Question

Solve simultaneously..? `x = 3^y` and `x = 1/2 (3 + 9y)`

Answer 1

This is the method I used in deducing the following simultaneously equation..

See steps below;

Explanation:

Solving simultaneously..

`x = 3^y - - - - - - eqn1`

`x = 1/2(3 + 9y) - - - - - - eqn2`

Look at the common value in both equations..

`x` is the common, hence we equate both together..

Having..

`3^y = 1/2(3 + 9y)`

`3^y = (3 + 9y)/2`

Cross multiplying..

`3^y/1 = (3 + 9y)/2`

`2xx 3^y = 3 + 9y`

`6^y = 3 + 9y`

Log both sides..

`log6^y = log(3 + 9y)`

Recall the law of logarithm `-> log6^y = x, ylog6 = x`

Therefore...

`ylog6 = log(3 + 9y)`

Divide both sides by `log6`

`(ylog6)/(log6) = log(3 + 9y)/(log6)`

`(ycancel(log6))/cancel(log6) = log(3 + 9y)/(log6)`

`y = (log(3 + 9y))/log(6)`

`y = (cancel (log)(3 + 9y))/(cancel (log)(6))`

`y = (3 + 9y)/6`

Cross multiplying..

`y/1 = (3 + 9y)/6`

`6 xx y = 3 + 9y`

`6y = 3 + 9y`

Collect like terms

`6y - 9y = 3`

`-3y = 3`

Divide both sides by `-3`

`(-3y)/(-3) = 3/-3`

`(cancel(-3)y)/cancel(-3) = 3/-3`

`y = -3/3`

`y = - 1`

Substitute the value of `y` into `eqn1` to get `x`

`x = 3^y - - - - - - eqn1`

`x = 3^-1`

Recall in indices, `x^-1 = 1/x`

`:. x = 1/3`

Hence the values are `rArr x = 1/3, y = -1`

Hope this helps!