Question

This problem addresses some common algebraic errors. For the equalities stated below assume that x and y stand for real numbers. Assume that any denominators are non-zero. Mark the equalities with T (true) if they are true for all values of x and y, and F (false) otherwise.

(x+y)^2 = x^2 + y^2.
(x+y)^2 = x^2 + 2xy+y^2.
\frac{x}{ x+y} = \frac{1}{y}.
x-(x+y) = y.
\sqrt{x^2} = x.
\sqrt{x^2} = |x|.
\sqrt{x^2+4} = x+2.
\frac{1}{x+y} = \frac{1}{x} + \frac{1}{y}.

Answer 1

i got
1=F 2=T 3=F 4=T 5=T 6=T 7=F 8=T

Answer 2

DONT NO WHICH ONES WRONG I SUBBED IN 2 FOR ALL THE X'S AND 4 FOR ALL THE Y'S

Answer 3

your answers for first 3 equalities are right

Answer 4

lets look at #4 :

` x-(x+y) = y`

Answer 5

distribute the negative for parenthesis on left hand side, what do u get ?

Answer 6

do you no multiply brackets by -??

Answer 7

i get -4 when i do that

Answer 8

yes you may think of :

x-(x+y)

as

x - 1(x+y)

Answer 9

distributing gives you :

x - 1x - 1y

x - x - y

0 - y

-y

which is NOT same as the right hand side y eh ?

Answer 10

woopsy your right that is false but i still i have another one or two of them wrong

Answer 11

yes lets see #5

Answer 12

\sqrt{x^2} = x

Answer 13

say x = -2

what do you get ?

Answer 14

sqrt((-2)^2) = -2

sqrt(4) = -2

Answer 15

can squareroot of something ever be negative ?

Answer 16

no so therefore it is false?

Answer 17

it has to be false
an equation has to work for all numbers

Answer 18

so what do we have so far ?

Answer 19

o i see where your coming from now

Answer 20

we have F T F F F

Answer 21

the next 3 i have T F T

Answer 22

yes #6 is T

Answer 23

#7 is F

Answer 24

#8 is also F

Answer 25

so overall only #2 and #6 are True
and everything else is False

Answer 26

thank you very helpful

Answer 27

np :)