Pure algebra

Answer: 11.25

Simultaneous equations by elimination  $$\quad a+b=7\\\underline{+\quad a-b=2\quad}\\

\quad 2a\quad=9$$ So $a=4.5$, $b=7-4.5=2.5$ so $ab=2.5\times4.5=11.25$

Simultaneous equations by substitution

Numbers with a difference of 2

$a-b=2$, so $a=b+2$

$a+b=7$, so $b+(b+2)=7\Rightarrow2b+2=7\Rightarrow2b=5$, so $b=2.5$

$a=b+2=2.5+2=4.5$ so $ab=2.5\times4.5=11.25$

Numbers which add up to 7

$a+b=7$ so $b=7-a$

$a-b=2$, so $a-(7-a)=2\Rightarrow 2a-7=2\Rightarrow 2a=9$, so $a=4.5$

$b=7-a=7-4.5=2.5$ so $ab=2.5\times4.5=11.25$

Finding $ab$ without finding $a$ and $b$  $$a+b=7\qquad\qquad\qquad\qquad a-b=2\qquad\qquad\qquad\qquad ab=?\\

(a+b)^2=a^2+b^2+2ab\quad\qquad (a-b)^2=a^2+b^2-2ab\quad\qquad \text{difference}=4ab\\

\ \ \\

\therefore 4ab = 7^2 - 2^2 = 45\qquad\qquad\qquad\quad\\

\Rightarrow ab=11.25\qquad\qquad\qquad\qquad\qquad$$

Trial methods

Looking at pairs of numbers with a difference of 2

1 and 3 have a difference of 2, but 1 + 3 = 4 not 7

2 and 4 have a difference of 2, but 2 + 4 = 6 not 7

3 and 5 have a difference of 2, but 3 + 5 = 8 not 7

So $a$ and $b$ must be (repectively) greater than 2 and 4 but less than 3 and 5.

2.5 and 4.5 have a difference of 2, and 2.5 + 4.5 = 7

So the numbers $a$ and $b$ are 2.5 and 4.5

2$\times$4.5 = 9, and 0.5$\times$4.5 = 2.25 (since multiplying by 0.5 is the same as dividing by 2).

So 2.5$\times$4.5 = 9 + 2.25 = 11.25

Looking at pairs of numbers which add up to 7

1 + 6 = 7 but 1 and 6 have a difference of 5, not 2

2 + 5 = 7 but 2 and 5 have a difference of 3, not 2

3 + 4 = 7 but 3 and 4 have a difference of 1, not 2

So the pair $a$ and $b$ must be between the pair 2 and 5 and the pair 3 and 4.

2.5 + 4.5 = 7 and 2.5 and 4.5 have a difference of 2

So the numbers $a$ and $b$ are 2.5 and 4.5

2$\times$4.5 = 9, and 0.5$\times$4.5 = 2.25 (since multiplying by 0.5 is the same as dividing by 2).

So 2.5$\times$4.5 = 9 + 2.25 = 11.25