hard question

4*7y+9+1=10

 

 10

-  1

-  9

--------

0

 

so then get 4*7y to 0 (so y = 0.....giving you 4*7y......4*(7*0).......4*(0)....0)

 

so y = 0

 

I could be wrong, but it may be right

\(4\times7y+9+1=10\)

 

\(28y+9+1=10\)

 

\(28y+10=10\)

 

\(28y+10-10=10-10\)

 

\(28y+0=10-10\)

 

\(28y=10-10\)

 

\(28y=0\)

 

\(\frac{28y}{28}=\frac{0}{28}\)

 

\(\frac{1y}{1}=\frac{0}{28}\)

 

\(1y=\frac{0}{28}\)

 

\(y=\frac{0}{28}\)

 

\(y=0\)

 

Now put the answer you came up with and put back into the original equation and solve to see if the answer you came up with is the correct answer.

 

\(4\times7y+9+1=10\)

 

\(4\times7\times0+9+1=10\)

 

\(28\times0+9+1=10\)

 

\(0+9+1=10\)

 

\(9+1=10\)

 

\(10=10\)

 

Because you came up with \(10=10\), that means that \(y=0\) is the correct answer.