7
10
9
8
To resolve the problem, we must determine the values of \(a\) and \(b\) that fulfill the provided equations and subsequently calculate \(2a+b\).
The given equations are:
\(a^2+ab+a=14\) (Equation 1)
\(b^2+ab+b=28\) (Equation 2)
Rearranging Equation 1 yields:
\[a(a+b+1)=14\]
Given that \(a\) is a natural number, and the factors of 14 are 1, 2, 7, and 14, we examine each possibility:
1. If \(a=1\), then \(1(b+2)=14\), implying \(b+2=14\), thus \(b=12\).
2. If \(a=2\), then \(2(b+3)=14\), implying \(b+3=7\), thus \(b=4\).
3. If \(a=7\), then \(7(b+8)=14\), which is not feasible for a natural number \(b\).
4. If \(a=14\), then \(14(b+15)=14\), which is not feasible as \(b+15\) cannot equal 1.
We now verify these potential solutions by substituting them into Equation 2:
Substituting \(a=2\) and \(b=4\) into Equation 2:
\[b^2+ab+b=28\]
\[4^2+2\times4+4=16+8+4=28\]
This pair of values satisfies both equations; therefore, \(a=2\) and \(b=4\).
Finally, we compute \(2a+b\):
\[2a+b=2(2)+4=4+4=8\]
The result is 8.