# Solve the system of equations given below,

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

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

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

Last updated date: 28th Mar 2023

•

Total views: 307.8k

•

Views today: 4.84k

Answer

Verified

307.8k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE