# How do you solve 15/7 = 60/ x?

Then teach the underlying concepts
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#### Explanation

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1
Mar 3, 2017

$28$

#### Explanation:

$\frac{15}{7} = \frac{60}{x}$

cross multiply

$\therefore 15 \times x = 7 \times 60$

$\therefore 15 x = 420$

$\therefore x = \frac{420}{15}$

$\therefore x = 28$

Then teach the underlying concepts
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Tony B Share
Jan 18, 2017

A slightly different approach

$x = 28$

#### Explanation:

$\textcolor{b l u e}{\text{Introduction to the idea of the method}}$

Treating it like a ratio

Given:$\text{ } \frac{15}{7} = \frac{60}{x}$

Notice that 5 is divisible into both 60 and 15.

If we can change the 15 into 60 and maintain proportionality then we can determine the value of $x$

To convert 15 into 60 first turn it into 1 then multiply by 60 to turn the 1 into 60 $\to 15 \times \frac{1}{15} \times 60 \text{ " ->" } 15 \times \frac{60}{15}$

But $\frac{60}{15} \text{ is the same as } 4$

To maintain proportionality what you do to the top you do to the bottom (for multiply and divide).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The actual calculation}}$

$\frac{15 \times 4}{7 \times 4} = \frac{60}{28} = \frac{60}{x}$

$x = 28$

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

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Dwight Share
Jan 18, 2017

Cross-multiply to show that $x = 28$

#### Explanation:

When you are faced with trying to solve a proportion like this, the method involves first doing a "cross-multiplication". What you are really doing is multiplying the left and right side by both denominators (I mention that just so it doesn't look like you are doing some sort of trick operation), with one of the multipliers used to cancel the denominator on that side of the equation.

$15 \times x = 60 \times 7$

Now, isolate the $x$ variable by dividing each side by 15

$x = \frac{60 \times 7}{15} = 28$

It is one of the truly useful "rules" of algebra, that you can perform any operation you like on an equation. As long as you do it identically to both sides, the equation remains true.

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