Question

How do you solve `\frac { 6y } { 2y + 6} + \frac { y + 18} { 3y + 9} = \frac { 7y + 17} { y + 3}`?

Answer 1

No real solution. Read the explanation carefully.

Explanation:

`(6y)/(2y+6)+(y+18)/(3y+9)=(7y+17)/(y+3)`

We can rewrite it as:-

`(6y)/[2(y+3)]+(y+18)/[3(y+3)]=(7y+17)/(y+3)`

`=>(9y+y+18)/(3(y+3))=(7y+17)/(y+3)`

Cross multiply;

`=>(10y+18)(y+3) = 3(7y+17) (y+3)`

`=>(10y+18)(y+3)-3(7y+17) (y+3)=0`

Taking `(y+3)` common;

`=>(y+3)(10y+18-21y-51)=0`

`=>(y+3)(-11y-33)=0`

`=>-11*(y+3)*(y+3)=0`

`=> (y+3)^2=0`

`=> y+3 = 0` `=> color(red)(y=-3)`

BUT there is a catch.

We might think that `y=-3` is the solution but try to substitute this value back into the original equation i.e.

`(6y)/(2y+6)+(y+18)/(3y+9)=(7y+17)/(y+3)`

We find that terms in the denominator become zero.
`2*(-3)+6 = -6+6=0`,
`3*(-3)+9=-9+9=0`,
`-3+3 = 0`.

That means the expressions `(6y)/(2y+6), (y+18)/(3y+9), (7y+17)/(y+3)` become undefined ** at `y=-3` since division by zero is not defined. Hence we cannot say for sure whether the left-hand-side equals the right-hand-side or not. So `y=-3` is not a solution** to this equation.

Where did we go wrong?

Remember the step where we cross multiplied `y+3`. Cross multiplication, or in fact any kind of multiplication on both sides of the equation is allowed only when the number(s) being multiplied is/are non-zero. Therefore cross multiplication in that step is only valid if `y+3!=0` or `y!=-3`. I.e. we are cross multiplying on the assumption that `y=-3` is not a solution and even if it turns out to be a solution after solving the subsequent equations, we are going to neglect it.

`therefore` we neglect `y=-3` and hence, the given equation has no real solution.

Just to be sure, I checked on wolfram.
![wolframalpha.com]()