2a+3b=c (1)
a+b+c=42 (2)
c-a=19 (3)
let's start by (3), we can say that a-c=-19 (The reverse) by just multiplying by -1
so let a-c=-19 (4)
Add (2) and (4)
a+b+c=42
a-c=-19
indeed,
2a+b=23
Relating it to (1) looks very similar, however, some steps are required.
2a+3b=c
2a+b=23
Let's before continuing this make some statements about c.
from (3)
we can say that
c=19+a
back to (1)
2a+3b=c, we can say 2a+3b=19+a
simplify this by subtracting an a
a+3b=19
Back to the system of equations,
a+3b=19
2a+b=23
Let's multiply the second equation by -3 so the b's cancel
a+3b=19
-6a-3b=-69
Again back to the system of equation
a+3b=19
-6a-3b=-69
let's add both of them
-5a= -50
divide by -5
a=10
Well we know that from (3) that c-a=19
so c-10=19
c=29
From (2) we know that a+b+c=42
so 10+b+29=42
39+b=42
b=3
Now we got the values of a,b,c
a=10
b=3
c=29
The question says find (c+1)/(ab)
(29+1)/10*3=30/30=1
Answer is =1
2a + 3b = c ⇒ 2a + 3b - c = 0 (1)
a + b + c = 42 ⇒ -3a -3b - 3c = -126 (2)
c - a = 19 (3)
Add (1) and (2)
-a - 4c = -126 ⇒ a + 4c = 126 (4)
Add (3) and (4)
5c = 145
c = 29
c - a = 19
29 - a = 19
a = 10
a + b + c =42
10 + b + 29 = 42
b = 3
So
(c + 1) / [ ab ] = (29 + 1) / [ 10 * 3 ] = 30 / 30 = 1