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then x =
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15 Apr 2018, 04:00
15 Apr 2018, 04:00
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Question Stats:
86% (01:26) correct
13% (02:04) wrong based on 29 sessions
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If \(\frac{x^2-4x+4}{x^2+x-6}=\)\(\frac{6x-12}{x^2+6x +9}\)
Drill 2
Question: 11
Page: 512
Question: 11
Page: 512
Show: :: OA
3
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Re: then x =
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01 Jan 2019, 09:07
01 Jan 2019, 09:07
2
sandy wrote:
If \(\frac{x^2-4x+4}{x^2+x-6}=\)\(\frac{6x-12}{x^2+6x +9}\)
Drill 2
Question: 11
Page: 512
Question: 11
Page: 512
Show: :: OA
3
GIVEN: \(\frac{x^2-4x+4}{x^2+x-6}=\)\(\frac{6x-12}{x^2+6x +9}\)
Factor top and bottom of each fraction to get: \(\frac{(x-2)(x-2)}{(x+3)(x-2)}=\)\(\frac{6(x-2)}{(x+3)(x+3)}\)
Simplify fraction on left side to get: \(\frac{(x-2)}{(x+3)}=\)\(\frac{6(x-2)}{(x+3)(x+3)}\)
Cross multiply to get: \((x-2)(x+3)(x+3) = 6(x-2)(x+3)\)
Divide both sides by \((x-2)\) to get: \((x+3)(x+3) = 6(x+3)\)
Divide both sides by \((x+3)\) to get: \(x+3 = 6\)
Solve: \(x = 3\)
Answer: 3
Cheers,
Brent
then x =
[#permalink]
03 Jan 2024, 01:02
03 Jan 2024, 01:02
1
\(\frac{x^2-4x+4}{x^2+x-6}=\)\(\frac{6x-12}{x^2+6x +9}\)
factorizing numerator and denominator on both sides of the equation we get
\(\frac{(x-2)(x-2)}{(x-2)(x+3)} = \frac{6(x-2)}{(x+3)(x+3)}\)
cancelling like terms on and across both sides
\(\frac{(x-2)}{1} = \frac{6(x-2)}{(x+3)}\)
cross-multiplying
\((x-2)(x+3) = 6(x-2)\)
cancelling like terms on both sides
\((x+3) = 6\)
rearranging constant terms
\(x = 6 - 3\)
\(x = 3\)
factorizing numerator and denominator on both sides of the equation we get
\(\frac{(x-2)(x-2)}{(x-2)(x+3)} = \frac{6(x-2)}{(x+3)(x+3)}\)
cancelling like terms on and across both sides
\(\frac{(x-2)}{1} = \frac{6(x-2)}{(x+3)}\)
cross-multiplying
\((x-2)(x+3) = 6(x-2)\)
cancelling like terms on both sides
\((x+3) = 6\)
rearranging constant terms
\(x = 6 - 3\)
\(x = 3\)
