adding and subtracting like terms

[zhyia]

can i solve this problem by 1/(2+X–3X + 3 + 2X -5)
adding the like terms
2+3-5=0
and
1x-3x+2x=0

would the final answer be zero?

 

[steve_b]

can i solve this problem by 1/(2+X–3X + 3 + 2X -5)
adding the like terms
2+3-5=0
and
1x-3x+2x=0

would the final answer be zero?

___________________

If the statement is 1 DIVIDED BY (2 + x - 3x + 3 + 2x -5), then you do get 0 in the denominator, so you have 1/0 which is undefined.

Steve

 

[stapel]

can i solve this problem by 1/(2+X–3X + 3 + 2X -5)
adding the like terms

One "solves" equations; you have posted an expression.

One is never allowed to divide by zero. The value of 1/0 is not zero.

What are the actual instructions for this exercise?

Eliz.

 

[zhyia]

the problem says evaluate: [(¾)² + (½)³ + π]° ÷ (2+X–3X + 3 + 2X -5)
I already know that this part [(¾)² + (½)³ + π]° =1
so what i am trying to figure out is how to evaluate or add and subtract like terms in the remaining part of this problem (2+X–3X + 3 + 2X -5)?

 

[pka]

Because the denominator is equal to 0, the answer is undefined!
We do not allow division by zero.