Question

How do you solve for $x$ in $\dfrac{4}{9}\left( x+3 \right)=g$?

Answer 1

Hint: We first assume the value of the $g$ both as variable and constant. We get an equation of line when we take $g$ as variable. The x coordinates of all the points on the line is the solution for $x$ in $\dfrac{4}{9}\left( x+3 \right)=g$. We then use the constant part to find a single point for $x$.

Complete step-by-step solution:
We have been given an equation of variable $x$ where $\dfrac{4}{9}\left( x+3 \right)=g$.
We can take the value $g$ as a variable and that gives a line equation of two variables.
We are taking the general equation of line to understand the slope and the intercept form of the line $\dfrac{4}{9}\left( x+3 \right)=g$. We assume the value of $g$ as variable $y$.
So, $\dfrac{4}{9}\left( x+3 \right)=y$ where $y=g$. The solution for $x$ in $\dfrac{4}{9}\left( x+3 \right)=g$ is $x$ in $x=\dfrac{9g}{4}-3$.
Simplifying we get $4x+12=9y\Rightarrow 4x-9y=-12$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be$p$ and $q$ respectively. The points will be $\left( p,0 \right),\left( 0,q \right)$. The given equation is $3x+y=7$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$
\[\begin{align}
  & 4x-9y=-12 \\
 & \Rightarrow \dfrac{4x}{-12}+\dfrac{-9y}{-12}=1 \\
 & \Rightarrow \dfrac{x}{-3}+\dfrac{y}{{}^{4}/{}_{3}}=1 \\
\end{align}\]
Therefore, the x intercept, and y intercept of the line $\dfrac{4}{9}\left( x+3 \right)=y$ is $-3$ and $\dfrac{4}{3}$ respectively. The axes intersecting points are $\left( -3,0 \right),\left( 0,\dfrac{4}{3} \right)$. These two values of $x=-3,0$ are two solutions of an infinite number of solutions.

Note: If we take $g$ as constant, we get the value of $x$ in $\dfrac{4}{9}\left( x+3 \right)=g$ as $x=\dfrac{9g}{4}-3$. This is a single point. Single solution for the single variable.
For example: if $g=4$, then $x=\dfrac{9g}{4}-3=\dfrac{9\times 4}{4}-3=9-3=6$.