Question

How do you solve the $x$ in $-ax+2b>8$?

Answer 1

Hint: We first assume the value of the $a,b$ as constants. We get a particular point when we take $a,b$ as constants. The x coordinates of all the points on the line is the solution for $x$ in $-ax+2b>8$. We then use the constant part to find a single point for $x$. We can take examples to understand the condition for $x$.

Complete step-by-step solution:
We have been given an inequation of variable $x$ where $-ax+2b>8$.
We take the values $a,b$ as constants and that gives a particular point of variable $x$.
The given inequation $-ax+2b>8$ is a linear inequation of $x$. We need to simplify the equation by solving the variables and the constants separately.
All the terms in the inequation of $-ax+2b>8$ are either variable of $x$ or a constant. We first separate the variables.
We take the constants all together to solve it.
There are two such constants which are 8 and $2b$ .
Now we apply the binary operation of subtraction to get
$\begin{align}
  & -ax+2b>8 \\
 & \Rightarrow -ax>8-2b \\
\end{align}$
Now we divide both sides of the equation with $-a$ to get
\[\begin{align}
  & -ax>8-2b \\
 & \Rightarrow \dfrac{-ax}{-a}<\dfrac{8-2b}{-a} \\
 & \Rightarrow x<\dfrac{2b-8}{a} \\
\end{align}\]
Therefore, the final solution becomes \[x<\dfrac{2b-8}{a}\].
The interval for $x$ will be \[x\in \left( -\infty ,\dfrac{2b-8}{a} \right)\].

Note: We can verify the result of the equation $-ax+2b>8$ by taking the value of as $a=2,b=1$. The inequation becomes $-2x+2>8$. The final solution becomes
\[\begin{align}
  & x<\dfrac{2\times 1-8}{2} \\
 & \Rightarrow x<-3 \\
\end{align}\].
The interval for $x$ will be \[x\in \left( -\infty ,-3 \right)\].