# Find the value of $p$ in the linear equation $4p + 2 = 6p + 10$

Answer

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**Hint:**First we have to define what the terms we need to solve the problem are.

$p$ is a linear equation which means $p$ is of degree one (first order equation) and $p$ is a straight line because it is linear. Linear equations are equations with first order.

**Complete step by step answer:**

Since as we know $p$is linear equation (of first order with degree one)

Now from the given question the equation is $4p + 2 = 6p + 10$

And we need to find the value of $p$ so all the equation must on a same side so that we can able to cancel $p$each other and find the value,

So by transposing $4p + 2$ on right hand side of the equation, so other side must equal to zero

Thus $4p + 2 = 6p + 10$turns to $6p + 10 - (4p + 2) = 0$

Now solving the above equation further,

Let $6p + 10 - (4p + 2) = 0$[now multiply $ - $ on the $4p + 2$ ]

$ \Rightarrow 6p + 10 - 4p - 2 = 0$

Now subtracting $6p - 4p = 2p$ and also subtracting $10 - 2 = 8$

Applying these values on the above equation we get

$ \Rightarrow (6p - 4p) + (10 - 2) = 0$

$ \Rightarrow 2p + 8 = 0$

Now to find $p$ from this equation we need to transpose $8$ into right hand side of the equation

Therefore $ \Rightarrow 2p = - 8$

Now dividing by $2$ on the both sides on the above equation we get

$ \Rightarrow \dfrac{2}{2}p = \dfrac{{ - 8}}{2}$ Further proceeding to find the conclusion

$ \Rightarrow p = - 4$ [Cancelling by $2$ on both sides of the equation]

Hence the final value is $p = - 4$

As we know $p = - 4$(degree one) is linear because it is of first order and it is straight also.

**Note:**Linear equations are equations with first order. An equation for a straight line is called a linear equation. The general representation of the straight-line equation is $y = mx + b$ where the slope of the equation is $m$.

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