How do you solve the linear equation $\dfrac{24}{5z+4}=\dfrac{4}{z-1}$?
Answer
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Hint: The equation $\dfrac{24}{5z+4}=\dfrac{4}{z-1}$, given in the above question is not looking simple to solve. This is due to the fractional variable terms on both sides of the given equation. So we must first simplify it by multiplying both sides of the given equation by the LCM of the denominator terms, which is equal to their product $\left( 5z+4 \right)\left( z-1 \right)$. Then using the distributive property of the algebraic multiplication $a\left( b+c \right)=ab+ac$ on both the sides of the resulting equation and using further algebraic operations, we will get the final solution.
Complete step by step solution:
The given equation is
$\Rightarrow \dfrac{24}{5z+4}=\dfrac{4}{z-1}$
Since the above equation involves the variable terms in the denominators, it is looking cumbersome. So we have to simplify it. For this, we multiply both sides of the above equation by the LCM of the denominator terms, which is equal to $\left( 5z+4 \right)\left( z-1 \right)$, to get
\[\begin{align}
& \Rightarrow \dfrac{24}{5z+4}\left( 5z+4 \right)\left( z-1 \right)=\dfrac{4}{z-1}\left( 5z+4 \right)\left( z-1 \right) \\
& \Rightarrow 24\left( z-1 \right)=4\left( 5z+4 \right) \\
\end{align}\]
Using the distributive property of the algebraic multiplication, given by $a\left( b+c \right)=ab+ac$, on both the sides of the above equation we get
$\Rightarrow 24z-24=20z+16$
Subtracting $20z$ from both the sides, we get
$\begin{align}
& \Rightarrow 24z-24-20z=20z+16-20z \\
& \Rightarrow 4z-24=16 \\
\end{align}$
Now, we add $24$ on both the sides to get
\[\begin{align}
& \Rightarrow 4z-24+24=16+24 \\
& \Rightarrow 4z=40 \\
\end{align}\]
Finally, dividing both the sides by \[4\] we get
$\begin{align}
& \Rightarrow \dfrac{4z}{4}=\dfrac{40}{4} \\
& \Rightarrow z=10 \\
\end{align}$
Hence, the solution of the given equation is $z=10$.
Note: For removing the variable terms from the given equation in order to simplify it, we can also take the reciprocal of both the sides. Then, on cross multiplying the numbers on the denominators of the resulting equation, the equation will get simplified. Also, after obtaining the final solution, we must check it by back substituting it into the given equation and confirm whether it is satisfying it or not.
Complete step by step solution:
The given equation is
$\Rightarrow \dfrac{24}{5z+4}=\dfrac{4}{z-1}$
Since the above equation involves the variable terms in the denominators, it is looking cumbersome. So we have to simplify it. For this, we multiply both sides of the above equation by the LCM of the denominator terms, which is equal to $\left( 5z+4 \right)\left( z-1 \right)$, to get
\[\begin{align}
& \Rightarrow \dfrac{24}{5z+4}\left( 5z+4 \right)\left( z-1 \right)=\dfrac{4}{z-1}\left( 5z+4 \right)\left( z-1 \right) \\
& \Rightarrow 24\left( z-1 \right)=4\left( 5z+4 \right) \\
\end{align}\]
Using the distributive property of the algebraic multiplication, given by $a\left( b+c \right)=ab+ac$, on both the sides of the above equation we get
$\Rightarrow 24z-24=20z+16$
Subtracting $20z$ from both the sides, we get
$\begin{align}
& \Rightarrow 24z-24-20z=20z+16-20z \\
& \Rightarrow 4z-24=16 \\
\end{align}$
Now, we add $24$ on both the sides to get
\[\begin{align}
& \Rightarrow 4z-24+24=16+24 \\
& \Rightarrow 4z=40 \\
\end{align}\]
Finally, dividing both the sides by \[4\] we get
$\begin{align}
& \Rightarrow \dfrac{4z}{4}=\dfrac{40}{4} \\
& \Rightarrow z=10 \\
\end{align}$
Hence, the solution of the given equation is $z=10$.
Note: For removing the variable terms from the given equation in order to simplify it, we can also take the reciprocal of both the sides. Then, on cross multiplying the numbers on the denominators of the resulting equation, the equation will get simplified. Also, after obtaining the final solution, we must check it by back substituting it into the given equation and confirm whether it is satisfying it or not.
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