Question

anybody know the correct way to solve for

Solve the system using the substitution method.
y-3x=2 and y=(x+1)^2-5

I keep getting x= -1.688 and y is a number alike to that but i don't get a whole number

Answer 1

Solve the left equation for y.
What do you get?
Then substitute that into the second equation.

Answer 2

would i substitute (x+1)^2-5 - 3x = 2 for y in the first equation and solve?

Answer 3

@mathstudent55 got it will do

Answer 4

If you need more help, show what you get, and we can work on it together.

Answer 5

y=3x+2 correct

Answer 6

@mathstudent55 i got y =3x +2

Answer 7

Good.
Now substitute y in the second equation with 3x + 2.
What do you get?

Answer 8

@mathstudent55 set it up as 3x+2= (x+1)^2-5 right

Answer 9

Great.
Now you have one equation in one variable. We need to solve it for x.
First square the binomial on the right sides.

Answer 10

@mathstudent55 what is the binomial (x+1)^2-5?

Answer 11

(x + 1)^2 is the binomial squared.
You need to square it.
Either do it by knowing the pattern of the squaring of a binomial
\((a + b)^2 = a^2 + 2ab + b^2\),
or use FOIL.

Answer 12

@mathstudent55 x2+2x+1?

Answer 13

Yes, (x+1)^2 = x^2 + 2x + 1
Put that in:
3x+2 = (x+1)^2- 5 and simplify

Answer 14

@aum so i just simplify 3x+2 = (x+1)^2- 5

Answer 15

3x+2 = (x+1)^2- 5
3x + 2 = x^2 + 2x + 1 - 5
3x + 2 = x^2 + 2x - 4
x^2 + 2x - 4 - 3x - 2 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or -2

Answer 16

got it thanks @aum

Answer 17

so how would i do this same equation using elimination method

Answer 18

y = (x+1)^2 - 5
y - 3x = 2
Subtract the 2 equations and that would eliminate y:
3x = (x+1)^2 - 5 - 2
3x = (x+1)^2 - 7
(x+1)^2 - 7 - 3x = 0
x^2 + 2x + 1 - 3x - 7 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = -3, -2

Answer 19

@aum so there is no solving for y in this one?

Answer 20

You have to put each of the x value in either one of the equations and solve for y. Choose y - 3x = 2 which is the easier one.

Answer 21

??

Answer 22

so what you just typed
Subtract the 2 equations and that would eliminate y:
3x = (x+1)^2 - 5 - 2
3x = (x+1)^2 - 7
(x+1)^2 - 7 - 3x = 0
x^2 + 2x + 1 - 3x - 7 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = -3, -2

Answer 23

thats solving for x
correct?

Answer 24

Yeah that solves for x. But you are given two equations in two unknowns: x and y. You have solved for x and now you will have to find the corresponding y.

Answer 25

@aum so how would i dove for y on the very first on using substitution ?

Answer 26

Oh, I just noticed a typo in two of my earlier replies. In both cases:
(x-3)(x+2) = 0
so x = 3; x = -2 (not -3).

Answer 27

which ones was that
1. Subtract the 2 equations and that would eliminate y:
3x = (x+1)^2 - 5 - 2
3x = (x+1)^2 - 7
(x+1)^2 - 7 - 3x = 0
x^2 + 2x + 1 - 3x - 7 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = -3, -2

or 2. 3x+2 = (x+1)^2- 5
3x + 2 = x^2 + 2x + 1 - 5
3x + 2 = x^2 + 2x - 4
x^2 + 2x - 4 - 3x - 2 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or -2

Answer 28

y-3x=2
or y = 3x + 2
put x = 3: y = 3(3) + 2 = 9 + 2 = 11
put x = -2: y = 3(-2) + 2 = -6 + 2 = -4

The solutions are: x = 3, y = 11; x = -2, y = -4
or (3,11) and (-2,-4)


I guess the mistake was in one place only
The one that has x = -3 should be x = 3.

Answer 29

and that is for substitution or both ?

Answer 30

@aum is that solving by substitution

Answer 31

never mind i get it

Answer 32

thank you i was confused now i know how to do this

Answer 33

For both. Once you have solved for x (either by substitution method or the elimination method) you are done with those methods.
After that you put x in the simpler of the two equations and compute y.

Answer 34

do you know about logirithmics ?

Answer 35

a little bit. This question has gotten too long. Close this one and post a new one.

Answer 36

ok i will