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Mathematics 14 Online
OpenStudy (anonymous):

Can you tell me how to do it? Don't give me answers...I want to know how to solve it... What is the value of x in the proportion (x-1)/5 = (4x + 2)/35?

OpenStudy (anonymous):

\[\frac{ (x-1) }{5}=\frac{ 4x+2 }{ 35 }\]

OpenStudy (anonymous):

cross multiply the top up

OpenStudy (anonymous):

What do you mean?

OpenStudy (jdoe0001):

\(\bf \cfrac{x-1}{5}=\cfrac{4x+2}{35}\quad \textit{times 35 each side}\\ \quad \\ {\color{red}{ \cancel{35}}}\cdot \cfrac{x-1}{\cancel{5}}={\color{red}{ \cancel{35}}}\cdot \cfrac{4x+2}{\cancel{35}} \implies {\color{red}{ 7}}\cdot (x-1)={\color{red}{ 1}}\cdot 4x+2\\ \quad \\\implies 7x-7=4x+2\)

OpenStudy (anonymous):

Ok i see but where did you get 7?

OpenStudy (jdoe0001):

7(x-1) <--- distribution

OpenStudy (jdoe0001):

hmm you meant the .... I see the 7 itself... hm ok

OpenStudy (anonymous):

ok i get it! so next 3x = 9? then you divided by 3 and i got 3.

OpenStudy (anonymous):

so the answer is 3?

OpenStudy (jdoe0001):

\(\bf \cfrac{x-1}{5}=\cfrac{4x+2}{35}\quad \textit{times 35 each side}\\ \quad \\ {\color{red}{ 35}}\cdot \cfrac{x-1}{5}={\color{red}{ 35}}\cdot \cfrac{4x+2}{35}\implies \cancel{\cfrac{{\color{red}{ 35}}}{5}}(x-1)=\cancel{\cfrac{{\color{red}{ 35}}}{35}}(4x+2)\)

OpenStudy (anonymous):

yeah i got that :)

OpenStudy (jdoe0001):

"so the answer is 3?" \(\checkmark\)

OpenStudy (anonymous):

Thanks! i understand now! can you help me another question? and i'll try solve it and you check it?

OpenStudy (jdoe0001):

ok

OpenStudy (anonymous):

What is the value of x in the proportion x+1/x+3 = 15/21?

OpenStudy (anonymous):

\[\frac{ x+1 }{ x+3 }=\frac{ 15 }{ 21 }\]

OpenStudy (anonymous):

Is it same as you solve it? or different?

OpenStudy (anonymous):

|dw:1389731685759:dw| is that correct so far?

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