Question

If \[\dfrac{(6x+1)}{3}+1=\dfrac{(x-3)}{6}\] , then $x=?$

Answer 1

Hint: First identify the unknown variable , then separate the variables and do manipulations and find value of variable.

Complete step by step answer:
Linear equations are the equations in which the degree of variables present in the equation is $1$ . Equations may or may have more than $1$ variable.
Equation having $1$ variable is the form of $x=c$ or $y=c$ where $c$ is a constant and equation having $2$ variables are in the form $Ax+By=C$ where$A,B$ and$C$ are constants.
When we graph the graph of equation $x=c$ in $X-Y$ plane we get a line parallel to $Y-axis$ and passes through the point $(c,0)$.
Graph of equation $y=c$ in$X-Y$ plane is line parallel to $x-axis$ passing through the point$(0,c)$ .
When we draw the graph of equation having $2$ variables i.e.$Ax+By=C$in $X-Y$ plane we get a line which is intercepted between the $2$ the axes having $X-\operatorname{int}ercept=\dfrac{-C}{A}$ and $Y-\operatorname{int}ercept=\dfrac{-C}{B}$ .
The Angle of inclination$(\theta )$ of line with $X-axis$ is given by $\operatorname{Tan}\theta =$ slope of line$=\dfrac{-A}{B}$
Now, we have $\dfrac{(6x+1)}{3}+1=\dfrac{(x-3)}{6}$ , in order to find the value of $x$ for which the equation holds.
Firstly, in L.H.S we add $1$ in $\dfrac{(6x+1)}{3}$ we get,
$\dfrac{(6x+1)}{3}+1=\dfrac{(6x+1)+3}{3}=\dfrac{6x+1+3}{3}=\dfrac{6x+4}{3}$
After manipulating L.H.S of equation we left with,
$\dfrac{(6x+4)}{3}=\dfrac{(x-3)}{6}$
Then we have to do cross-multiplication i.e. we have to multiply L.H.S by$6$ and R.H.S by $3$ then we get,
$\begin{align}
  & 6*(6x+4)=3*(x-3) \\
 & (36x+24)=(3x-9) \\
\end{align}$
After that we just separate variable $x$ and constants means at L.H.S has variable $x$ and R.H.S has constants term
$\begin{align}
  & 36x+24=3x-9 \\
 & 36x-3x=-9-24 \\
 & 33x=-33 \\
 & x=\dfrac{-33}{33} \\
 & x=-1 \\
\end{align}$
Hence, the value of $x$ such that the equation $\dfrac{(6x+1)}{3}+1=\dfrac{(x-3)}{6}$ satisfies is $x=-1$.

Note: Since in $X-Y$ plane $Ax+By=C$ is an equation of line, we have other form which represent same equation of line i.e. $(y-{{y}_{\circ }})=m(x-{{x}_{\circ }})$ where $m$ is slope of line and $({{x}_{\circ }},{{y}_{\circ }})$ is point lie on line this equation is known Slope-Point form.