Solve 1-9+0^2/9+1-3-1=1

$$\frac { 1 - 9 + 0 ^ { 2 } } { 9 + 1 } - 3 - 1 = 1$$

Verify

Solution Steps

Subtract 9 from 1 to get -8.

$$\frac{-8+0^{2}}{9+1}-3-1=1$$

Calculate 0 to the power of 2 and get 0.

$$\frac{-8+0}{9+1}-3-1=1$$

Add -8 and 0 to get -8.

$$\frac{-8}{9+1}-3-1=1$$

Add 9 and 1 to get 10.

$$\frac{-8}{10}-3-1=1$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$-\frac{4}{5}-3-1=1$$

Convert 3 to fraction \frac{15}{5}.

$$-\frac{4}{5}-\frac{15}{5}-1=1$$

Since -\frac{4}{5} and \frac{15}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-4-15}{5}-1=1$$

Subtract 15 from -4 to get -19.

$$-\frac{19}{5}-1=1$$

Convert 1 to fraction \frac{5}{5}.

$$-\frac{19}{5}-\frac{5}{5}=1$$

Since -\frac{19}{5} and \frac{5}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-19-5}{5}=1$$

Subtract 5 from -19 to get -24.

$$-\frac{24}{5}=1$$

Convert 1 to fraction \frac{5}{5}.

$$-\frac{24}{5}=\frac{5}{5}$$

Compare -\frac{24}{5} and \frac{5}{5}.

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