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Mathematics
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OpenStudy (anonymous):
Solve for x: 2 over quantity x plus 2 plus 1 over 5 equals 6 over quantity x plus 5.
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OpenStudy (anonymous):
@Whitemonsterbunny17 @bre4h @shenandoah123 @vera_ewing @yomamabf
OpenStudy (zzr0ck3r):
\[\dfrac{2}{x+2}+\dfrac{1}{5}=\dfrac{6}{x+5}\] ?
OpenStudy (anonymous):
Yes
OpenStudy (zzr0ck3r):
multiply all three terms by \((x+5)(x+2)\) and tell me what you have.
OpenStudy (anonymous):
ok hold on:)
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OpenStudy (anonymous):
x=-13?
OpenStudy (zzr0ck3r):
\(\dfrac{2}{x+2}+\dfrac{1}{5}=\dfrac{6}{x+5}\\(x+2)(x+5)\dfrac{2}{x+2}+(x+2)(x+5)\dfrac{1}{5}=(x+2)(x+5)\dfrac{6}{x+5}\\
\cancel{(x+2)}(x+5)\dfrac{2}{\cancel{(x+2)}}+(x+2)(x+5)\dfrac{1}{5}=(x+2)\cancel{(x+5)}\dfrac{6}{\cancel{(x+5)}}\\
(x+5)2+(x+5)(x+2)\frac{1}{5}=(x+2)6\)
OpenStudy (zzr0ck3r):
\(
2(x+5)+(x+5)(x+2)\frac{1}{5}=6(x+2)\)
OpenStudy (anonymous):
ohh so it would be x=0, x=13
OpenStudy (anonymous):
Yes, no, maybe so? Lol
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OpenStudy (anonymous):
maybe ask zzr0ck3r
OpenStudy (anonymous):
Ok:) She/he is offline :(
OpenStudy (anonymous):
@Mertsj What are your thoughts?
OpenStudy (mertsj):
\[5(x+2)(x+5)(\frac{2}{x+2})+5(x+2)(x+5)(\frac{1}{5})=5(x+2)(x+5)(\frac{6}{x+5})\]
OpenStudy (mertsj):
10x+50+x^2+7x+10=30x+60
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OpenStudy (mertsj):
x^2-13x=0
x=0,13
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