If you were to use the substitution method to solve the following system write the new equation after the expression equilavent to x from the first equation is substituted into the second equation x+y=6 3×-y=6

Subtract y from the fist equation to get an expression for x.

... x + y - y = 6 - y

... x = 6 - y

Assuming your multiplication symbol (×) in the second equation is supposed to be the variable x, performing the substitution gives

... 3(6-y) - y = 6 . . . . . . equation after substitution

... 18 - 4y = 6 . . . . . . . . eliminate parentheses

... 12 - 4y = 0 . . . . . . . . subtract 6

... 3 - y = 0 . . . . . . . . . . divide by 4

... 3 = y . . . . . . . . . . . . .add y

Now, we can use the above expression for x to find x

... x = 6 - y = 6 - 3

... x = 3

The solution of the system of equations is (x, y) = (3, 3).

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step-by-step explanation: