Solve 01x+05/02-1<03x+02/03

$$\frac{ 01x+05 }{ 02 } -1 < \frac{ 03x+02 }{ 03 }$$

Evaluate

Solution Steps

Multiply 0 and 1 to get 0. Multiply 0 and 5 to get 0.

$$\frac{0x+0}{0\times 2}-1<\frac{0\times 3x+0\times 2}{0\times 3}$$

Anything times zero gives zero.

$$\frac{0+0}{0\times 2}-1<\frac{0\times 3x+0\times 2}{0\times 3}$$

Add 0 and 0 to get 0.

$$\frac{0}{0\times 2}-1<\frac{0\times 3x+0\times 2}{0\times 3}$$

Multiply 0 and 2 to get 0.

$$\frac{0}{0}-1<\frac{0\times 3x+0\times 2}{0\times 3}$$

Multiply 0 and 3 to get 0. Multiply 0 and 2 to get 0.

$$\frac{0}{0}-1<\frac{0x+0}{0\times 3}$$

Anything times zero gives zero.

$$\frac{0}{0}-1<\frac{0+0}{0\times 3}$$

Add 0 and 0 to get 0.

$$\frac{0}{0}-1<\frac{0}{0\times 3}$$

Multiply 0 and 3 to get 0.

$$\frac{0}{0}-1<\frac{0}{0}$$

Subtract \frac{0}{0} from both sides.

$$\frac{0}{0}-1-\frac{0}{0}<0$$

Subtract \frac{0}{0} from \frac{0}{0} to get 0.

$$-1<0$$

Compare -1 and 0.

$$\text{true}$$