Question

Select the correct solution in each column of the table.
Solve the following equation.

Table included in image below:

Answer 1

Answer:

No of real solutions =1

No of  extraneous solution =2

Real solution: x =3

Step-by-step explanation:

$\frac{3}{x}-\frac{x}{x+6}=\frac{18}{x^2+6x}$

solving:

Taking LCM of x, x+6 and x^2+6 we get x(x+6)

Multiply the equation with LCM

$\frac{3}{x}*x(x+6)-\frac{x}{x+6}*x(x+6)=\frac{18}{x^2+6x}*x(x+6)\\3(x+6)-x*x=\frac{18}{x(x+6)}*x(x+6)\\3(x+6)-x*x=18\\3x+18-x^2=18\\-x^2+3x+18-18=0\\-x^2+3x=0\\x^2-3x=0\\x(x-3)=0\\x=0 \,\,and\,\, x =3$

Checking for extraneous solution

for extraneous solution we check the points where the solution is undefined

The solution will be undefined. if, x=0 or x=-6 so both are extraneous solutions

Putting x =3

$\frac{3}{3}-\frac{3}{3+6}=\frac{18}{(3)^2+6(0)}$

$\frac{3}{3}-\frac{3}{3+6}=\frac{18}{(3)^2+6(3)}\\1-\frac{3}{9}=\frac{18}{9+18}\\1-\frac{1}{3}=\frac{18}{27}\\\frac{3-1}{3}=\frac{2}{3}\\\frac{2}{3}=\frac{2}{3}$

So, x=3 is real solution.

Now, Selecting answers from tables

No of real solutions =1

No of  extraneous solution =2

Real solution: x =3

Answer 2

Answer:

Number of                Number of                         Real solutions

real solution            Extraneous solution

     1                                        1                                     x=3

Step-by-step explanation:

Extraneous solution--

It is a solution which is obtained on solving the equation but it does not satisfies the equation i.e. after it is put back to the equation it does not occur as a valid solution.

True solution or real solution--

It is the solution which is obtained on solving the equation and is also a valid solution to the equation.

The equation is:

$\dfrac{3}{x}-\dfrac{x}{x+6}=\dfrac{18}{x^2+6x}$

On taking lcm in the left hand side of the equation we get:

$\dfrac{3\times (x+6)-x\times x}{x(x+6)}=\dfrac{18}{x^2+6x}\\\\i.e.\\\\\dfrac{3x+18-x^2}{x(x+6)}=\dfrac{18}{x(x+6)}\\\\i.e.\\\\\dfrac{3x+18-x^2}{x(x+6)}-\dfrac{18}{x(x+6)}=0\\\\i.e.\\\\\dfrac{3x+18-x^2-18}{x(x+6)}=0\\\\i.e.\\\\\dfrac{3x-x^2}{x(x+6)}=0\\\\i.e.\\\\3x-x^2=0\\\\i.e.\\\\x(3-x)=0\\\\i.e.\\\\x=0\ or\ x=3$

When we put x=0 back to the equation we observe that the first term of the left hand side of the equation becomes undefined.

Hence, x=0 is the extraneous solution.

whereas x=3 is a valid solution to the equation.