Question

How do you solve \[3(p-2)+7=32-4(2p+5)\]?

Answer 1

Hint: The degree of an equation is the highest power to which the variable in the equation is raised. If the degree of the equation is one, then it is a linear equation. To solve a linear equation, we have to take all the variable terms to one side of the equation, and leave constants to the other side. By this, we can find the solution value of the equation.

Complete step by step solution:
We are given the equation \[3(p-2)+7=32-4(2p+5)\], we have to solve it. The highest power of the variable of the equation is 1, so the degree of the equation is also one. Hence, it is a linear equation. As we know to solve a linear equation, we have to take all the variable terms to one side of the equation and leave constants to the other side.
\[3(p-2)+7=32-4(2p+5)\]
Using the distributive property, \[a\left( b+c \right)=ab+ac\] to expand the brackets on both sides, we get
\[\begin{align}
  & \Rightarrow 3p-6+7=32-8p-20 \\
 & \Rightarrow 3p+1=12-8p \\
\end{align}\]
Subtracting 1 from both sides of the above expression, we get
\[\Rightarrow 3p=11-8p\]
Adding \[8p\] to both sides of above equation, we get
\[\Rightarrow 11p=11\]
Dividing by 11 to both sides of the above equation and cancelling out common factor, we get
\[\therefore p=1\]
Hence, the solution of the given equation is \[p=1\].

Note: We can check if the answer is correct or not by substituting the value in the given equation. From the given equation, we get the left-hand side as \[3(p-2)+7\], and right-hand side as \[32-4(2p+5)\]. Substituting \[p=1\] in both sides of the equation, we get LHS as \[3(1-2)+7=-3+7=4\], and RHS as \[32-4(2(1)+5)=32-28=4\]. As \[LHS=RHS\], the solution is correct.