Find the solution to the system of equations 2x+3y+z=20 3y+5z=12 y-5z=4

x =4 ,y =4 and z =0 is the solution.

Step-by-step explanation:

We shall solve the equations using elimination  method

In equation 2 and 3 we can see that  coefficient of z in both equations  are same with opposite sides , adding equations (2) and (3)

3y+5z = 12

y -5z = 4     adding the equations

___________

4y      = 16

dividing both sides by 4

  y =

 y = 4

Plugging the value of y in equation 2 or equation 3 ,we get

3(4) +5z = 12

12+5z = 12

  5z = 12-12 or 5z =0 gives z =0

plugging y =4 and z =0 in equation 1

2x+3(4)+ 0 = 20

2x+12 = 20

2x =20-12

2x = 8

x = 8 divided by 2 gives

x =4

therefore solution of the system is given by

x=4 , y =4 and z =0

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