I assume that you mean 1/(x+5) + 3/(x+4) = -1(x^2+9x+20)
Taking the left hand side...we make one fraction by using a common denominator.
[ 1(x+4) + 3(x+5) ] / (x+5)(x+4) = -1(x^2+9x+20)
-->
(4x+19)/(x^2 + 9x + 20) = -1(x^2+9x+20)
Equating numerators since denominators are the same...
4x + 19 = -1
4x = -20
x = -5
Please note that even though -5 solves the equation, when substituted back into the original equation, you will get undefined terms ( 1/0 = infinity) on both sides.
2006-11-24 00:03:36
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answer #1
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answered by ludacrusher 4
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LCD of the LHS=(x+5)(x+4)
multiplying by the LCD
(x+4)+(x+5)=2x+9
so the LHS=(2x+9)/(x+5)(x+4)
The Dr of the RHS=x^2+9x+20
=x^2+5x+4x+20
=x(x+5)+4(x+5)
=(x+5)(x+4)
so the RHS=-1/(x+5)(x+4)
equatiing
2x+9/(x+5)(x+4)=-1/(x+5)(x+4)
the Drs get cancelled out
comparing the Nrs
2x+9=-1
adding -9
2x=-10
dividing by 2
x=-5
2006-11-23 23:54:46
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answer #2
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answered by raj 7
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the first member can be written
(4x+19)/ (x+5) (x+4) =(4x+19)/ (x^2 +9x+20)
The denominatror are the same and you have
multiplying by (x^2 +9x+20)
4x+19 = -1 hence 4x = -20
The result x = -5
2006-11-24 00:10:03
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answer #3
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answered by maussy 7
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by multipling the denominator is same so
x+4+3x+15=-1
4x+19=-1
4x=-20
x=-5
2006-11-24 01:53:51
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answer #4
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answered by riya s 2
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woman: please word this might nicely be a quadratic expression, x^2 - 2x - 3 = 0 fixing for the roots, you will get (it extremely is carried out in general by utilising trial and mistake ... except you choose for to apply the quadratic equation) (x-3)(x+a million) = x^2 + x -3x - 3 =x^2 - 2x - 3 venture is now solved. stable success.
2016-11-26 19:50:37
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answer #5
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answered by barreda 4
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You need to know the laws of elementary algebra, not have people solve equations for you.
If you know all the rules, with practice you will get it!
2006-11-24 00:24:01
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answer #6
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answered by aperson 3
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thatz ur home work
2006-11-23 23:56:42
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answer #7
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answered by Anonymous
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