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Mathematics 6 Online
OpenStudy (anonymous):

Solve for x: 2 over quantity x plus 2 plus 1 over 5 equals 6 over quantity x plus 5.

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:)

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

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

OpenStudy (mertsj):

x^2-13x=0 x=0,13

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