# Solve the system of equations given below,

\[x+y+z=11\],

\[2x-6y-z=0\],

\[3x+4y+2z=0\].

Answer

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Hint: Firstly do the adjustments in the equation \[x+y+z=11\] and eliminate ‘z’ from other two equations with the help of it. After elimination you will get the system of equations with two variables which can be solved by eliminating one of its variables so that you can get the value of one variable. And solve there after

Complete step-by-step answer:

As they have not mentioned any particular method to solve the equations therefore we will solve the system of equations by elimination method and for that we will first rewrite the equations given in the problem,

\[x+y+z=11\] ……………………………………………. (1)

\[2x-6y-z=0\]………………………………………… (2)

\[3x+4y+2z=0\]……………………………………….. (3)

If we observe the above equations then we can say that in equation (1) the adjustments can be easily done so that we can eliminate ‘z’ from other two equations.

Therefore we are going to eliminate ‘z’ from the equations to get the solution,

To eliminate ‘z’ from equation (2) we will add equation (1) in equation (2) therefore we will get,

\[\begin{align}

& \text{ }x+y+z=11 \\

& +\underline{2x-6y-z=0} \\

& \text{ }3x-5y+0=11 \\

& \Rightarrow 3x-5y=11...............(4) \\

\end{align}\]

Also to eliminate ‘z’ from equation (3) we will first multiply equation (1) by 2 therefore we will get,

\[\therefore 2\times \left( x+y+z \right)=2\times \left( 11 \right)\]

\[\therefore 2x+2y+2z=22\] ………………………………… (5)

We will now subtract equation (5) from equation (3) therefore we will get,

\[\begin{align}

& \text{ }3x+4y+2z=0 \\

& \underline{-\left( 2x+2y+2z=22 \right)} \\

& \text{ }x+2y+0=-22 \\

& \Rightarrow x+2y=-22.............(6) \\

\end{align}\]

Equation (4) and (6) are now reduced in the form of system of equations with two variables and we can easily solve them by doing simple operations,

We will multiply equation (6) by 3 so that we can eliminate ‘x’ easily,

\[\therefore 3\times \left( x+2y \right)=3\times \left( -22 \right)\]

\[\therefore 3x+6y=-66\] ……………………………………………… (7)

Now, we will subtract equation (4) from equation (7) therefore we will get,

\[\begin{align}

& \text{ }3x+6y=-66 \\

& \underline{-\left( 3x-5y=11 \right)} \\

& \text{ }0+11y=-77 \\

& \Rightarrow 11y=-77 \\

& \Rightarrow y=\dfrac{-77}{11}=-7 \\

\end{align}\]

Now if we put the value of ‘y’ in equation (6) we will get,

\[\therefore x+2\times \left( -7 \right)=-22\]

\[\therefore x-14=-22\]

\[\therefore x=-22+14\]

\[\therefore x=-8\]

Now we will put the value of ‘x’ and ‘y’ in equation (1) we will get,

\[x+y+z=11\]

\[\therefore -8-7+z=11\]

\[\therefore -15+z=11\]

\[\therefore z=11+15\]

\[\therefore z=26\]

The solution of the given system of equations is, x = -8, y = -7, and z = 26.

Note: Always use the equation like \[x+y+z=11\] so that we can do the adjustments easily which ultimately saves our time, also if you have practice and want a quick answer then you can use Cramer’s Rule also. In problems like this we should recheck our answer by substituting the values in equations which we haven’t used for calculating them.

Complete step-by-step answer:

As they have not mentioned any particular method to solve the equations therefore we will solve the system of equations by elimination method and for that we will first rewrite the equations given in the problem,

\[x+y+z=11\] ……………………………………………. (1)

\[2x-6y-z=0\]………………………………………… (2)

\[3x+4y+2z=0\]……………………………………….. (3)

If we observe the above equations then we can say that in equation (1) the adjustments can be easily done so that we can eliminate ‘z’ from other two equations.

Therefore we are going to eliminate ‘z’ from the equations to get the solution,

To eliminate ‘z’ from equation (2) we will add equation (1) in equation (2) therefore we will get,

\[\begin{align}

& \text{ }x+y+z=11 \\

& +\underline{2x-6y-z=0} \\

& \text{ }3x-5y+0=11 \\

& \Rightarrow 3x-5y=11...............(4) \\

\end{align}\]

Also to eliminate ‘z’ from equation (3) we will first multiply equation (1) by 2 therefore we will get,

\[\therefore 2\times \left( x+y+z \right)=2\times \left( 11 \right)\]

\[\therefore 2x+2y+2z=22\] ………………………………… (5)

We will now subtract equation (5) from equation (3) therefore we will get,

\[\begin{align}

& \text{ }3x+4y+2z=0 \\

& \underline{-\left( 2x+2y+2z=22 \right)} \\

& \text{ }x+2y+0=-22 \\

& \Rightarrow x+2y=-22.............(6) \\

\end{align}\]

Equation (4) and (6) are now reduced in the form of system of equations with two variables and we can easily solve them by doing simple operations,

We will multiply equation (6) by 3 so that we can eliminate ‘x’ easily,

\[\therefore 3\times \left( x+2y \right)=3\times \left( -22 \right)\]

\[\therefore 3x+6y=-66\] ……………………………………………… (7)

Now, we will subtract equation (4) from equation (7) therefore we will get,

\[\begin{align}

& \text{ }3x+6y=-66 \\

& \underline{-\left( 3x-5y=11 \right)} \\

& \text{ }0+11y=-77 \\

& \Rightarrow 11y=-77 \\

& \Rightarrow y=\dfrac{-77}{11}=-7 \\

\end{align}\]

Now if we put the value of ‘y’ in equation (6) we will get,

\[\therefore x+2\times \left( -7 \right)=-22\]

\[\therefore x-14=-22\]

\[\therefore x=-22+14\]

\[\therefore x=-8\]

Now we will put the value of ‘x’ and ‘y’ in equation (1) we will get,

\[x+y+z=11\]

\[\therefore -8-7+z=11\]

\[\therefore -15+z=11\]

\[\therefore z=11+15\]

\[\therefore z=26\]

The solution of the given system of equations is, x = -8, y = -7, and z = 26.

Note: Always use the equation like \[x+y+z=11\] so that we can do the adjustments easily which ultimately saves our time, also if you have practice and want a quick answer then you can use Cramer’s Rule also. In problems like this we should recheck our answer by substituting the values in equations which we haven’t used for calculating them.

Last updated date: 26th Sep 2023

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Total views: 362.1k

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