Question

find the solution of the system
y=x^2+5
y=2x+4

Answer 1

same thing as before yes?

Answer 2

x^2+5 =2x+4
x^2-2x+1
(x-1)(x-1)
ok?
x=1

Answer 3

understand?

Answer 4

nope.

Answer 5

would it look like this http://go.hrw.com/math/midma/gradecontent/manipulatives/GraphCalc/graphCalc.html

Answer 6

@fontez , pretty much did most of the work for you... all you need to do now is find the y-coordinate of the solution... plug in that x value into one of those equations to get the y-value.

Answer 7

i cant read it

Answer 8

ok i have the choices

(1,6) and (-1,6)
(-1,6)
(1,6)
(2,9)

Answer 9

you'll need to learn this sometime... why not now?
plug in x=1 into any one of those equations in your system
to get the y-value.

Answer 10

i dont understand how they got x=1

Answer 11

x^2+5 =2x+4
x^2-2x+1 =0
(x-1)(x-1) =0
x=1

which step do you get lost on?

Answer 12

the beginning i dont understand why they are set equal to each other

Answer 13

because they both equal y....you are substituting (replacing) 2x+4 in for y

Answer 14

oh okay then what next?

Answer 15

because its a quadratic, you need to set one side equal to 0

Answer 16

ok so move one over to the other side to make one side 0?

Answer 17

right, what they did was subtract 2x and subtract 4 to other side

Answer 18

x^2-2x+1=0 right?

Answer 19

yep, then next step they factored...its like backwards FOIL
what factors of 1 add up to -2 ?

Answer 20

you could also use the quadratic formula here if you get stuck factoring

Answer 21

how would i do the quadratic way?

Answer 22

\[x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]
for
ax^2 +bx +c = 0

Answer 23

then i plug in the values correct?

Answer 24

yes

Answer 25

ok \[-(-2)\pm \sqrt{-2^2-4(1)(1)}\div2(1)\] like that?

Answer 26

yes, make sure you square the negative as well....b^2 will always be positive

Answer 27

\[2\pm \sqrt{4-4} \div 2\]

Answer 28

like that ?

Answer 29

correct

Answer 30

ok \[2\pm \sqrt{0}\div2\] how is that right

Answer 31

the 0 goes away
--> 2/2 = 1

Answer 32

oh okay

Answer 33

then i plug in the 1

Answer 34

yep, and it doesn't matter which equation

Answer 35

y=6

Answer 36

correct

Answer 37

thank you so much

Answer 38

your welcome