In the system of equations 1-x-1-y-5-6- 1-y-1-z-7-12 and 1-z-1-x-3-4- values of x- y and z will be

The correct option is C 2, 3 and 4Given equations are 1/x+1/y=5/6 6x+6y=5xy nbsp; nbsp; nbsp;Equation no1 1/y+1/z=7/12 12z+12y=7zy ...

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Standard X

Mathematics

Elimination Method to Find the Solution of Pair of Linear Equations

In the system...

Question

In the system of equations $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}, \frac{1}{y} + \frac{1}{z} = \frac{7}{12}$ and $\frac{1}{z} + \frac{1}{x} = \frac{3}{4}$ , values of x, y and z will be

4, 3 and 2

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3, 2 and 4

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2, 3 and 4

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3, 4 and 2

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Solution

The correct option is C 2, 3 and 4
Given equations are
$\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$
$6 x + 6 y = 5 x y$ Equation no1
$\frac{1}{y} + \frac{1}{z} = \frac{7}{12}$
$12 z + 12 y = 7 z y$ Equation No 2
$\frac{1}{z} + \frac{1}{x} = \frac{3}{4}$
$4 z + 4 x = 3 z x$ Equation No 3
From equation 3
$z = \frac{4 x}{(3 x - 4)}$ Equation no 4
Substituting the value of z in equation 2, we get
$12 y + 12 (\frac{4 x}{3 x - 4}) = 7 y \times (\frac{4 x}{3 x - 4})$
$36 x y - 48 y + 48 x = 28 x y$
$48 x - 48 y = 28 x y - 36 x y$
$6 x - 6 y = - x y$ Equation No 5
Solving equation 1 and 5 simultaneously
$12 x = 4 x y$
$y = 3$
Substituting the value of y in Equation 1
$6 x + 18 = 15 x$
$9 x = 18$
$x = 2$
Substituting the value of x in Equation 4
z=4
Value of x=2, y=3 and z=4
Answer is Option C

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Similar questions

Q. In the system of the equations

\frac{1}{x} + \frac{1}{y} = \frac{5}{6}, \frac{1}{y} + \frac{1}{z} = \frac{7}{12}

and

\frac{1}{z} + \frac{1}{x} = \frac{3}{4}

the values of x, y and z are correctly denoted in which of the following options?

Q. (i) If

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii)

A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]

, find A⁻¹ and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii)

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]

, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If

A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given

A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]

, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi) If

A = (\begin{array}{rrr} 2 & 3 & 1 \\ 1 & 2 & 2 \\ - 3 & 1 & - 1 \end{array})

, find A^–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.
(vii) Use product

[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}] [\begin{matrix} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{matrix}]

to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

Q. In the system of equation

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 6, \frac{2}{x} - \frac{3}{y} + \frac{4}{z} = 8

and

\frac{3}{x} - \frac{4}{y} + \frac{5}{z} = 10

, values of

x

y

and

z

which satisfy the above three equations, will be

Q. Find the angle between the following pairs of lines:
(i)

\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

(ii)

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

(iii)

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

(iv)

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

(v)

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

(vi)

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

Q. Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii)

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

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In the system of equations 1x+1y=56,1y+1z=712 and 1z+1x=34, values of x, y and z will be

In the system of equations $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}, \frac{1}{y} + \frac{1}{z} = \frac{7}{12}$ and $\frac{1}{z} + \frac{1}{x} = \frac{3}{4}$ , values of x, y and z will be