The values of x, y and z for the system of equations $x + 2y + 3z = 6,3x - 2y + z = 2$ and $4x + 2y + z = 7$are respectively
A. $1,1,1$
B. $1,2,3$
C. $1,3,2$
D. $2,3,1$
Last updated date: 29th Mar 2023
•
Total views: 310.5k
•
Views today: 7.87k
Answer
310.5k+ views
Hint: Eliminate one variable and make an equation in two variables then solve them to get an answer.
We have been given three equations in three variables and by seeing options that a solution exists of them
So, we will eliminate y in the above equations to get two equations in x and z.
Let
$x + 2y + 3z = 6$ -Equation (1)
$3x - 2y + z = 2$ -Equation (2)
$4x + 2y + z = 7$ -Equation (3)
When we add Equation (1) and Equation (2), we get
$4x + 4z = 8$
$ \Rightarrow x + z = 2$ -Equation (4)
Now, if we add Equation (2) and Equation (3), we get
$7x + 2z = 9$ -Equation (5)
Now when we multiply Equation (4) with 2, we get
$2x + 2z = 4$ -Equation (6)
Subtracting Equation (6) from Equation (5), we get
$5x = 5$$ \Rightarrow x = 1$
Putting this value of x in Equation (6), we get
$2 + 2z = 4$$ \Rightarrow z = 1$
Using $x = 1$ and $z = 1$ in Equation (1), we get
$1 + 2y + 3 = 6$$ \Rightarrow y = 1$
So Option A is correct.
Note: In this question firstly we eliminate the variable y as it is easiest to eliminate it first, then we get two equations in two variables, solving which we get values of x and z. Now, when we put values of x and z in any given equation we also get y. We could have also solved this question by using options as it is clearly visible that the equations satisfy the first option which is a faster method to use in an exam.
We have been given three equations in three variables and by seeing options that a solution exists of them
So, we will eliminate y in the above equations to get two equations in x and z.
Let
$x + 2y + 3z = 6$ -Equation (1)
$3x - 2y + z = 2$ -Equation (2)
$4x + 2y + z = 7$ -Equation (3)
When we add Equation (1) and Equation (2), we get
$4x + 4z = 8$
$ \Rightarrow x + z = 2$ -Equation (4)
Now, if we add Equation (2) and Equation (3), we get
$7x + 2z = 9$ -Equation (5)
Now when we multiply Equation (4) with 2, we get
$2x + 2z = 4$ -Equation (6)
Subtracting Equation (6) from Equation (5), we get
$5x = 5$$ \Rightarrow x = 1$
Putting this value of x in Equation (6), we get
$2 + 2z = 4$$ \Rightarrow z = 1$
Using $x = 1$ and $z = 1$ in Equation (1), we get
$1 + 2y + 3 = 6$$ \Rightarrow y = 1$
So Option A is correct.
Note: In this question firstly we eliminate the variable y as it is easiest to eliminate it first, then we get two equations in two variables, solving which we get values of x and z. Now, when we put values of x and z in any given equation we also get y. We could have also solved this question by using options as it is clearly visible that the equations satisfy the first option which is a faster method to use in an exam.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths 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

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
