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# help

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If 2a + 3b = c, a + b + c = 42, c - a = 19, find (c + 1)/(ab).

Dec 26, 2019

#1
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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)

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

-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

Dec 26, 2019
#2
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2a + 3b  = c   ⇒        2a + 3b  - c  =  0          (1)

a + b + c   =  42  ⇒   -3a -3b - 3c =  -126        (2)

c - a  = 19     (3)

-a - 4c  = -126       ⇒   a + 4c =  126    (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   Dec 27, 2019