Question

Solve: 2x +y+4z= 15, x-2y+3z=13, 3x+y-z=2​

Answer 1

Answer:

hope this answer helpful.

Answer 2

The values of x,y and z from the given equations are 2,-1 and 3 respectively

∴ x=2, y=-1 and z=3

Step-by-step explanation:

Given equations are $2x+y+4z=15\hfill (1)$

$x-2y+3z=13\hfill (2)$ and

$3x+y-z=2\hfill (3)$

Now to solve the given equations by elimination method :

Taking equations (1) and (2)

Multiply the equation (1) into 2 we get $4x+2y+8z=30\hfill (4)$

Now adding the equations (2) and (4) we have

$4x+2y+8z=30$

$x-2y+3z=13$

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$5x+11z=43\hfill (5)$

• Now taking the equations (2) and (3)

Multiply the equation (3) into 2 we get

$6x+2y-2z=4\hfill (6)$

Adding the equations (2) and (6)

$x-2y+3z=13$

$6x+2y-2z=4$

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$7x+z=17\hfill (7)$

• Now taking the equations (5) and (7)

Multiply the equation (7) into 11 we get $77x+11z=187\hfill (8)$

Now subtracting the equations (5) and (8) we have

$5x+11z=43$

$77x+11z=187$

_(-)__(-)___(-)_____

$-72x=-144$

$x=\frac{144}{72}$

Therefore x=2

Now substituting the value of x=2 in the equation (5)

$5(2)+11z=43$

$11z=43-10$

$z=\frac{33}{11}$

Therefore z=3

Substitute the values of x=2 and z=3 in the equation (1)

2(2)+y+4(3)=15

4+y+12=15

16+y=15

y=15-16

Therefore y=-1

Therefore the values of x,y and z from the given equations are 2,-1 and 3 respectively

∴ x=2, y=-1 and z=3