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Question: How do you solve \[3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right)\]?...

How do you solve 3(4y−1)=2(5y+12)3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right)?

Explanation

Solution

We separate the variables and the constants of the equation 3(4y−1)=2(5y+12)3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right). We have in total two multiplications. We multiply the constants. Then we apply the binary operation of addition and subtraction for both variables and constants. The solutions of the variables and the constants will be added at the end to get the final answer to equate with 0. Then we solve the linear equation to find the value of y.

Complete step by step solution:
The given equation 3(4y−1)=2(5y+12)3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right) is a linear equation of y. We need to simplify the equation by completing the multiplication of the constants separately.
All the terms in the equation of 3(4y−1)=2(5y+12)3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right) are either variables of y or a constant. We break the multiplication by multiplying 3 with (4y−1)\left( 4y-1 \right) and 2 with (5y+12)\left( 5y+\dfrac{1}{2} \right).
So, 3(4y−1)=12y−33\left( 4y-1 \right)=12y-3 and 2(5y+12)=10y+12\left( 5y+\dfrac{1}{2} \right)=10y+1.
The equation becomes 12y−3=10y+112y-3=10y+1.
We take all the variables and the constants on one side and get 12y−3−10y−1=012y-3-10y-1=0.
There are two variables which are 12y,10y12y,10y.
The binary operation between them is subtraction which gives us 12y−10y=2y12y-10y=2y.
Now we take the constants.
There are two such constants which are −3,−1-3,-1.
The binary operation between them is addition which gives us −3−1=−4-3-1=-4.
The final solution becomes
12y−3−10y−1=0 ⇒2y−4=0 \begin{aligned} & 12y-3-10y-1=0 \\\ & \Rightarrow 2y-4=0 \\\ \end{aligned}.
Now we take the variable on one side and the constants on the other side.

& 2y-4=0 \\\ & \Rightarrow y=\dfrac{4}{2}=2 \\\ \end{aligned}$$ Therefore, the solution is $y=2$. **Note:** We can verify the result of the equation $$3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right)$$ by taking the value of x as $y=2$. Therefore, the left-hand side of the equation becomes $$3\left( 4y-1 \right)=3\left( 4\times 2-1 \right)=3\left( 8-1 \right)=3\times 7=21$$ The right-hand side of the equation becomes $$2\left( 5y+\dfrac{1}{2} \right)=2\left( 5\times 2+\dfrac{1}{2} \right)=2\left( 10+\dfrac{1}{2} \right)=2\times \dfrac{21}{2}=21$$ Thus, verified for the equation $$3\left( 4y-1 \right)=2\left( 5y+\dfrac{1}{2} \right)$$ the solution is $y=2$.