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mel-nik [20]
3 years ago
15

When 8 is subtracted from Ollie’s age, the result is the same as subtracting 16 from 3 times his age.

Mathematics
2 answers:
Scilla [17]3 years ago
5 0
X - Ollie's age

8 is subtracted from Ollie's age = x-8
the result is the same as
subtracting 16 from 3 times his age = 3x-16

x-8=3x-16 \\
x-3x=-16+8 \\
-2x=-8 \\
x=4

Ollie is 4 years old.
Allushta [10]3 years ago
3 0
X - his age

x-8=3x-16\\
x-3x=-16+8\\
-2x=-8\\
x=4

He's 4 years old.
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Which expressions are equivalent to 2[3/4x+7]-3(1/2x-5)? Check all that apply
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Answer:

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Step-by-step explanation:

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3 years ago
A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses
mash [69]

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for x days.

So, for x days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = \frac{360}{x}

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = (x+4) days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= (\frac{360}{x}-3)

Total expense in dollars can now be given as:  (x+4)(\frac{360}{x}-3)

Simplifying by using distribution (FOIL).

(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))

360-3x+\frac{1440}{x}-12

348-3x+\frac{1440}{x}

We know total expense remains the same which is = $360.

So, we have the equation as:

348-3x+\frac{1440}{x}=360

Multiplying each term with x to remove fractions.

348x-3x^2+1440=360x

Subtracting 348x both sides

348x-348x-3x^2+1440=360x-348x

-3x^2+1440=12x

Dividing each term with -3.

\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}

x^2-480=-4x

Adding 4x both sides.

x^2+4x-480=-4x+4x

x^2+4x-480=0

Solving using quadratic formula.

For a quadratic equation: ax^2+bx+c=0

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Plugging in values from the equation we got.

x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}

x=\frac{-4\pm\sqrt{16+1920}}{2}

x=\frac{-4\pm\sqrt{1936}}{2}

x=\frac{-4\pm44}{2}

So, we have

x=\frac{-4+44}{2}   and x=\frac{-4-44}{2}

x=\frac{40}{2}   and x=\frac{-48}{2}

∴ x=20           and x=-24

Since number of days cannot be negative, so we take x=20 as the solution for the equation.

Thus, the original duration of the tour = 20 days

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