Solve using elimination. x+4y+3z=3 2x-5y-z=5 3x+2y-2z=-3

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to decide upon the variable you want to eliminate. Supposing that you want to eliminate the variable z you need to consider the first and the second equations and you need to do the following step. You should multiply the second equation by 3 and then you need to add it to the first equation such that:

`3*(2x-5y-z) + x+4y+3z = 3*5 + 3`

Opening the brackets yields:

`6x - 15y - 3z + x + 4y + 3z = 15 + 3`

`7x - 11y = 18`

You need to eliminate the same variable z, hence either you consider the first and third equations, or the second and third equations.

Supposing that you select the second and the third equations, you need to multiply the second equation by -2 and then you need to add it to the third equation such that:

`-2*(2x-5y-z) + 3x+2y-2z = -2*5 - 3`

Opening the brackets yields:

`-4x + 10y + 2z + 3x + 2y - 2z = -13`

`-x + 12y = -13`

You need to consider the new equations `7x - 11y = 18 ` and `-x + 12y = -13`  and you should decide upon the variable you want to eliminate. Supposing that you eliminate x, you need to multiply by 7 the equation`x + 12y = -13`  such that:

`7(-x + 12y )= -7*13`

`-7x + 84y = -91`

You need to add this equation to `7x - 11y = 18`  such that:

`7x - 11y - 7x + 84y = 18-91`

`73y=-73 =gt y = -1`

You need to substitute -1 for y in equation `7x - 11y = 18`  such that:

`7x + 11 = 18 =gt 7x = 18-11 =gt 7x = 7`

`x = 1`

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

`1 - 4 + 3z = 3 =gt -3 + 3z = 3 =gt -1 + z = 1 =gt z = 2`

Hence, the solution to simultaneous equations is `x=1 , y=-1 , z=2.`

Top Answer

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The following system of equations has to be solved using elimination.

x+4y+3z=3 ...(1)

2x-5y-z=5 ...(2)

3x+2y-2z=-3 ...(3)

(2) - 2*(1)

=> -5y - 8y - z - 6z = 5 - 6

=> 13y + 7z = 1 ...(4)

(3) - 3*(1)

=> 3x - 3x + 2y - 12y - 2z - 9z = -3 - 9

=> 10y + 11z = 12 ...(5)

10*(4) - 13*(5)

=> 130y + 70z - 130y - 143z = -146

=> -73z = -146

=> z = 2

11*(4) - 7*(5)

143y + 77z - 70y - 77z = 11 - 84

=> 73y = -73

=> y = -1

The value for y can also be arrived at by elimination but it is easier to substitute z = 2 and y = 1 in x+4y+3z=3

=> x -4 + 6 = 3

=> x = 1

The solution of the set of equations is x = 1, y = -1 and z = 2

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