Let a and b be natural numbers. If \(a^2+ab+a=14\) and \(b^2+ab+b=28\), then (2a+b) equals
7
10
9
8
The Correct Option is D
Approach Solution - 1
To solve the problem, we need to find the values of \(a\) and \(b\) that satisfy the given equations and then compute \(2a+b\).
We have the following two equations:
\(a^2+ab+a=14\) (Equation 1)
\(b^2+ab+b=28\) (Equation 2)
Let's rewrite Equation 1:
\[a(a+b+1)=14\]
Since \(a\) is a natural number and the factors of 14 are 1, 2, 7, and 14, we will test these values:
1. If \(a=1\), then the equation becomes \(1(b+2)=14\) which gives \(b+2=14\), so \(b=12\).
2. If \(a=2\), then the equation becomes \(2(b+3)=14\) which gives \(b+3=7\), so \(b=4\).
3. If \(a=7\), then the equation becomes \(7(b+8)=14\), which is not possible since \(b\) must be a natural number.
4. If \(a=14\), then the equation becomes \(14(b+15)=14\), which is not possible as \(b+15\) cannot be 1.
Now, let's check these values by substituting into Equation 2:
Substitute \(a=2, b=4\) into Equation 2:
\[b^2+ab+b=28\]
\[4^2+2\times4+4=16+8+4=28\]
This solution satisfies both equations, so \(a=2\) and \(b=4\).
Finally, calculate \(2a+b\):
\[2a+b=2(2)+4=4+4=8\]
The answer is 8.
Approach Solution -2
\(a^2 + ab + a = 14\) — (1)
\(b^2 + ab + b = 28\) ------(2)
Eq 2 – Eq 1
\(b^2 - a^2 + b – a = 14\)
\((b-a)(b+a) + (b-a) = 14\)
\((b-a)(b+a+1) = 14\)
Now, \(14 = 1×14\ or \ 2×7\)
But \(a < 4\) since \(a^2 < 14\)
And \(b < 5\) since \(b^2 + b < 28\)
So,\( a+b+1 < 4+5+1 = 10\)
So, \(a+b+1 = 7\) and \(b-a = 2\)
Solving, \(b = 4\) and \(a = 2\)
So, \(2a+b = 2(2) + 4 = 8\)
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