Question

How do you solve the system of equations $2x-3y=6$ and $3y-2x=-6$ ?

Answer 1

Hint: In order to find a solution to this problem, we will use a substitution method. The solution to the system of equations can be found using the substitution method or Gaussian elimination method. Since Substitution is the easiest one we will find out by substitution method.

Complete step by step answer:
We have system of equations as:
$2x-3y=6\to \left( 1 \right)$ and
$3y-2x=-6$
We can write this as,
$2x-3y=6\to \left( 2 \right)$
Since we have a system of equations, we will use a substitution method.
First we will use on equation $\left( 1 \right)$,
Therefore, we will start by isolating$x$ for $2x-3y=6$.
Now we will add $3y$ on both sides, we get:
$\Rightarrow 2x-3y+3y=6+3y$
On simplifying, we get:
$\Rightarrow 2x=6+3y$
Now on dividing both sides by $2$:
$\Rightarrow \dfrac{2x}{2}=\dfrac{6}{2}+\dfrac{3y}{2}$
On simplifying, we get:
$\Rightarrow x=3+\dfrac{3y}{2}$
As we can see that we have a value of $x$ that is $x=3+\dfrac{3y}{2}$. Therefore, now we can find our solution by substituting the value of $x$ in the equation $\left( 2 \right)$.
Therefore, we get:
$\left[ 2\left( 3+\dfrac{3y}{2} \right)-3y=6 \right]$
$2\left( 3+\dfrac{3y}{2} \right)-3y=6$
Now on simplifying left hand side, we get:
$\Rightarrow 2\left( 3+\dfrac{3y}{2} \right)-3y$
On expanding the above expression, we get:
$\Rightarrow 6+\dfrac{6y}{2}-3y$
$\Rightarrow 6+\dfrac{6y}{2}-\dfrac{6y}{2}$
On simplifying, we get:
$\Rightarrow 6$
With this now, we will write left hand side and right hand side simultaneously. So, we get:
$\Rightarrow 6=6$
As we can see that by substituting the value of x from one equation to another we are getting an equation free of variables.
Therefore, we can say that both the equations overlap each other and so they have infinite solutions.

Note: To find whether the value of $x$ is correct, we will substitute it in the given equations and equate it.
$2x-3y=6\to \left( 1 \right)$ and
$2x-3y=6\to \left( 2 \right)$
On substituting $x=3+\dfrac{3y}{2}$ in the left-hand side in equation $\left( 1 \right)$ we get:
$\Rightarrow 2\left( 3+\dfrac{3y}{2} \right)-3y$
On expanding we get:
$\Rightarrow 6+\dfrac{6y}{2}-3y$
$\Rightarrow 6+\dfrac{6y}{2}-\dfrac{6y}{2}$
On simplifying, we get:
$\Rightarrow 6$
Since equation $\left( 1 \right)$ is same as equation $\left( 2 \right)$ and since the left-hand side equals to the right-hand side in both equation $\left( 1 \right)$ and $\left( 2 \right)$.
Therefore, we can say that both the equations overlap each other and so they have infinite solutions.