Moving Along A Linear Function

Let the linear function is $f(x)=ax+b$ , so we have

$f(2013)-f(2001)=12a=100$

$a=\frac{100}{12}$

Now, $f(2031)-f(2013)=18a=150$

(This an answer that goes along with an understanding of Straight lines, rather than linear function) It is mentioned, the F(), is a liner function, which basically means, its a straight line when plotted o a graph.

So, crucially, the slope is constant at all points of this slope.

$Here,\quad we\quad will\quad assume\quad { y }_{ 1 }=f\left( 2001 \right) \quad and\quad { y }_{ 2 }=f\left( 2013 \right) \\ \\ Also,\quad { x }_{ 1 }=2001\quad and\quad { x }_{ 2 }=2013\\ \\ \\ f\left( 2013 \right) -f\left( 2001 \right) =100\\ \\ { y }_{ 2 }-{ y }_{ 1 }=100\\ \\ Now,\quad to\quad the\quad slope\quad formula\\ \\ \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =slope\\ \\ \frac { 100 }{ 2013-2001 } =slope\\ \\ \frac { 100 }{ 12 } =slope.\\ \\ Now,\quad the\quad second\quad half\quad of\quad the\quad question,\quad \\ { y }_{ 1 }=f\left( 2013 \right) \quad and\quad { y }_{ 2 }=f\left( 2031 \right) \\ \\ \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =slope\\ \\ \frac { { y }_{ 2 }-{ y }_{ 1 } }{ 2031-2013 } =\frac { 100 }{ 12 } \\ \\ \frac { { y }_{ 2 }-{ y }_{ 1 } }{ 6 } =\frac { 25 }{ 1 } \\ \\ \frac { { y }_{ 2 }-{ y }_{ 1 } }{ 1 } =\frac { 25\quad \times \quad 6 }{ 1 } \\ \\ { y }_{ 2 }-{ y }_{ 1 }\quad =\quad 150\\ \\ f\left( 2031 \right) -f\left( 2013 \right) =150\\ \\ \\ \\$