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$x+2y-z=11$

${x^2} - 4{y^2} + {z^2} = 37$

$xz=24$

A. x=2,-5; y=2; z=2,-4

B. x=-8,3; y=3; z=-3,8

C. x=-3 5; y=4; z=2, 5

Answer

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Hint: Since we have three equations here, we will take one equation and express it in terms of any other equation and try to solve it.

Complete step-by-step answer:

The equation given to us are

${x^2} - 4{y^2} + {z^2} = 37$----------(i)

$xz=24$---------(ii)

$ \Rightarrow z = \dfrac{{24}}{x}$

$x+2y-z=11$------(iii)

Now, let us consider eq (iii)

x+2y-z=11

This, can be written as

x-z=11-2y

Now, let us square both the sides so that we can express this in terms of eq(i)

So, we get ${(x - z)^2} = {(11 - 2y)^2}$

${x^2} + {z^2} - 2xz = 121 + 4{y^2} - 44y$

${x^2} + {z^2} - 4{y^2} - 2xz = 121 - 44y$

But, already from eq(i), we had ${x^2} - 4{y^2} + {z^2} = 37$

So, let us substitute this value here

So, we get

37-2(24)=121-44y

-44y=-132

From this, we get y=3

To get the value of x , let us substitute this value of y in eq(iii),

So, we get

x+2y-z=11

x+2(3)-z=11

x+6-z=11

From eq (ii), we get $z = \dfrac{{24}}{x}$

So, substituting this value in the above equation ,we get

$x - \dfrac{{24}}{x} = 5$

On solving this further, we get

$

{x^2} - 24 = 5x \\

\Rightarrow {x^2} - 5x - 24 = 0 \\

$

On factorising this, we get

$

{x^2} - 3x + 8x - 24 = 0 \\

\Rightarrow x(x - 3) + 8(x - 3) = 0 \\

\Rightarrow x = 3, - 8 \\

$

Putting these values of x in $z = \dfrac{{24}}{x}$, we get

In the first case, let us consider the value of x=3,so z=$\dfrac{{24}}{3} = 8$

In the second case, let us consider the value of x=-8, so we get z=$\dfrac{{24}}{{ - 8}} = - 3$

So, z=8,-3

So, from this , we can write x=-8,3 ; y=3; z=-3,8

So, option B is the correct answer for this question.

Note: In these type of questions, first try to manipulate a specific equation and try to express in terms of another equation which is given in the data so that we can easily find out the required values(in this case x,y,z) using those equations.

Complete step-by-step answer:

The equation given to us are

${x^2} - 4{y^2} + {z^2} = 37$----------(i)

$xz=24$---------(ii)

$ \Rightarrow z = \dfrac{{24}}{x}$

$x+2y-z=11$------(iii)

Now, let us consider eq (iii)

x+2y-z=11

This, can be written as

x-z=11-2y

Now, let us square both the sides so that we can express this in terms of eq(i)

So, we get ${(x - z)^2} = {(11 - 2y)^2}$

${x^2} + {z^2} - 2xz = 121 + 4{y^2} - 44y$

${x^2} + {z^2} - 4{y^2} - 2xz = 121 - 44y$

But, already from eq(i), we had ${x^2} - 4{y^2} + {z^2} = 37$

So, let us substitute this value here

So, we get

37-2(24)=121-44y

-44y=-132

From this, we get y=3

To get the value of x , let us substitute this value of y in eq(iii),

So, we get

x+2y-z=11

x+2(3)-z=11

x+6-z=11

From eq (ii), we get $z = \dfrac{{24}}{x}$

So, substituting this value in the above equation ,we get

$x - \dfrac{{24}}{x} = 5$

On solving this further, we get

$

{x^2} - 24 = 5x \\

\Rightarrow {x^2} - 5x - 24 = 0 \\

$

On factorising this, we get

$

{x^2} - 3x + 8x - 24 = 0 \\

\Rightarrow x(x - 3) + 8(x - 3) = 0 \\

\Rightarrow x = 3, - 8 \\

$

Putting these values of x in $z = \dfrac{{24}}{x}$, we get

In the first case, let us consider the value of x=3,so z=$\dfrac{{24}}{3} = 8$

In the second case, let us consider the value of x=-8, so we get z=$\dfrac{{24}}{{ - 8}} = - 3$

So, z=8,-3

So, from this , we can write x=-8,3 ; y=3; z=-3,8

So, option B is the correct answer for this question.

Note: In these type of questions, first try to manipulate a specific equation and try to express in terms of another equation which is given in the data so that we can easily find out the required values(in this case x,y,z) using those equations.