Question

How do you solve the following system of equations: $3x+2y=1$ , $2x+3y=12$?

Answer 1

Hint: Here we have been given a system of equations and we have to solve it. Firstly we will make the coefficient of any one unknown variable the same in both the equations by multiplying each equation with a certain number. Then we will subtract both the equations and get a new equation in from of one unknown variable. Finally we will get the value of one unknown variable and substitute it in any of the equations to get our desired answer.

Complete step by step solution:
We will solve the given equations by elimination method:
$3x+2y=1$ …..$\left( 1 \right)$
$2x+3y=12$….$\left( 2 \right)$
Now as we have to make the coefficient of one variable same in both the equation we will multiply equation (1) by $2$ as follows:
$\Rightarrow 2\left( 3x+2y \right)=1\times 2$
$\Rightarrow 6x+4y=2$…..$\left( 3 \right)$
Multiply equation (2) by $3$ as follows:
$\Rightarrow 3\left( 2x+3y \right)=12\times 3$
$\Rightarrow 6x+9y=36$…..$\left( 4 \right)$
Subtract equation (4) from equation (3) as follows:
$\Rightarrow \left( 6x+4y \right)-\left( 6x+9y \right)=2-36$
$\Rightarrow \left( 6x-6x \right)+\left( 4y-9y \right)=-34$
Simplifying further we get,
$\Rightarrow 0-5y=-34$
Dividing both sides by $-5$ we get,
$\Rightarrow \dfrac{-5y}{-5}=\dfrac{-34}{-5}$
$\Rightarrow y=\dfrac{34}{5}$
Putting $y=\dfrac{34}{5}$ in equation (3) we get,
$\Rightarrow 6x+4\times \dfrac{34}{5}=2$
$\Rightarrow 6x=2-\dfrac{136}{5}$
Taking LCM on right hand side we get,
$\Rightarrow 6x=\dfrac{10-136}{5}$
$\Rightarrow x=\dfrac{-126}{6\times 5}$
So,
$\Rightarrow x=\dfrac{-21}{5}$
So we get the value as $x=\dfrac{-21}{5}$ and $y=\dfrac{34}{5}$ .
Hence the solution of equation $3x+2y=1$ and $2x+3y=12$ is $x=\dfrac{-21}{5}$ and $y=\dfrac{34}{5}$.

Note:
The systems of equations given are Linear equations in two variables. When the highest degree of each variable is $1$ the equation is known as a linear equation. We can solve two linear equations by using elimination method, substitution method of and cross multiplication method as well. As we have eliminated the variable $y$ in this question it is not necessary to do that we can also eliminate the variable $x$. We just have to keep in mind which variable cost elimination is less and eliminate that variable accordingly.