anonymous
  • anonymous
Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
GoldPhenoix
  • GoldPhenoix
You want to solve for x, right?
anonymous
  • anonymous
yes @GoldPhenoix

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GoldPhenoix
  • GoldPhenoix
If so, you want to use to FOIL method. |dw:1370192591176:dw|
GoldPhenoix
  • GoldPhenoix
Now combine like term for: \[\large 18x + 90 - 9x - 45 = 0\]
GoldPhenoix
  • GoldPhenoix
Do you understand what I did?
anonymous
  • anonymous
No :( @GoldPhenoix
GoldPhenoix
  • GoldPhenoix
http://www.algebrahelp.com/lessons/simplifying/foilmethod/pg2.htm . This link explain well on how to use the FOIL method.
anonymous
  • anonymous
Okay
GoldPhenoix
  • GoldPhenoix
Or, if you don't feel like reading the link, you can look at this pic. http://upload.wikimedia.org/wikipedia/commons/0/0c/MonkeyFaceFOILRule.JPG
GoldPhenoix
  • GoldPhenoix
My bad. I mest up.
anonymous
  • anonymous
okay thanks
GoldPhenoix
  • GoldPhenoix
3(2x-1)(x+5) = 0 You want to solve X. So you want to distribute first. Do you know how to use the distributive property for: 3(2x-1) ?
anonymous
  • anonymous
yea I'll try it
Jhannybean
  • Jhannybean
Still need help solving this?...
reemii
  • reemii
I think that when the equation is in the form \(3(2x-1)(x+5) = 0\), it is better not to distribute. If a product of numbers is equal to 0, that means one of the numbers is zero. Here, that means \(2x-1=0\) or \(x+5\). the solutions of both equations give the solutions of the initial equation.
anonymous
  • anonymous
yes i need help still @Jhannybean @reemii
reemii
  • reemii
have yo found some solutions already?
reemii
  • reemii
what I said before is , to solve an equation of the form \[f(x)g(h)h(x)=0\] means to "find the \(x\)'s". A product of numbers equals zero if ONE OF the factors is zero. tehrefore, solve separately the equations \(f(x)=0\), \(g(x)=0\), \(h(x)=0\) (etc). Collect all the solutions you find for those equations, and this gives the solutions of the first equation. In your case, \(3(2x-1)(x+5)=0\) You solve separately: \(2x-1=0\) and \(x+5=0\). For the first equation, the solutions is x = 1/2. For the second equation, the solution is x = -5. Collect them all: the solutions of your equation are 1/2 and -5.
reemii
  • reemii
In general, when facing an equation, people try to factorize, not to distribute. Factorizing allows to break the initial equation in smaller ones. Just as we (I) did.
reemii
  • reemii
@breja
Jhannybean
  • Jhannybean
\[\large 3(2x-1)(x+5)=0\]\[\large 3(2x-1)=0\]divide 3 on both sides of the equation \[\large \frac{\cancel3(2x-1)}{\cancel 3}=\frac{0}{3}\]\[\large 2x-1 = 0\]\[\large 2x=1\]divide both sides by 2, you'll have 1 answer. \[\large x+3=0\]subtract -3 from both sides.\[\large x+3-3=0-3\] simplify nd you'll have another answer for x.

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