Ok so my problem is -3(3x+15)-(10+x)=35 After doing most of the problem I ended up getting -10x=105. I need to have a whole number as an answer but I can't understand what to do.

i don't know how you got 105. it should be 90.

so,

-10x=90

divide by -10

Hmmmm

-9x-45-10-x=35
-10x=90
x=-9

check that.

Oh thank you both! I for some reason thought 15x3 was 60!

To solve the equation -3(3x+15)-(10+x)=35, let's first simplify both sides of the equation.

Start by applying the distributive property on the left side of the equation:
-3(3x+15) = -9x - 45

Now, rewrite the equation with the simplified left side:
-9x - 45 - (10+x) = 35

Next, simplify the expression within the parentheses:
-9x - 45 - 10 - x = 35

Combine like terms:
-9x - x - 55 = 35

Combine the x terms:
-10x - 55 = 35

To isolate the variable x, you need to move the constant term (-55) to the other side of the equation.

Adding 55 to both sides:
-10x - 55 + 55 = 35 + 55

Simplifying:
-10x = 90

Now, to get x alone, divide both sides of the equation by -10:
-10x / -10 = 90 / -10

Simplifying:
x = -9

So the value of x that satisfies the equation is -9.

However, you mentioned that you need a whole number as an answer. In this case, since -9 is an integer but not a whole number (whole numbers are non-negative integers), we need to go back and check if we made any mistakes during the simplification process.

Let's double-check the equation:
-3(3x+15)-(10+x)=35

Starting from where we simplified the left side of the equation:

-9x - 45 - (10 + x) = 35

Applying the distributive property:

-9x - 45 - 10 - x = 35

Combining like terms:

-10x - 55 = 35

To get a whole number as an answer, we can try adding the constant term 55 to both sides before dividing by -10:

-10x - 55 + 55 = 35 + 55

Simplifying:

-10x = 90

Dividing both sides by -10:

-10x / -10 = 90 / -10

Simplifying:

x = -9

Since we arrived at the same result again, -9 is indeed the solution to the equation. Although it is not a whole number, it is the correct answer based on the given equation.