Let xx cm be the length of the shorter piece. Since the longer piece is 2 cm longer than the shorter piece, then the longer piece is x+2 cm.

The combined length of the two pieces is then x+(x+2)=2x+2 cm. Since the length of the wire before it was cut was 22 cm, then 2x+2=22.

To solve 2x+2=22, we need to isolate the variable term of 2x and then isolate the variable of xx using inverse operations.

Since 2 is being added to the 2x, we need to subtract 2 on both sides to isolate the 2x because subtraction is the inverse operation of addition. Subtracting 2 on both sides gives:

2x+2−2=22−2

2x=20

Since the xx is being multiplied by 2, we need to divide both sides by 2 to isolate the xx because division is the inverse operation of multiplication. Dividing both sides by 2 gives:

\(\displaystyle{\frac{{{2}{x}}}{{{2}}}}={\frac{{{20}}}{{{2}}}}\)

x=10

The length of the shorter piece is then x=10 cm and the length of the longer piece is x+2=10+2=12 cm

The combined length of the two pieces is then x+(x+2)=2x+2 cm. Since the length of the wire before it was cut was 22 cm, then 2x+2=22.

To solve 2x+2=22, we need to isolate the variable term of 2x and then isolate the variable of xx using inverse operations.

Since 2 is being added to the 2x, we need to subtract 2 on both sides to isolate the 2x because subtraction is the inverse operation of addition. Subtracting 2 on both sides gives:

2x+2−2=22−2

2x=20

Since the xx is being multiplied by 2, we need to divide both sides by 2 to isolate the xx because division is the inverse operation of multiplication. Dividing both sides by 2 gives:

\(\displaystyle{\frac{{{2}{x}}}{{{2}}}}={\frac{{{20}}}{{{2}}}}\)

x=10

The length of the shorter piece is then x=10 cm and the length of the longer piece is x+2=10+2=12 cm