
How do you solve \[5x-6=x+12\]?
Answer
547.8k+ views
Hint: To solve a linear equation in one variable, we have to find the value of the variable which satisfies the equation. This can be easily done by taking all the variable terms in the equation to one side and other constant terms to another side. By this, we can find the solution to the equation.
Complete answer:
We are given the equation \[5x-6=x+12\], as we can see that the degree of the equation is 1 and the only variable present in the equation is \[x\]. So, this is a linear equation in \[x\]. We are asked to solve the equation, which means we have to find the value of \[x\], which satisfies the given equation. The linear equation in one variable can be solved by taking the variable terms to one side and constants on another side of the equation. So, we will use this method to solve the equation.
The given equation is \[5x-6=x+12\]
Adding 6 to both sides of the equation we get,
\[\begin{align}
& \Rightarrow 5x-6+6=x+12+6 \\
& \Rightarrow 5x=x+18 \\
\end{align}\]
Subtracting \[x\] from both sides of the equation we get,
\[\begin{align}
& \Rightarrow 5x-x=x+18-x \\
& \Rightarrow 4x=18 \\
\end{align}\]
Dividing both sides of the equation by 4 we get,
\[\begin{align}
& \Rightarrow \dfrac{4x}{4}=\dfrac{18}{4} \\
& \Rightarrow x=\dfrac{18}{4} \\
\end{align}\]
Converting the fraction to simplest form,
\[\therefore x=\dfrac{9}{2}\]
Hence, the solution of the equation is \[x=\dfrac{9}{2}\].
Note: We can check if our solution is correct or not by substituting the value we get in the equation.
Substituting \[x=\dfrac{9}{2}\] in the LHS of the equation we get, \[5\times \dfrac{9}{2}-6=\dfrac{45}{2}-6=\dfrac{45-12}{2}=\dfrac{33}{2}\].
Substituting \[x=\dfrac{9}{2}\] in the RHS of the equation we get, \[\dfrac{9}{2}+12=\dfrac{9+24}{2}=\dfrac{33}{2}\]. As LHS = RHS the solution is correct.
Complete answer:
We are given the equation \[5x-6=x+12\], as we can see that the degree of the equation is 1 and the only variable present in the equation is \[x\]. So, this is a linear equation in \[x\]. We are asked to solve the equation, which means we have to find the value of \[x\], which satisfies the given equation. The linear equation in one variable can be solved by taking the variable terms to one side and constants on another side of the equation. So, we will use this method to solve the equation.
The given equation is \[5x-6=x+12\]
Adding 6 to both sides of the equation we get,
\[\begin{align}
& \Rightarrow 5x-6+6=x+12+6 \\
& \Rightarrow 5x=x+18 \\
\end{align}\]
Subtracting \[x\] from both sides of the equation we get,
\[\begin{align}
& \Rightarrow 5x-x=x+18-x \\
& \Rightarrow 4x=18 \\
\end{align}\]
Dividing both sides of the equation by 4 we get,
\[\begin{align}
& \Rightarrow \dfrac{4x}{4}=\dfrac{18}{4} \\
& \Rightarrow x=\dfrac{18}{4} \\
\end{align}\]
Converting the fraction to simplest form,
\[\therefore x=\dfrac{9}{2}\]
Hence, the solution of the equation is \[x=\dfrac{9}{2}\].
Note: We can check if our solution is correct or not by substituting the value we get in the equation.
Substituting \[x=\dfrac{9}{2}\] in the LHS of the equation we get, \[5\times \dfrac{9}{2}-6=\dfrac{45}{2}-6=\dfrac{45-12}{2}=\dfrac{33}{2}\].
Substituting \[x=\dfrac{9}{2}\] in the RHS of the equation we get, \[\dfrac{9}{2}+12=\dfrac{9+24}{2}=\dfrac{33}{2}\]. As LHS = RHS the solution is correct.
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