Solve the system of equations:  2x - 5y + 3z = -1 x + 4y - 2z = 9 x - 2y - 4z = -5

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve the following equations by elimination

2x-5y+3z=-1 ...(1)

x+4y-2z=9 ...(2)

x-2y-4z=-5 ...(3)

Let's eliminate x and create 2 equations with y and z.

(2) - (3)

=> 4y - 2z + 2y + 4z = 9+5

=> 6y + 2z = 14

=> 3y + z = 7 ...(4)

(1) - 2*(2)

=> 2x - 5y + 3z - 2x - 8y + 4z = -1 - 18

=> -13y + 7z = -19 ...(5)

Now 7*(4) - (5)

=> 21y + 7z + 13y - 7z = 49 + 19

=> 34y = 68

=> y = 2

substitute in (4)

3y + z = 7

=> 6 + z = 7

=> z = 1

substitute y and z in (2)

=> x + 8 - 2 = 9

=> x = 3

Therefore x = 3, y= 2 and z = 1.

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neela | High School Teacher | (Level 3) Valedictorian

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To solve the system of equations:

2x - 5y + 3z = -1...(1)

x + 4y - 2z = 9......(2)

x - 2y - 4z = -5......(3).

(1)-(2)*2 = 2x-5y+3z-2(x+4y-2z) = -1-2*9 = -19.

=> -13y+7z = -19....(4).

(2)-(3): (x+4y-2z) -(x-2y-4z) = 9-(-5) = 14

=> 6y+2z = 14. We divide both sides by  (2):

3y+z = 7...(5):

5*7-(4) = 7(3y+z)-(-13y+7z) = 7*7 -(-19) = 68

34y = 68. So y = 68/34 = 2.

Substitute y = 2 in (5): 3y+z= 7.  Or z = 7-3y = 7-3*2 = 1. So z = 1.

We put y = 2, z= 1 in (1): 2x-5y+3z = 1. Or 2x-5*2+3*1 = -1.

2x = -1+10-3 = 6. Or 2x= 6, So x = 6/2 = 3. 

So x= 3, y = 2 and z = 1.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

It would be much more easier using matrix.

Till then, we'll apply elimination method.

We'll eliminate the unknown x from the 2nd and 3rd equations.

(2) - (3):

x+4y-2z-x+2y+4z=9+5

We'll combine and eliminate like terms:

6y + 2z = 14

We'll divide by 2:

3y + z = 7 (4)

We'll eliminate the unknown x from the 1st and 5th equations. For this reason, we'll multiply (3) by -2:

-2x + 4y + 8z = 10 (5)

(1) + (5):

2x - 5y + 3z - 2x + 4y + 8z = -1 + 10

We'll combine and eliminate like terms:

-y + 11z = 9 (6)

3y + z = 7 (4)

We'll multiply (6) by 3:

-3y + 33z = 27 (7)

(7) + (4):

-3y + 33z + 3y + z = 27 + 7

34z = 34

z = 1

3y + z = 7 <=> 3y + 1 = 7

3y = 7 - 1

3y = 6

y = 2

We'll substitute y and z in the 2nd equation:

x+ 8 - 2  = 9

x + 6 = 9

x = 3

The solution of the system, solved by elimination method is (3 ; 2 ; 1).

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