Question

# Solve the following system of equations by using the method of subtraction:\begin{align} & x+2y=-1 \\ & 2x-3y=12 \\ \end{align}

Hint: From the 2 equations, choose any 1 equation. We need to eliminate either x or y to solve the equations. Modify one equation in terms of x and y and substitute in the 2nd equation of the chosen variable. Solve it and find the value of x and y.

We have been given 2 equations,
\begin{align} & x+2y=-1-(1) \\ & 2x-3y=12-(2) \\ \end{align}
Let us work to solve these 2 equations.
The tricky thing is that there are 2 variables x and y. We need to get rid of one of the variables. So, we can find the value of x and y.
Let us consider equation (1),$x+2y=-1$
Take 2y to the LHS, the equation becomes,
\begin{align} & x=-1-2y \\ & x=-\left( 2y+1 \right)-(3) \\ \end{align}
Let us mark the equation (3). Now we got an equation which can be substituted in equation (2) in the place of x. Thus we get rid of the variable â€˜xâ€™.
Now substitute the value of x in equation (2).
\begin{align} & 2x-3y=12 \\ & -2\left( 2y+1 \right)-3y=12 \\ \end{align}
Now we got an equation, with only the variable â€˜yâ€™. Simplify the above expression to get the value of y.
\begin{align} & -4y-2-3y=12 \\ & -7y=14 \\ & \Rightarrow y=\dfrac{14}{-7}=-2 \\ \end{align}
Hence, we got the value of y = -2.
Substitute this value in equation (3), to get the value of â€˜xâ€™.
\begin{align} & x=-2y-1 \\ & x=-2\left( -2 \right)-1=4-1=3 \\ \end{align}
Hence, we got x = 3 and y = -2.
So the solution to the system of equations is (3, -2).

Note: Itâ€™s a good idea to always check the results we get. Substitute (3, -2) in equation (1) and equation (2) and check if you get the answer equal to the LHS of the equation.
\begin{align} & x+2y=3+2\times \left( -2 \right)=3-4=-1 \\ & \therefore x+2y=-1 \\ \end{align}
Similarly, check the 2nd equation,
\begin{align} & 2x-3y=2\times 3-3\times \left( -2 \right) \\ & 2x-3y=6+6=12 \\ & \therefore 2x-3y=12 \\ \end{align}
Hence, we got the same value as that of LHS. So the solution is correct.