Answer:
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 =[tex]\frac{16}{4}[/tex]
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
Answer:
(4,4,0) is solution of given systems of equation.
Step-by-step 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).
