
Solve the equation $3y+4=5y-4$ .
Answer
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Hint: To solve the equation $3y+4=5y-4$ , we have to collect the terms in y on the LHS and constants on the RHS. Then, we have to add the like terms. Then, we have to move the coefficient of y to the other side and find the value of y.
Complete step by step solution:
We have to solve the equation $3y+4=5y-4$ . Let us first collect the terms in y on the LHS and constants on the RHS. For this, we have to move 5y to the LHS. This term will be -5y in the LHS.
$\begin{align}
& \Rightarrow 3y+4=5y-4 \\
& \Rightarrow 3y-5y+4=-4 \\
\end{align}$
Now, let us move 4 from the LHS to RHS. This term will become negative in the THS.
$\Rightarrow 3y-5y=-4-4$
Now, we have to add the like terms.
$\Rightarrow -2y=-8$
We have to take -2 to the RHS. This term will be the divisor in the RHS.
$\Rightarrow y=\dfrac{-8}{-2}$
Let us cancel the negative sign
$\Rightarrow y=\dfrac{\require{cancel}\cancel{-}8}{\require{cancel}\cancel{-}2}$
We can write the result of the above simplification as
$\Rightarrow y=\dfrac{8}{2}$
Now, let us divide 8 by 2.
$\Rightarrow y=4$
Therefore, the solution of the equation $3y+4=5y-4$ is 4.
Note: Students must thoroughly understand the concept of solving algebraic expressions and the rules associated with it. The positive (or negative) term will become negative (or positive) when moved from one side to the other. When a coefficient is moved to the other side, it becomes the divisor and when a divisor is from to the other side, it gets multiplied with the terms on the other side. Similarly, when we add, substract, multiply or divide one side by an additional term, we have to do the similar operation on the opposite side. We can also solve $3y+4=5y-4$ in an alternate way.
We have to subtract 3y from both the sides.
$\Rightarrow 3y-3y+4=5y-3y-4$
Let us add the like terms.
$\Rightarrow 4=2y-4$
We have to take 4 from the RHS to the LHS.
$\begin{align}
& \Rightarrow 4+4=2y \\
& \Rightarrow 2y=8 \\
\end{align}$
Let us take 2 to the RHS.
$\begin{align}
& \Rightarrow y=\dfrac{8}{2} \\
& \Rightarrow y=4 \\
\end{align}$
Complete step by step solution:
We have to solve the equation $3y+4=5y-4$ . Let us first collect the terms in y on the LHS and constants on the RHS. For this, we have to move 5y to the LHS. This term will be -5y in the LHS.
$\begin{align}
& \Rightarrow 3y+4=5y-4 \\
& \Rightarrow 3y-5y+4=-4 \\
\end{align}$
Now, let us move 4 from the LHS to RHS. This term will become negative in the THS.
$\Rightarrow 3y-5y=-4-4$
Now, we have to add the like terms.
$\Rightarrow -2y=-8$
We have to take -2 to the RHS. This term will be the divisor in the RHS.
$\Rightarrow y=\dfrac{-8}{-2}$
Let us cancel the negative sign
$\Rightarrow y=\dfrac{\require{cancel}\cancel{-}8}{\require{cancel}\cancel{-}2}$
We can write the result of the above simplification as
$\Rightarrow y=\dfrac{8}{2}$
Now, let us divide 8 by 2.
$\Rightarrow y=4$
Therefore, the solution of the equation $3y+4=5y-4$ is 4.
Note: Students must thoroughly understand the concept of solving algebraic expressions and the rules associated with it. The positive (or negative) term will become negative (or positive) when moved from one side to the other. When a coefficient is moved to the other side, it becomes the divisor and when a divisor is from to the other side, it gets multiplied with the terms on the other side. Similarly, when we add, substract, multiply or divide one side by an additional term, we have to do the similar operation on the opposite side. We can also solve $3y+4=5y-4$ in an alternate way.
We have to subtract 3y from both the sides.
$\Rightarrow 3y-3y+4=5y-3y-4$
Let us add the like terms.
$\Rightarrow 4=2y-4$
We have to take 4 from the RHS to the LHS.
$\begin{align}
& \Rightarrow 4+4=2y \\
& \Rightarrow 2y=8 \\
\end{align}$
Let us take 2 to the RHS.
$\begin{align}
& \Rightarrow y=\dfrac{8}{2} \\
& \Rightarrow y=4 \\
\end{align}$
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