The Domain of an Algebraic Expression: More than one unknown

Let's insert the given data into the expression:

8*50-0(7+50) =
400-0*57 =
400-0 =
400

Note that we have two unknowns, a and b, and we are also given values for them,

Therefore, let's start by substituting these values in the equation instead of the unknowns:

8*2-1/3*(7+2)=

When there is a number before parentheses, it's like having a multiplication sign between them.

Let's start solving according to the order of operations, beginning with the parentheses:

8*2-1/3*(9)=

Now let's continue with multiplication and division:

16-9/3=
16-3=

‎‎‎‎‎‎‎13

And that's the solution!

We have the given exercise, and it has two variables, X and Y.

In this case, we are given the values of these variables,

Therefore, what we need to do is substitute them in the relevant place in the exercise:

8(x-7)+4(6-2y)=

We know that x=0, so we will replace every X in the exercise with 0:

8(0-7)+4(6-2y)=
8(-7)+4(6-2y)=
-56+4(6-2y)=

We'll do the same thing with y, knowing that it equals -1

-56+4(6-2*(-1))=
-56+4(6-(-2)))=

-56+4(8)=

-56+32=

-24

And that's the solution!