Question 1080935
Twice one integer plus 3 times a second integer equals 9.
2a + 3b = 9
 Five times the first iteger plus 4 times the second integer equals 5.
5a + 4b = 5
5a = -4b + 5
divide by 5
a = {{{-4/5}}}b + 1
In the first equation, replace a with {{{-4/5}}}b + 1
2({{{-4/5}}}b + 1) + 3b = 9
{{{-8/5}}}b + 2 + 3b = 9
{{{-8/5}}}b + 3b = 9 - 2 
{{{-8/5}}}b + 3b = 7
multiply by 5
-8b + 15b = 35
7b = 35
b = 35/7
b = 5
Find a
a = {{{-4/5}}}(5) + 1
a = -4 + 1
a = -3
:
;
See if that checks out in the first equation
2(-3) + 3(5) = 
-6 + 15 = 9
:
 What are the numbers? -3 and 5