Solve the systems algebraically:X+4Y+3Z=10 4X+2Y-2Z=-2 3X-Y+Z=11

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to decide what variable you wish to eliminate first. Selecting the variable z the one to be eliminated first, you need to consider the first and the second equations such that:

`{(x+4y+3z = 10),(4x+2y-2z = -2):}`

Multiplying the first equation by 2 and the second equation by 3 yields:

`{(2x+8y+6z = 20),(12x+6y-6z = -6):}`

Adding the equations yields:

`14x + 14y = 14 => x + y = 1`

Considering the second and the third equations yields:

`{(4x + 2y - 2z = -2),(3x - y + z = 11) :}`

Multiplying the third equation by 2 yields:

`{(4x + 2y - 2z = -2),(6x - 2y + 2z = 22):}`

Adding the equations yields:

`10x = 20 => x = 2`

Substsituting 2 for x in equation `x + y = 1`  yields:

`2+ y = 1 => y = -1`

You need to substitute 2 for x and -1 for y in `x+4y+3z = 10`  such that:

`2-4+3z = 10 => 3z = 10+2 => 3z = 12 => z = 4`

Hence, evaluating the solutions to the given system of equations yields `x = 2 , y= -1 , z = 4` .

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najm1947 | Elementary School Teacher | (Level 1) Valedictorian

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X+4Y+3Z=10 ...............(1)

4X+2Y-2Z=-2 ..............(2)

3X-Y+Z=11 .................(3)

multiplying equation (3) with 3 we get:

 9X-3Y+3Z = 33, now subtracting eaquation (1) from this we get:

9X-X -3Y-4Y +3Z-3Z = 33-10

=> 8X -7Y = 23 ............(4)

again multiplying equation (3) with 2 we get:

6X-2Y+2Z = 22, adding equation (2) to it we get:

6X+4X -2Y+2Y +2Z-2Z = 22-2

=> 10X = 20

=> X = 2

substituting this value of X in (4) we get:

8*2 -7Y = 23

=> -7Y = 23-16 = 7

=> Y = -1

now substituting the value of X and Y in equation (3) we get:

3*2 -(-1) +Z = 11

=> Z = 11-7 = 4

The solution of the three equations is:

X = 2

Y = -1

Z = 4

 

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