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If the system of equations

\(\begin{align*} 3x+y&=a,\\ 2x+5y&=2a, \end{align*}\)

has a solution\((x,y)\)when\(x=2\), 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+4077 
+1

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+4077 
+1
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+99237 
0

Excellent, tertre  !!!

 

cool cool cool

CPhill  Mar 17, 2019
 #3
avatar+27653 
+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

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