# 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. $2,3,1$

Answer

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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.

Last updated date: 19th Sep 2023

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