In 35, 3 is in the tens place and 5 is in the ones place, that is,
place value of 3: 30
place value of 5: 5
This means that 35 can be expressed as 30 + 5.
Substituting 35 = 30 + 5 into the multiplication given, we get:
[tex]\begin{gathered} 2\cdot35=2\cdot(30+5) \\ \text{Distributing the multiplication over the addition:} \\ 2\cdot35=(2\cdot30)+(2\cdot5) \\ \text{ Solving each multiplication on the right} \\ 2\cdot35=60+10 \\ 2\cdot35=70 \end{gathered}[/tex]
