Question

How do you solve $\dfrac{x-1}{2}+\dfrac{y+2}{3}=4$ and $x-2y=5$?

Answer 1

Hint: Simplify the first equation and try to find the value of ‘x’ in terms of ‘y’. They put the value of ‘x’ obtained earlier to get the value of ‘y’. After getting the value of ‘y’ put that value in any of the equations to get the value of ‘x’.

Complete step by step solution:
Solving simultaneous equations: First we have to take any one of the two equations and simplify them if necessary. Then we have to find the value of ‘x’ in terms of ‘y’. Putting that value of ‘x’ in the second equation we can get the value of ‘y’. Again putting that value of ‘y’ in any of the equations we can get the value of ‘x’.
Considering the first equation
$\begin{align}
  & \dfrac{x-1}{2}+\dfrac{y+2}{3}=4 \\
 & \Rightarrow \dfrac{3\left( x-1 \right)+2\left( y+2 \right)}{6}=4 \\
 & \Rightarrow \dfrac{3x-3+2y+4}{6}=4 \\
 & \Rightarrow 3x+2y+1=4\times 6 \\
 & \Rightarrow 3x+2y=24-1 \\
 & \Rightarrow 3x+2y=23 \\
\end{align}$
From here ‘x’ can be obtained as
$\begin{align}
  & \Rightarrow 3x=23-2y \\
 & \Rightarrow x=\dfrac{23-2y}{3} \\
\end{align}$
Now considering the second equation $x-2y=5$
It can be written as
$\Rightarrow x=5+2y$
Putting the value of ‘x’ we got earlier in the above equation, we get
$\begin{align}
  & \Rightarrow \dfrac{23-2y}{3}=5+2y \\
 & \Rightarrow 23-2y=3\left( 5+2y \right) \\
 & \Rightarrow 23-2y=15+6y \\
 & \Rightarrow -2y-6y=15-23 \\
 & \Rightarrow -8y=-8 \\
 & \Rightarrow y=\dfrac{-8}{-8} \\
 & \Rightarrow y=1 \\
\end{align}$
Putting the value of ‘y’ in the second equation $x-2y=5$, we get
$\begin{align}
  & x-2y=5 \\
 & \Rightarrow x-2\times 1=5 \\
 & \Rightarrow x-2=5 \\
 & \Rightarrow x=5+2 \\
 & \Rightarrow x=7 \\
\end{align}$
Hence the solution of two simultaneous equation $\dfrac{x-1}{2}+\dfrac{y+2}{3}=4$ and $x-2y=5$ is $x=7$ and $y=1$.

Note: There is another method for solving simultaneous equations. At first we will always have to do the simplification if necessary. After simplification both the equations we got can be solved by equating the coefficient of either ‘x’ or ‘y’. For example
The simplified form of the given first equation we obtained as $3x+2y=23$……….(1)
And the second equation which is already in simplified form is $x-2y=5$……….(2)
To get both ‘x’ coefficient same we have to multiply the equation (1) by ‘1’ and equation (2) by ‘3’
$eq(1)\times 1\Rightarrow 3x+2y=23$……….(3)
$eq(2)\times 3\Rightarrow 3x-6y=15$……….(4)
Subtracting equation (4) from equation (3), we get
$\begin{align}
  & \left( 3x+2y \right)-\left( 3x-6y \right)=23-15 \\
 & \Rightarrow 3x+2y-3x+6y=8 \\
 & \Rightarrow 8y=8 \\
 & \Rightarrow y=\dfrac{8}{8} \\
 & \Rightarrow y=1 \\
\end{align}$
Putting the value of ‘y’ in any of the above equations we can get $x=7$.
This is the alternative method.