Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
+1
842
3
avatar+5 

If the system of equations

3x+y=a,2x+5y=2a,

has a solution(x,y)whenx=2, compute a

 Mar 17, 2019
edited by ilikepizza  Mar 17, 2019
edited by ilikepizza  Mar 17, 2019
edited by ilikepizza  Mar 17, 2019
edited by ilikepizza  Mar 17, 2019

Best Answer 

 #1
avatar+4624 
+2

First, we plug the value of 2 for x.

 

This leads to 3(2)+y=a and 2(2)+5y=2a 

or 6+y=a and 4+5y=2a

 

Rearranging, we have a-y=6 and 2a-5y=4

Multiplying the first expression by 5, we get 5a-5y=30 and 2a-5y=4.

3a=26 and a=26/3.

 

smileysmiley

 Mar 17, 2019
edited by tertre  Mar 18, 2019
 #1
avatar+4624 
+2
Best Answer

First, we plug the value of 2 for x.

 

This leads to 3(2)+y=a and 2(2)+5y=2a 

or 6+y=a and 4+5y=2a

 

Rearranging, we have a-y=6 and 2a-5y=4

Multiplying the first expression by 5, we get 5a-5y=30 and 2a-5y=4.

3a=26 and a=26/3.

 

smileysmiley

tertre Mar 17, 2019
edited by tertre  Mar 18, 2019
 #2
avatar+130466 
0

Excellent, tertre  !!!

 

cool cool cool

CPhill  Mar 17, 2019
 #3
avatar+33654 
+1

We can immediately multiply the first equation by 5, so we have the two equations:

 

15x + 5y = 5a

2x + 5y = 2a

 

Subtract the second from the first:

13x = 3a

 

Rearrange and divide by 3:

a = (13/3)*x

 

Substiute 2 for x:

a = (13/3)*2  or a = 26/3

 

(tertre added when he/she should have subtracted).

 Mar 17, 2019
edited by Alan  Mar 17, 2019

0 Online Users