The number 'in front of' an unknown is called the Coefficient. When eliminating an unknown, look to see whether the coefficients of x are the same in both equations, or if the coefficients of y are the same. In this case, the y coefficients are both 1, and so we can 'get rid of' y from our equations.

Adding the equations together we get; 2x + 3x + y - y = 7 + 8 ---> 5x + 0 = 15 ---> x = 3 The y's cancel out.

Then subsitute x = 3 into either equation, we'll use the first; 2(3) + y = 7 ---> 6 + y = 7 ---> y = 1

By substitution. Rearrange one of the equations to get x = ... or y = ... . If we rearrange the first equation we get y = 7 - 2x.

Now we can substitute y = 7 - 2x into the second equation; 3x - (7 - 2x) = 8. Use brackets here so that we don't get confused with signs.

Expand this out; 3x - 7 + 2x = 8 ---> 5x = 15 ---> x = 3

Then substitute x = 3 into our 'y' equation; y = 7 - 2(3) ---> y = 7 - 6 ---> y = 1

We get x = 3 and y = 1 with both methods.

Let f(x)= x/5 + 1 . Find f^-1(x)