Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you solve $\dfrac{5}{y-3}=\dfrac{y+7}{2y-6}+1$ and find any extraneous solutions?

Last updated date: 19th Jul 2024
Total views: 379.2k
Views today: 6.79k
Verified
379.2k+ views
Hint: The equation given in the above question is a linear equation in one variable, that is y. The question says that we have to solve the given equation in y. Extraneous solutions are those solutions which satisfy the simplified equation but do not satisfy the original equation

Complete step-by-step solution:
The equation in the given question says that $\dfrac{5}{y-3}=\dfrac{y+7}{2y-6}+1$ ….. (i)
Let the analyse the above equation and try to find the value of x for which the equation is valid or it holds true. We can begin this by simplifying the summation on the right hand side of the equation.
The right hand side of the equation is given as $\dfrac{y+7}{2y-6}+1$ and we can simplify this expression by making a common denominator for both the terms.
With this we get that $\dfrac{y+7}{2y-6}+1=\dfrac{y+7}{2y-6}+\dfrac{2y-6}{2y-6}=\dfrac{y+7+2y-6}{2y-6}$
$\Rightarrow \dfrac{y+7}{2y-6}+1=\dfrac{3y+1}{2y-6}$
With this, the equation (i) changes to $\dfrac{5}{y-3}=\dfrac{3y+1}{2y-6}$ …. (ii).
Now, we can cross multiply the denominator such that equation becomes $5(2y-6)=(3y+1)(y-3)$
Let us open up the brackets.
With this, the equation will change to $10y-30=3{{y}^{2}}+y-9y-3$
$\Rightarrow 3{{y}^{2}}-18y+27=0$
On dividing the above equation by 3 we get that ${{y}^{2}}-6y+9=0$ …. (iii)
The above equation is a quadratic equation and we can see that the equation is a perfect square such that ${{y}^{2}}-6y+9={{(y-3)}^{2}}=0$.
Then,
$\Rightarrow y-3=0$, which means that $y=3$.
This means that $y=3$ satisfies the equation (iii). However, if we see the original equation (i), we can understand that y cannot be equal to 3 as it will result in an undetermined value.
Therefore, the extraneous solution to the given solution is $y=3$ and there is no solution to the given equation.

Note: Note that we use change or transform equations, sometimes the conditions on the equation may change. Most common reason for this is that the denominator of a fraction cannot be equal to zero. That is why we always check whether the final solutions satisfy the original equation.