Solve {l}{(4-4)(4+4)*2=8-4(110-4)}{(4^2-16)=8-40+44}

$$\left. \begin{array} { l } { ( 4 - 4 ) ( 4 + 4 ) \times 2 = 8 - 4 ( 110 - 4 ) } \\ { ( 4 ^ { 2 } - 16 ) = 8 - 40 + 44 } \end{array} \right.$$

Verify

Solution Steps

Subtract 4 from 4 to get 0.

$$0\left(4+4\right)\times 2=8-4\left(110-4\right)\text{ and }4^{2}-16=8-40+44$$

Add 4 and 4 to get 8.

$$0\times 8\times 2=8-4\left(110-4\right)\text{ and }4^{2}-16=8-40+44$$

Multiply 0 and 8 to get 0.

$$0\times 2=8-4\left(110-4\right)\text{ and }4^{2}-16=8-40+44$$

Multiply 0 and 2 to get 0.

$$0=8-4\left(110-4\right)\text{ and }4^{2}-16=8-40+44$$

Subtract 4 from 110 to get 106.

$$0=8-4\times 106\text{ and }4^{2}-16=8-40+44$$

Multiply 4 and 106 to get 424.

$$0=8-424\text{ and }4^{2}-16=8-40+44$$

Subtract 424 from 8 to get -416.

$$0=-416\text{ and }4^{2}-16=8-40+44$$

Compare 0 and -416.

$$\text{false}\text{ and }4^{2}-16=8-40+44$$

Calculate 4 to the power of 2 and get 16.

$$\text{false}\text{ and }16-16=8-40+44$$

Subtract 16 from 16 to get 0.

$$\text{false}\text{ and }0=8-40+44$$

Subtract 40 from 8 to get -32.

$$\text{false}\text{ and }0=-32+44$$

Add -32 and 44 to get 12.

$$\text{false}\text{ and }0=12$$

Compare 0 and 12.

$$\text{false}\text{ and }\text{false}$$

The conjunction of \text{false} and \text{false} is \text{false}.

$$\text{false}$$