Question

How do you use cross products to solve `2/t=5/(t-6)`?

Answer 1

`t=-4`

Explanation:

Cross multiply the denominator with numerator like:

`2/t=5/(t-6)`

`2xx(t-6) = 5 xx t`

`2t-12 = 5t`

Add `12` both sides:
`2t-12+12` = 5t+12#

`2t cancel(-12+12)` = 5t+12#

`2t = 5t+12`
`2t-5t = 12` -----> making `t` the subject by subtracting `-5t` both sides:

`-3t=12`

`t=-12/3`

`t=-4`

Answer 2

`t=-4`

Explanation:

`"using the method of "color(blue)"cross-products"`

`•color(white)(x)a/b=c/drArrbc=ad`

`rArr5t=2(t-6)`

`rArr5t=2t-12`

`"subtract "2t" from both sides"`

`5t-2t=cancel(2t)cancel(-2t)-12`

`rArr3t=-12`

`"divide both sides by 3"`

`(cancel(3) t)/cancel(3)=(-12)/3`

`rArrt=-4`

`color(blue)"As a check"`

Substitute this value into the equation and if both sides are equal then it is the solution.

`"left "=2/(-4)=-1/2`

`"right "=5/(-4-6)=5/(-10)=-1/2`

`rArrt=-4" is the solution"`