systems of equations-substitution method: y=x^2-79^2, y=2x-1

A

atw

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Apr 8, 2008
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Hello my name is atw.

I need help with this problem and I'm stuck on trying to factor but it is not coming out right. Here is the problem.

y=x^2-(79)^2
y=2x-1

my equation was that I got everything on one side and now I'm stuck. My equation is.

y=x^2-2x-6241+1 or y=x^2-2x-6240 and I have to do substitution method and I'm stuck here. can you help me.

thanks,

atw
 
Re: systems of equations-substitution method

78 X 80 = ???
 
Re: systems of equations-substitution method

(x-78)(x+80)-2x now I need to know how to solve for the y value. that's what is confusing to me I got lost after this point
 
Re: systems of equations-substitution method

Are you pulling my leg?

y=x^2-(79)^2
y=2x-1

2x-1=x^2-79[sup:11wu9n42]2[/sup:11wu9n42] <<< There us no y in the equation. Get everything on the left side of the equation. The right side will be a zero. On the left side will be a trinomial. Factor that using FOIL (or any other way). Set each factor = to zero and solve for x. You will have two values for x. Plug each back into one of the original equations to get the corresponding value of y. Be sure to check each pair of answers to see if they tell the truth.
 
Re: systems of equations-substitution method

so it will be -x^2+2x+6240=0
 
Re: systems of equations-substitution method

or... multiply both sides by -1?
 
Re: systems of equations-substitution method

x^2-2x-6240=0
(x-78)(x+80)=0

x=78 x=-80

for the y value do I multiply -2(78)=-156 and -2(-80)=160
 
ARE YOU GUESSING, IN GENERAL?

atw said:
x^2 - 2x - 6240 = 0

(x - 78) (x + 80) ? This is not the correct factorization. FOIL shows why.

(x - 78) * (x + 80) = x[sup:2sh80768]2[/sup:2sh80768] + 80x - 78x - 6240 = x[sup:2sh80768]2[/sup:2sh80768] + 2x - 6240



... for the y [values] do I multiply -2(78) ... and -2(-80) ... NO!

The value of y is not given as 2x. Look again. (Perhaps, you need to slow down.)
Click to expand...
 
x^2-(79)^2 or x^2-6241 (x-79)(x+79)

x=79 x=-79

2x-1

2(79)-1=157 2(-79)-1=-160
 
atw said:
x^2-(79)^2 or x^2-6241 (x-79)(x+79)

x=79 x=-79
Click to expand...
On what basis did you decide that the value of y should be zero...?

Please be complete. Thank you! :D

Eliz.
 
atw said:
x^2-(79)^2 or x^2-6241 (x-79)(x+79)

x=79 x=-79

2x-1

2(79)-1=157 2(-79)-1=-160
Click to expand...


This post looks like garbage to me. I wonder why you typed it. I also wonder whether or not you are intentionally goofing around with us. I also wonder if you have major issues understanding English or following instructions.

You have a formula for finding the value of y.

y = 2x - 1

Therefore, if you know values of x, then you can substitute them into this formula to calculate the corresponding values of y.

You tried to do this in your last post of November 15th, but you made two mistakes.

Mistake #1: you changed the formula for y from 2x - 1 to 2x. I asked you to look at the formula for y again.

It is not y = 2x.

It is y = 2x - 1.

Mistake #2: you did not properly factor x^2 - 2x - 6240.

I showed you the sign error that you made. Look at my post again. The sign error is displayed IN BLUE.

Fix the signs in your factors, so that their product is x^2 - 2x - 6240.

After you do this, then you will be able to find the two CORRECT values of x.

After you discover these values, substitute them into the formula y = 2x - 1 to find the corresponding values of y.

Once you have two pairs of x and y values, check them to make sure that each pair satisfies both equations in the original system.

If you do not know what you are doing, then it is better to ASK US QUESTIONS, rather than guess and type garbage.

Cheers,

~ Mark :)

 
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