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sjsmith21
11.12.2019 •
Mathematics
Find the solution to the system of equations 2x+3y+z=20 3y+5z=12 y-5z=4
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Ответ:
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 =![\frac{16}{4}](/tpl/images/0412/7075/0d471.png)
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
Ответ:
40 I think
Step-by-step explanation:
75+40x=40 I think