Question

How do you use cross products to solve `21/56=z/8`?

Answer 1

`z=3`

Explanation:

It may seem innocuous and pedantic but you really mean 'cross multiplying' because a 'cross product' is a technique involving vectors and is not applicable here.

Anyway, on with the question. When we cross multiply all we are doing is multiplying both sides of an equation by the LCM of the denominators. We often skip some steps and just say that we 'move' the denominator up to the other side. ie:

`21/56xx56 = z/8xx56`

`21/cancel56xxcancel56 = z/cancel8xxcancel56^7`

`21 = 7z`

`21/7 = (7z)/7`

`z = 3`

Answer 2

`z = 3`

Explanation:

The simplified version of cross-multiplying is a quick and easy way of getting rid of fractions in an equation. However, it can only be used under certain conditions.

1. Must be an equation

2. There may only be one term on each side, at least one must be fraction.

The result of cross-multiplying is the simplified version of having multiplied both sides by both denominators.

`" "color(red)(21)/color(blue)(56) = color(blue)(z)/color(red)(8)`

Multiply with the combination which will give a positive variable on the left.

`" "color(blue)(56) xx color(blue)(z) = color(red)(21) xxcolor(red)(8)`

`" "z = (21 xx8)/56 " "z = (cancel21^3 xxcancel8)/cancel56^(cancel7)`

`" "z = 3`

Answer 3

z = 3

Explanation:

An alternative approach is.

Consider the following `color(blue)"equivalent fractions"` in ratio form.

`color(blue)(1)/color(red)(2)=color(red)(2)/color(blue)(4)`

Now if we (X)`color(magenta)"cross-multiply"` That is multiply the blue on opposite sides of the X and multiply the red on opposite sides of the X.

`rArrcolor(blue)(1xx4)" and " color(red)(2xx2)` we obtain 4 = 4 a true syatement.

Try this with other equivalent pairs. This 'fact' can also be applied to algebraic fractions.

`rArrcolor(blue)(21)/color(red)(56)=color(red)(z)/color(blue)(8)`

Now apply method of `color(magenta)" cross-multiplication"`

`rArrcolor(red)(56z)=color(blue)(21xx8)=168rArrz=3`