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How do you solve 2y + 8 = 12 – 6 (y + 7)?

Last updated date: 26th Feb 2024
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IVSAT 2024
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Hint: We are asked to find the solution of 2y + 8 = 12 – 6 (y + 7). Firstly, we learn what the solution of the equation means is and then we will learn what a linear equation in 1 variable term is. We use a hit and trial method to find the value of ‘y’. In this method, we put the value of ‘y’ one by one by hitting arbitrary values and looking for needed values. Once we work with the hit and trial method we will try another method where we apply algebra. We subtract, add or multiply terms to get to our final term and get our required solution. We will also learn that doing the questions using an algebraic tool makes them easy.

Complete step by step answer:
We are given that we have 2y + 8 = 12 – 6 (y + 7) and we are asked to find the value of ‘y’, or we are asked how we will be able to solve this expression. The solution of any problem is that value which when put into the given problem then the equation is satisfied. Now we learn about the equation on one variable. One variable simply represents the equation that has one variable (say x, y, or z) and another one constant. For example, x + 2 = 4, 2 – x = 2, 2x, 2y, etc.
Our equation 2y + 8 = 12 – 6 (y + 7) also has just one variable ‘y’. We have to find the value of ‘y’ which will satisfy our given equation. Firstly we try by the method of hit and trial. In which we will put a different value of ‘y’ and take which one fits the solution correctly.
Let y = 0, we put y = 0 in 2y + 8 = 12 – 6 (y + 7). So, we get,
\[2\times 0+8=12-6\left( 0+7 \right)\]
So, we get,
\[\Rightarrow 8=12-42\]
\[\Rightarrow 8=-30\]
So, this is not true. So, y = 0 is not the solution to our problem.
Let y = – 1, we put y = – 1 in 2y + 8 = 12 – 6 (y + 7). So, we get,
\[2\times -1+8=12-6\left( -1+7 \right)\]
So, we get,
\[\Rightarrow 6=-24\]
So, this is not true. So, y = – 1is not the solution to our problem.
Let y = – 4, we put y = – 4 in 2y + 8 = 12 – 6 (y + 7). So, we get,
\[2\times -4+8=12-6\left( -4+7 \right)\]
So, we get,
\[\Rightarrow -8+8=12-18\]
\[\Rightarrow 0=-6\]
So, this is not true. So, y = – 4 is not the solution to our problem.
As we can see that the difference between the terms on the left side to the right side has decreased means the solution will lie on this side, but this method surely takes a lot of time. So we will use an alternate method which will reduce our time.
We will use an Algebraic tool to solve our problem. Now we have
\[2y+8=12-6\left( y+7 \right)\]
Opening the brackets, we get,
\[\Rightarrow 2y+8=12-6y-42\]
On simplifying, we get,
\[\Rightarrow 2y+8=-30-6y\]
Now we add 6y to both the sides, so we get,
\[\Rightarrow 2y+8+6y=-30-6y+6y\]
On simplifying, we get,
\[\Rightarrow 8y+8=-30\]
Now, we subtract 8 on both the sides, so we get,
\[\Rightarrow 8y+8-8=-30-8\]
On simplifying, we get,
\[\Rightarrow 8y=-38\]
Now dividing both the sides by 8, we get,
\[\Rightarrow \dfrac{8y}{8}=\dfrac{-38}{8}\]
On simplifying, we get,
\[\Rightarrow y=\dfrac{-19}{4}\]
\[\Rightarrow y=-4.75\]

Hence, the solution in y is \[\dfrac{-19}{4}\] or – 4.75.

Note: Remember that, we cannot add the variable to the constant. Usual mistakes like this where one adds constants with variables usually happen. For example 3x + 6 = 9x. here, one added ‘6’ with 3 of x and made it 9x. This is wrong. We cannot add constants and variables at once. Only the same variables are added to each other. When we add the variable the only constant part is added or subtracted variable remains the same. That is a 2x + 2x = 4x error like doing it \[2x+2x=4{{x}^{2}}\] may happen. So, be careful there. Remember when we divide a positive term by a negative value the solution we get is a negative term this may happen if we skip the negative sign. We need to choose the best techniques as per the question if the question is defined using variable only time so hit and trials would be ok but if the equation has more time variable then we go for the arithmetic tool.