Question

Let a and b be natural numbers. If \(a^2+ab+a=14\) and \(b^2+ab+b=28\), then (2a+b) equals

• A. 7
• B. 10
• C. 9
• D. 8

Answer 1

The Correct Option is D

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.

Answer 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, \(2a+b = 2(2) + 4 = 8\)