
How do you solve $\dfrac{n}{3}-8=-2$?
Answer
558k+ views
Hint: We solve the given linear equation by simplifying the equation. We separate the variable and the constants on different sides of the equality. We cross-multiply the equations. The process of cross-multiplication is equivalent to taking the L.C.M of the denominators and multiplying with that. Then we apply the multiplication to get the value of h. We also take the L.C.M to multiply both sides and verify the answer.
Complete step by step answer:
The given equation $\dfrac{n}{3}-8=-2$ is a linear equation of n. We first take all the constants on the right-hand side and keep the variables on the left-hand side.
So, $\dfrac{n}{3}-8=-2\Rightarrow \dfrac{n}{3}=-2+8=6$
we assume the right-hand side number 6 as $\dfrac{6}{1}$.
We apply cross-multiplication to multiply n with 1 and 3 with 6.
\[\begin{align}
& \dfrac{n}{3}=\dfrac{6}{1} \\
& \Rightarrow 1\left( n \right)=3\times 6 \\
\end{align}\]
The multiplication value of 1 with n gives us value n. Similarly, the multiplication value of 3 with 6 gives us value \[3\times 6=18\]
We complete the multiplication to get \[n=18\].
Therefore, the solution of the equation $\dfrac{n}{3}-8=-2$ is \[n=18\].
The L.C.M of 3 and 1 gives 3. So, multiplying both sides of \[\dfrac{n}{3}=\dfrac{6}{1}\] with 3, we get
$\begin{align}
& \dfrac{n}{3}\times 3=\dfrac{6}{1}\times 3 \\
& \Rightarrow n=18 \\
\end{align}$.
So, the correct answer is “n=18 ”.
Note: The process of cross-multiplication comes from the L.C.M of the denominators. The L.C.M gets multiplied with the both sides of the equation which in turn gives the cross multiplication. In the case of denominators being co-primes, both processes are the same.
For example, if we take $\dfrac{x}{36}=\dfrac{y}{27}$, we can cross multiply to get $27x=36y$ and then we divide both side with their G.C.D value 9.
That gives us
$\begin{align}
& \dfrac{27x}{9}=\dfrac{36y}{9} \\
& \Rightarrow 3x=4y \\
\end{align}$
We can also use the L.C.M of the denominators which is 108 and multiply both sides of $\dfrac{x}{36}=\dfrac{y}{27}$ to get the final answer directly.
\[\begin{align}
& \dfrac{x}{36}\times 108=\dfrac{y}{27}\times 108 \\
& \Rightarrow 3x=4y \\
\end{align}\].
Complete step by step answer:
The given equation $\dfrac{n}{3}-8=-2$ is a linear equation of n. We first take all the constants on the right-hand side and keep the variables on the left-hand side.
So, $\dfrac{n}{3}-8=-2\Rightarrow \dfrac{n}{3}=-2+8=6$
we assume the right-hand side number 6 as $\dfrac{6}{1}$.
We apply cross-multiplication to multiply n with 1 and 3 with 6.
\[\begin{align}
& \dfrac{n}{3}=\dfrac{6}{1} \\
& \Rightarrow 1\left( n \right)=3\times 6 \\
\end{align}\]
The multiplication value of 1 with n gives us value n. Similarly, the multiplication value of 3 with 6 gives us value \[3\times 6=18\]
We complete the multiplication to get \[n=18\].
Therefore, the solution of the equation $\dfrac{n}{3}-8=-2$ is \[n=18\].
The L.C.M of 3 and 1 gives 3. So, multiplying both sides of \[\dfrac{n}{3}=\dfrac{6}{1}\] with 3, we get
$\begin{align}
& \dfrac{n}{3}\times 3=\dfrac{6}{1}\times 3 \\
& \Rightarrow n=18 \\
\end{align}$.
So, the correct answer is “n=18 ”.
Note: The process of cross-multiplication comes from the L.C.M of the denominators. The L.C.M gets multiplied with the both sides of the equation which in turn gives the cross multiplication. In the case of denominators being co-primes, both processes are the same.
For example, if we take $\dfrac{x}{36}=\dfrac{y}{27}$, we can cross multiply to get $27x=36y$ and then we divide both side with their G.C.D value 9.
That gives us
$\begin{align}
& \dfrac{27x}{9}=\dfrac{36y}{9} \\
& \Rightarrow 3x=4y \\
\end{align}$
We can also use the L.C.M of the denominators which is 108 and multiply both sides of $\dfrac{x}{36}=\dfrac{y}{27}$ to get the final answer directly.
\[\begin{align}
& \dfrac{x}{36}\times 108=\dfrac{y}{27}\times 108 \\
& \Rightarrow 3x=4y \\
\end{align}\].
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