
Solve the following equation: - \[\dfrac{3y}{2}+\dfrac{y+4}{4}=5-\dfrac{y-2}{4}\].
Answer
556.8k+ views
Hint: Multiply both sides of the equation with 4 to get rid of the fractional terms. Now, take the terms containing ‘y’ to the left – hand side. Apply simple addition and subtraction to simplify the terms. Find the value of ‘y’ to get the answer.
Complete step by step answer:
We have been provided with the equation: - \[\dfrac{3y}{2}+\dfrac{y+4}{4}=5-\dfrac{y-2}{4}\]. We have to solve this equation, that means we have to find the value of y.
As we can see that, this is a linear equation in one variable, which is y. Therefore, we have,
Multiplying both sides of the equation with 4, we get,
\[\begin{align}
& \Rightarrow \left( 2\times 3y \right)+\left( y+4 \right)=5\times 4-\left( y-2 \right) \\
& \Rightarrow 6y+y+4=20-y+2 \\
& \Rightarrow 7y+4=22-y \\
\end{align}\]
Taking the terms containing ‘y’ to the L.H.S and the constant terms to the R.H.S, we get,
\[\begin{align}
& \Rightarrow 7y+y=22-4 \\
& \Rightarrow 8y=18 \\
& \Rightarrow y=\dfrac{18}{8} \\
\end{align}\]
Cancelling the common factors, we get,
\[\Rightarrow y=\dfrac{9}{4}\]
Hence, the value of y is \[\dfrac{9}{4}\].
Note: One may note that we have been provided with a single equation only. The reason is that, we have to find the value of only one variable, that is y. So, if we have to solve an equation having ‘n’ number of variables then we should be provided with ‘n’ equations. Now, one can check the answer by substituting the obtained value of ‘y’ in the equation provided in the question. We have to determine the value of L.H.S and R.H.S separately and if they are equal then our answer is correct.
Complete step by step answer:
We have been provided with the equation: - \[\dfrac{3y}{2}+\dfrac{y+4}{4}=5-\dfrac{y-2}{4}\]. We have to solve this equation, that means we have to find the value of y.
As we can see that, this is a linear equation in one variable, which is y. Therefore, we have,
Multiplying both sides of the equation with 4, we get,
\[\begin{align}
& \Rightarrow \left( 2\times 3y \right)+\left( y+4 \right)=5\times 4-\left( y-2 \right) \\
& \Rightarrow 6y+y+4=20-y+2 \\
& \Rightarrow 7y+4=22-y \\
\end{align}\]
Taking the terms containing ‘y’ to the L.H.S and the constant terms to the R.H.S, we get,
\[\begin{align}
& \Rightarrow 7y+y=22-4 \\
& \Rightarrow 8y=18 \\
& \Rightarrow y=\dfrac{18}{8} \\
\end{align}\]
Cancelling the common factors, we get,
\[\Rightarrow y=\dfrac{9}{4}\]
Hence, the value of y is \[\dfrac{9}{4}\].
Note: One may note that we have been provided with a single equation only. The reason is that, we have to find the value of only one variable, that is y. So, if we have to solve an equation having ‘n’ number of variables then we should be provided with ‘n’ equations. Now, one can check the answer by substituting the obtained value of ‘y’ in the equation provided in the question. We have to determine the value of L.H.S and R.H.S separately and if they are equal then our answer is correct.
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