# How do you use cross products to solve 3/4=x/(x+3)?

Aug 5, 2016

$x = 9$

#### Explanation:

$\frac{3}{4} = \frac{x}{x + 3}$
or
$4 x = 3 x + 9$
or
$4 x - 3 x = 9$
or
$x = 9$

Aug 5, 2016

x = 9

#### Explanation:

To use cross products or $\textcolor{b l u e}{\text{cross multiplication}}$ as it is also named.

$\frac{\textcolor{red}{3}}{\textcolor{b l u e}{4}} = \frac{\textcolor{b l u e}{x}}{\textcolor{red}{x + 3}}$

now multiply the terms in $\textcolor{b l u e}{\text{blue")" and "color(red)("red}}$ (X) and equate them.

$\Rightarrow \textcolor{b l u e}{4 x} = \textcolor{red}{3 \left(x + 3\right)}$

distribute the bracket : 4x = 3x + 9

subtract 3x from both sides to solve for x

$4 x - 3 x = \cancel{3 x} + 9 \cancel{- 3 x} \Rightarrow x = 9$

Aug 5, 2016

This is why the cross product works!!!

#### Explanation:

The cross product is a shortcut that bypasses some stages in solving by first principles. I will use first principles so you can see where the shortcut takes over.

A fraction is split up into two parts. Using descriptive but $\underline{\text{unconventional names}}$ we have $\left(\text{count")/("size indicator}\right)$

When you wish to ul(directly") compare quantities the "size indicators" have to be the same. This is also true for fractional addition and subtraction. You can not $\underline{\text{directly}}$ apply addition or subtraction unless the "size indicators" are the same.

" "("count")/("size indicator") ->("numerator")/("denominator")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving your question}}$

Using a 'common denominator' of $4 \left(x + 3\right)$

color(brown)([3/4xx1]=[x/(x+3)xx1]color(blue)(""->""[3/4xx(x+3)/(x+3) ]=[x/(x+3)xx4/4]

$\frac{3 \left(x + 3\right)}{4 \left(x + 3\right)} = \frac{4 x}{4 \left(x + 3\right)}$

If you look at the numerators you will see the result you get by the short cut

Multiply both sides by $4 \left(x + 3\right)$ and you end up with

$3 \left(x + 3\right) = 4 x \text{ "larr" the consequences of the shortcut}$