Question

I really need the answer now!! Thanks !

Answer 1

Answer:

The winning ticket is: A. $x^2-3xy+4-2x^2=-x^2-2xy-xy+4$.

Step-by-step explanation:

We have been given that Tomas’s math class held a raffle. The student who picked the ticket with a pair of equivalent equations on it would win.

Let us check our given pairs of equations one by one.

A. $x^2-3xy+4-2x^2=-x^2-2xy-xy+4$

Let us combine like terms on the both sides of our equation.

$(x^2-2x^2)-3xy+4=-x^2-(2xy+xy)+4$

$(-x^2)-3xy+4=-x^2-(3xy)+4$

$-x^2-3xy+4=-x^2-3xy+4$

We can see that both sides of our equation are equal, therefore, option A is the correct choice.

B. $7y^2-yz+4-2yz=7y^2-2yz+4yz$

Let us combine like terms on the both sides of our equation.

$7y^2-(yz+2yz)+4=7y^2-(2yz-4yz)$  

$7y^2-(3yz)+4=7y^2-(-2yz)$

$7y^2-3yz+4\neq 7y^2+2yz$  

Since the both sides of our equation are not equal, therefore, option B is not a correct choice.

C. $3a^2-ab+3-3ab=6a^2-2ab+3-2ab$

Upon combining like terms we will get,

$3a^2-(ab+3ab)+3=6a^2-(2ab+2ab)+3$

$3a^2-(4ab)+3=6a^2-(4ab)+3$

$3a^2-4ab+3\neq 6a^2-4ab+3$

Since the both sides of our equation are not equal, therefore, option C is not a correct choice.

D. $9s^2-st+5-3st=6s^2-3st+5+3s^2$

Upon combining like terms we will get,

$9s^2-(st+3st)+5=(6s^2+3s^2)-3st+5$

$9s^2-(4st)+5=(9s^2)-3st+5$

$9s^2-4st+5\neq 9s^2-3st+5$

Since the both sides of our equation are not equal, therefore, option D is not a correct choice.

Answer 2

When you collect terms, for the first one you get
  -x² -3xy +4 = -x² -3xy +4 . . . . . . equivalent equations.