Question

x+y+z=5;2x-y+z=9;x-2y+3z=16​

Answer 1

i) x + y + z = 5 ; 2x − y + z = 9 ; x − 2y + 3z = 16

Solution :

x + y + z = 5 -------(1)

2x − y + z = 9 -------(2)

x − 2y + 3z = 16 -------(3)

Let us add (1) and (2)


Multiply the first equation by 2 and add by (3)


Let 3x + 2z = 14 ----(4) and 3x + 5z = 26 ---(5)

(4) - (5)

3x + 2z = 14

3x + 5z = 26

(-) (-) (-)

----------------

-3z = -12 ==> z = 4

By applying z = 4 in (4), we get

3x + 2(4) = 14

3x + 8 = 12

3x = 14 - 8

3x = 6 ==> x = 2

By applying x 2 and z = 4 in (1), we get

2 + y + 4 = 5

6 + y = 5

y = 5 - 6

y = -1

Hence the required solution is x = 2, y = -1 and z = 2.

Answer 2

Step-by-step explanation:

$hello \: user......... \\ we \: will \: solve \: this \: problem \: \\ using \: cramers \: rule = >$

$x + y + z = 5........(1) \\ 2x - y + z = 9.......(2) \\ x - 2y + 3z = 16.....(3)$

$according \: to \: cramers \: rule = >$

D = | 1 1 1|

| 2 -1 1|

| 1 -2 3|

$= 1( - 3 + 2) \: - 1(6 - 1) \: + 1( - 4 + 1)$

$= \: - 1 - 5 - 3 \\ = \: - 9$

Dx = | 1 1 5|

| -1 1 9|

| -2 3 16|

$= 1(16 - 27) - 1( - 16 + 18) + 5( - 3 + 2)$

$= \: - 11 \: - 2 \: - 5 \\ = - 18$

Dy = | 1 1 5|

| 2 1 9|

| 1 3 16|

$= \: 1(16 - 27) \: - 1(32 - 9) \: + 5(5)$

$= \: - 11 \: - 23 \: + 25$

$= - 9$

Dz = | 1 1 5|

| 2 -1 9|

| 1 -2 16|

$= \: 1( - 16 + 18) \: - 1(32 - 9) \: + 5( - 4 + 1)$

$= \: 2 \: - 23 \: - 15$

$= - 36$

$we \: know \: that \: = > \\ x \: = \frac{dx}{d} = \frac{ - 18}{ - 9} = 2$

$y = \frac{dy}{d} = \frac{ - 9}{ - 9} = 1$

$z = \frac{dz}{d} = \frac{ - 36}{ - 9} = 4$

$so \: we \: finally \: got \: solution \: of \: \\ following \: equations \: as \\ x = 2 \\ y = 1 \\ z = 4$

$hope \: it \: helps \: uh.....$