Find two numbers whose sum is $15$ and difference is $3$ .

13

and

2

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7

and

8

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9

and

6

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3

and

12

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Solution

The correct option is C $9$ and $6$
Let the two numbers be $x$ and $y$ .

It is given that the sum of the numbers is $15$ that is

$x + y = 15.......... (1)$

Also the difference of the numbers is $3$ that is

$x - y = 3.......... (2)$ .
Add equations 1 and 2 to eliminate $y$ , because the coefficients of $y$ are opposite. So, we get

$(x + x) + (y - y) = 15 + 3$

i.e. $2 x = 18$

$⟹ x = 9$

Substituting this value of $x$ in (1), we get

$9 + y = 15$

i.e. $y = 15 - 9 = 6$

Hence, the two numbers are $9$ and $6$ .

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Similar questions

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$