Question

which ordered pair is in the solution set of the system of equations, y = -x + 1 and y = x^2 + 5x + 6?

A. (-5,-1)
B. (-5,6)
C. (5,-4)
D. (5,2)

Answer 1

you have a choice
you can
a) solve it or
b) plug in each pair of numbers and see which one fits both equations
what do you choose?

Answer 2

hint: solving is probably easier

Answer 3

also u can plot a graph

Answer 4

solve it

Answer 5

since \(y=y\) you can solve
\[-x+1=x^2+5x+6\]

Answer 6

add \(x\) subtract \(1\) and solve
\[x^2+6x+5=0\] by factoring

Answer 7

you good from there?

Answer 8

6x^2 + 5 = 0
is what i got from factoring

Answer 9

that is not factoring
as a matter of fact, i am not sure what that is
but \[x^2+6x+5\neq 6x^2+5\] in any way

Answer 10

factoring means to write as a product, like
\[x^2+2x+1=(x+1)(x+1)\] for example

Answer 11

ah i see..

Answer 12

can you factor
\[x^2+6x+5\]? you do not have too many choices

Answer 13

ok i am going to try one sec..

Answer 14

(x+5)(6x+0) is the correct answer?

Answer 15

no

Answer 16

\(6x+0=6x\) that is not right

Answer 17

the \(x+5\) part is right

Answer 18

i think the correct answer to the original question is D. (5,2)

Answer 19

it is not

Answer 20

damn, i should of paid attention to my teacher :(

Answer 21

you really have to factor \(x^2+6x+5\) there is no avoiding it

Answer 22

do you mind teaching me how to factor, like the steps?

Answer 23

what two numbers multiply to get 5 and add to 6?

Answer 24

3 and 2

Answer 25

do you mean two numbers multiply to get 6, and add to get 5?

Answer 26

no i do not

Answer 27

5 and 1

Answer 28

you have
\[x^2+\color{red}6x+\color{blue}5\] two numbers that multiply to \(\color{blue}5\) and add to \(\color{red}6\)

Answer 29

5 and 1 are two numbers that when multiplied make 5 and added make 6

Answer 30

yeah 5 and 1
that means \[x^2+6x+5=(x+1)(x+5)\]

Answer 31

if you multiply it out, you will see it is right
now set each factor equal to zero and solve
\[x+1=0\\
x=?\]

Answer 32

you want me to solve the question above?

Answer 33

yes

Answer 34

ok, what do you mean set each factor equal to zero? i dont understand what that means sorry

Answer 35

the factors of \((x+1)(x+5)\) are \(x+1\) and \(x+5\)
setting them equal to zero means... set them equal to zero
\[\huge x+1=0\] solve for \(x\)

Answer 36

oh i see, x + 1 = 0 and x + 5 = 0

Answer 37

right
now solve for \(x\)

Answer 38

x + 1 = 0

x=1

Answer 39

no \(1+1=2\) not \(1+1=0\)
try again
takes one step but if \(x+1=0\) then \(x\neq 1\)

Answer 40

oh its

x = -1

Answer 41

i had a feeling it was -1, i had a feeling it was not right.

Answer 42

the correct answer for the original question is A. (-5,-1)

i see now! sweet!

Answer 43

yes, that is it

Answer 44

you're better than my math teacher .. TY!

Answer 45

yw