+0

+1
198
3

# 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

#1
+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.  Mar 17, 2019
edited by tertre  Mar 18, 2019

#1
+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.  tertre Mar 17, 2019
edited by tertre  Mar 18, 2019
#2
0

Excellent, tertre  !!!   CPhill  Mar 17, 2019
#3
+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