Question

two numbers differ by 3 the sum of the greater number and twice the second number is 15 find the smaller number ​

Answer 1

let greater no. be x

and smaller no. be x-3

ATQ,

=> x+2(x-3)=15

=> x+2x-6=15

=> 3x-6=15

=> 3x=15+6

=> 3x=21

=> x= 7

hence, smaller no.= 7-3

=4

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{The difference between the numbers = 3}} \\ \\ \text{ \pink{The sum of the greater number and twice }} \\ \text{ \pink{the smaller number = 15}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \: \: \: \: \: \: \: \: \text {\red{The greater number = x}} \\ \\ \: \: \: \: \: \: \: \text{ \red{The smaller number = y}}$

$\bold{ \underline{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf{ \blue{x - y = 3 \: \: \: ------> (1)}} \\ \\ \bold{And, } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf{ \blue{x + 2y = 15 \: \: -----> (2)}} \\ \\ \underline{ \bold{ \: Now, \: }} \\ \\ \sf{ \blue{ (1) \implies x = 3 + y -----> (3)}}$

$\underline{ \text{ \: Putting the value of \pink{x} from eq. (3) in eq.(2), we get, \: }} \\ \\ \tt{(2) \implies x + 2y = 15} \\ \\ \: \: \: \: \: \: \: \tt{ \implies (3 + y) + 2y = 15} \\ \\ \: \: \: \: \: \: \: \tt{ \implies 3 + y + 2y = 15} \\ \\ \: \: \: \: \: \: \:\tt{ \implies 3 + 3y = 15 } \\ \\ \: \: \: \: \: \: \:\tt{\implies 3y = 15 - 3 } \\ \\ \: \: \: \: \: \: \: \tt{\implies 3y = 12} \\ \\ \: \: \: \: \: \: \: \tt{ \implies y = \frac{12}{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\therefore \: \: \red{ y = 4}} \\ \\ \therefore \: \: \text{ \red{The smallest number, y = \underline {\: 4 \:} }}$

$\underline{\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}\\ \\ \underline{ \mathfrak{ \: \: Again, \: \: }} \\ \\ \underline{ \text{ \: Putting the value of \pink{y} in eq. (3), we get, \: }} \\ \\ \tt{(3) \implies x = 3 + y } \\ \\ \: \: \: \: \: \: \: \tt{\implies x = 3 + 4 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\therefore \: \: \red{ x = 7}} \\ \\ \therefore \: \: \text{ \red{The greater number = \underline {\: 7 \: }}}$