Question

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

Answer 1

(4,4,0) is solution of given systems of equation.

Explanation:

We have given a system of equations.

2x+3y+z = 0 eq(1)

3y+5z = 12 eq(2)

y-5z = 4 eq(3)

Adding 5z to both sides of eq(3) ,we get

y-5z+5z = 4+5z

y+0 = 4+5z

y = 4+5z eq(4)

Putting above value of y in eq (2), we get

3(4+5z)+5z = 12

12+15z+5z = 12

Adding -12 to both sides of above equation,we get

-12+12+15z+5z = -12+12

Adding like terms,we get

0+20z = 0

20z = 0

z = 0

putting the value of z in eq(4), we get

y = 4+5(0)

y = 4+0

y = 4

Putting the value of y and z in eq(1), we get

2x+3(4)+0 = 20

2x +12 = 20

adding -12 to both sides of above equation,we get

2x+12-12 = 20-12

2x = 8

dividing by 2, we get

2x/2 = 8 / 2

x = 4

Hence, the solution is (4,4,0).

Answer 2

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

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 =
`(16)/(4)`

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