Question

Solve the following pair of equations:
$x+y=3.3;\,\dfrac{0.6}{3x-2y}=-1;\,3x-2y\ne 0$

Answer 1

Hint: We will multiply these equations with a power of ten according to the decimal number it contains to change them into a whole number. We will rearrange the equations if necessary.

Complete step by step solution:
Let us consider the given system of equations,
$\begin{align}
  & x+y=3.3.......\left( 1 \right) \\
 & \dfrac{0.6}{3x-2y}=-1,\,3x-2y\ne 0.......\left( 2 \right) \\
\end{align}$
We are asked to solve this system. As we can see, both equations contain a term that is a decimal number.
So, now, we will multiply the equation $\left( 1 \right)$ with $10$ to get $10x+10y=33.......\left( 3 \right)$
As the next step, we will transpose the denominator in the equation $\left( 2 \right)$ from the left-hand side to the right-hand side.
We will get $0.6=-3x+2y.$
We need to multiply this equation with $-10$ to get $30x-20y=-6.......\left( 4 \right)$
Let us multiply the equation $\left( 3 \right)$ with $2$ and add the equation $\left( 4 \right)$ to it.
We will get,
$\begin{align}
  & \underline{\begin{align}
  & 20x+20y=66+ \\
 & 30x-20y=-6 \\
\end{align}} \\
 & 50x+0=60 \\
\end{align}$
We will get the equation $50x=60.$
From this equation, we can find the value of $x$ and thus the value of $y$ by putting the value of $x$ in any of the given equations.
Now, when we transpose $50$ from the left-hand side to the right-hand side, we will get $x=\dfrac{60}{50}.$
Now, we will get $x=\dfrac{6}{5}$
When we substitute this value in any of the given equations, we will be able to find the value of $y.$
So, let us replace $x$ in the equation $\left( 1 \right)$ with $\dfrac{6}{5}=1.2$
Then, we will get $1.2+y=3.3.$
Now, we will transpose $1.2$ from the left-hand side to the right-hand side of the equation, we will get $y=3.3-1.2=2.1.$
Hence the solution of the given pair of equations is $x=1.2$ and $y=2.1.$

Note: We can convert the pair of equations into the form $AX=b,$ where $A$ is the coefficient matrix of the equations, $X$ is the column matrix that contains the unknowns and $b$ is the column matrix that contains the RHS of the equations. And then using $X={{A}^{-1}}b,$ we can solve the equations.