Answer

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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.

Complete step-by-step answer:

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 ‘x’.

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 ‘y’. 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 ‘x’.

\[\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: It’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 2

\[\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.

Complete step-by-step answer:

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 ‘x’.

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 ‘y’. 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 ‘x’.

\[\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: It’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 2

^{nd}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.

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