+0  
 
0
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avatar+564 

I solved this earlier but Im not sure if Im doing it right. Could someone help guide me through this?

 

Solve for y:

 

\(x = 7y\)

\(x+y+z=80\)

\(2x-y+z=105\)

 Mar 2, 2016

Best Answer 

 #1
avatar+2592 
+5

x=7y

x+y+z=80

2x-y+z=105

Isolate and substitute

 

(7y)+y+z=80

8y+z=80

z=80-8y

 

2(7y)+y+(80-8y)=105

14y+y-8y=25

15y-8y=25

7y=25

y=25/7

 Mar 2, 2016
 #1
avatar+2592 
+5
Best Answer

x=7y

x+y+z=80

2x-y+z=105

Isolate and substitute

 

(7y)+y+z=80

8y+z=80

z=80-8y

 

2(7y)+y+(80-8y)=105

14y+y-8y=25

15y-8y=25

7y=25

y=25/7

SpawnofAngel Mar 2, 2016
 #2
avatar+98128 
+5

x = 7y

x + y + z = 80

2x - y + z = 105  →  z  = 105 - 2x + y     (3)

 

Subbing the first equation into the second, we have

 

7y + y + z  = 80   →  8y + z = 80      (4)

 

And subbing the first into (3) we have

 

z = 105 - 2(7y) + y

 

z = 105 - 13y      (5)

 

Put (5)  into (4)

 

8y + 105 - 13y  = 80

 

-5y  = -25    →  y =  5

 

And  z = 105 - 13(5)  = 105 - 65  = 40

 

And x = 7y = 7(5)  = 35

 

So (x, y, z)  = (35, 5, 40)

 

 

 

cool cool cool

 Mar 2, 2016
 #3
avatar+564 
0

Thanks both of you!

 Mar 2, 2016

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