# Solve the following equations and choose the correct answer from the below given options:

$x+2y-z=11$

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

$xz=24$

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

Verified

327k+ views

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.

Last updated date: 03rd Jun 2023

â€¢

Total views: 327k

â€¢

Views today: 5.84k

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