Question

How do you solve the system of linear equations $4x+6y=12$ for $x$?

Answer 1

Hint: Change of form of the given equation will give the x-intercept and y-intercept of the line $4x+6y=12$. As there is only one equation to solve two unknowns, we will get an infinite number of solutions. We change the equation to the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x intercept, and y intercept of the line as $p$ and $q$ respectively. Then we place the points on the axes and from there we draw the line on the graph. All the points on the line are solutions for $x$ in $4x+6y=12$.

Complete step-by-step solution:
We are taking the general equation of line to understand the slope and the intercept form of the line $4x+6y=12$. We have to find the x-intercept, and y-intercept of the line $4x+6y=12$
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 $4x+6y=12$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$
$\begin{align}
  & 4x+6y=12 \\
 & \Rightarrow \dfrac{4x}{12}+\dfrac{6y}{12}=1 \\
 & \Rightarrow \dfrac{x}{3}+\dfrac{y}{2}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $4x+6y=12$ is 3 and 2 respectively. The axes intersecting points are $\left( 3,0 \right),\left( 0,2 \right)$. These two values of $x=3,0$ are two solutions of an infinite number of solutions.

Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance. Hence, we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $\infty $.