Question & Answer
QUESTION

Solve: $\dfrac{10{{x}^{2}}+15x+63}{5{{x}^{2}}-25x+12}=\dfrac{2x+3}{x-5}$.

ANSWER Verified Verified
Hint: In this question, from the left side we take 5x common from both the numerator and denominator which is having variable x in it. After that, we let a = 2x+3 and b = x -5. Now, solving further and putting the value of a and b, we get the value of x.

Complete step by step answer:
According to the question, we are given an expression with variable x: $\dfrac{10{{x}^{2}}+15x+63}{5{{x}^{2}}-25x+12}=\dfrac{2x+3}{x-5}$. Considering this equation, we proceed to solving for x.
$\dfrac{10{{x}^{2}}+15x+63}{5{{x}^{2}}-25x+12}=\dfrac{2x+3}{x-5}$
Now, we take 5x common from numerator and denominator the term having variable x in it and leaving the constant apart in the left-hand side, we get
$\dfrac{5x\left( 2x+3 \right)}{5x\left( x-5 \right)}=\dfrac{2x+3}{x-5}$
To solve the question easily we let, $2x+3=a$ and $x-5=b$.
$\dfrac{5ax+63}{5bx+12}=\dfrac{a}{b}$
Now, we cross multiply both left-hand side and right-hand side with each other to simplify the expression.
$\begin{align}
  & \left( 5ax+63 \right)\left( b \right)=\left( 5bx+12 \right)\left( b \right) \\
 & 5axb+63b=5axb+12a \\
\end{align}$
Now, $5axb$ is present on both sides with the same sign, so they cancel out each other.
Remaining term will be,
$63b=12a\ldots (1)$
Now, we divide the whole equation by 3 to simplify the equation (1) and renaming the equation as equation (2):
 $21b=4a\ldots (2)$
Putting the value of a and b in the above equation (2), we get:
$\begin{align}
  & 21\left( x-5 \right)=4\left( 2x+3 \right) \\
 & 21x-105=8x+12 \\
 & 21x-8x=12+105 \\
 & 13x=117 \\
 & x=\dfrac{117}{3} \\
 & x=9 \\
\end{align}$
Hence, the value of x in this question is 9.

Note: The key concept for solving this problem is the rewriting of the same variables as some other variable to reduce the exponent of the variable. This problem can be alternatively solved if we simply cross multiply both the terms. But this procedure will take more time to solve. Hence, to save time and effort students must adopt the above method.