To determine the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that the equations
[tex]\[ 2x + 3y = 7 \][/tex]
and
[tex]\[ 2ax + (a + b)y = 28 \][/tex]
represent coincident lines, we need to have these equations proportional to each other. This means that each corresponding coefficient of the first equation must be a multiple of the corresponding coefficient in the second equation, all by the same factor [tex]\( k \)[/tex].
Let's consider the general form of the equations with the factor [tex]\( k \)[/tex]:
1. The first equation is:
[tex]\[ 2x + 3y = 7 \][/tex]
2. The second equation, when written as a line that is coincident with the first, must be of the form:
[tex]\[ 2k \cdot x + 3k \cdot y = 7k \][/tex]
Given the second equation's form:
[tex]\[ 2ax + (a + b)y = 28 \][/tex]
we equate the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and the constants term by term to find [tex]\( k \)[/tex], [tex]\( a \)[/tex], and [tex]\( b \)[/tex].
First, by equating the constants on the right-hand side, we get:
[tex]\[ 28 = 7k \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{28}{7} = 4 \][/tex]
Now, equate the coefficient of [tex]\( x \)[/tex]:
[tex]\[ 2a = 2k \][/tex]
[tex]\[ 2a = 2 \cdot 4 \][/tex]
[tex]\[ 2a = 8 \][/tex]
[tex]\[ a = \frac{8}{2} = 4 \][/tex]
Next, equate the coefficient of [tex]\( y \)[/tex]:
[tex]\[ a + b = 3k \][/tex]
[tex]\[ 4 + b = 3 \cdot 4 \][/tex]
[tex]\[ 4 + b = 12 \][/tex]
[tex]\[ b = 12 - 4 \][/tex]
[tex]\[ b = 8 \][/tex]
Thus, we have:
[tex]\[ a = 4 \][/tex]
[tex]\[ b = 8 \][/tex]
The sum [tex]\( a + b \)[/tex] is therefore:
[tex]\[ a + b = 4 + 8 = 12 \][/tex]
So, the value of [tex]\( a + b \)[/tex] is:
[tex]\[
\boxed{12}
\][/tex]
