$

Multiply 3 and 3 to get 9.

$$2k\left(9-\frac{2\times 3}{2}-6\right)=0$$

Cancel out 2 and 2.

$$2k\left(9-3-6\right)=0$$

Subtract 3 from 9 to get 6.

$$2k\left(6-6\right)=0$$

Subtract 6 from 6 to get 0.

$$2k\times 0=0$$

Multiply 2 and 0 to get 0.

$$0k=0$$

Anything times zero gives zero.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any k.

$$k\in \mathrm{R}$$