Question

prove that the equation (a^(2)-5ab)-(7-3ab)+(2ab-a^(2)) is equivilent to -7 please show it step by step! Tysm!

Answer 1

Answer:

Step-by-step explanation:

you can bring -(7-3ab) to the other side of the equation . then you will have this : (a^2 - 5ab) + (2ab - a^2) = -7 + (7 - 3ab)

now look at this part : -7 + (7 - 3ab) (because adding is a associative procedure) you can get rid of -7 and 7 . then you will have only -3ab

so now this would look like this :  (a^2 - 5ab) + (2ab - a^2) = -3ab

now take a look at the left of the equation : just like the last part we can simplify a^2 and - a^2 . and -5ab + 2ab would result in -3ab

because if you subtract -5ab from a^2 and subtract a^2 from 2ab ... then add then to each other would give you the same result

at the end you will have -3ab = -3ab which is a true conclusion.

Answer 2

Answer:

Step-by-step explanation:

(a^2 - 5ab) - (7 - 3ab) + (2ab - a^2) = -7

a^2 - 5ab -7 +3ab + 2ab -a^2 = -7 [combine like terms]

a^2 - a^2 = 0

-5ab + 3ab + 2ab = 0

-7 = -7