Question 172994
formula to work with is:
rate * time = number of units
the number of units is 1 room.
it takes the painter x - 3 hours to paint a room.
it take the apprentice x hours to paint the same room.
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let Rp = rate of the painter.
let Ra = rate of the apprentice.
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since rate * time = number of units,
for the painter:
Rp * (x-3) = 1
for the apprentice:
Ra * (x) = 1
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if they are both working together, it takes 2 hours to paint the room.
working together their rates are additive.
since rate * time = number of units,
for both:
2 * (Ra + Rp) = 1
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for the apprentice:
Ra * x = 1
which means:
Ra = 1/x
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for the painter:
Rp * (x-3) = 1
which means:
Rp = 1/(x-3)
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substituting for Ra and Rp, the formula:
2 * (Ra + Rp) = 1
becomes:
{{{2 * (1/x + 1/(x-3)) = 1}}}
which becomes:
{{{2/x + 2/(x-3) = 1}}}
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if we multiply both sides of the equation by x*(x-3), we get:
{{{2*(x-3) + 2*x = x*(x-3)}}}
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this becomes:
{{{2*x - 6 + 2*x = x*(x-3)}}}
which becomes:
{{{4*x - 6 = x^2 - 3*x}}}
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if we subtract 4*x from both sides of the equation, and if we add 6 to both sides of the equation, we get:
{{{0 = x^2 - 7*x + 6}}}
which is the same as:
{{{x^2 - 7*x + 6 = 0}}}
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this factors out to be:
{{{(x-1)*(x-6) = 0}}}
which give us:
x = 1
or:
x = 6
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x cannot be 1, since x-3 would be negative which is not possible.
x must be 6.
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if x is 6, then
x-3 is 3.
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since x * Ra = 1
then Ra must be 1/6
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since (x-3)*Rp = 1
then Rp must be 1/3
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to prove this is true, we substitute in the original equations to see if they are true.
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first equation:
x * Ra = 1
6 * 1/6 = 1 is true.
second equation:
(x-3)*Rp = 1
(6-3)*1/3 = 1
3*1/3 = 1 is true.
third equation:
2 * (Ra + Rp) = 1
2 * (1/6 + 1/3) = 1
2 * (1/6 + 2/6) = 1
2 * (3/6) = 1
6/6 = 1 which is true.
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answer of x = 6 looks good.
apprentice takes 6 hours to paint a room.
painter takes 6-2 = 3 hours to paint the same room.
together they can paint the room in 2 hours.
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