Question

2x + 3y = 17,3x - 2y = 6 then x-y
A) First prime number
C) First odd prime number
B) First composite number
D) neither prime nor composite number​

Answer 1

Given

⇒2x + 3y = 17        (i)

⇒3x - 2y = 6            (ii)

To Find

⇒x - y

Now using Substitution method , So take (ii)

⇒3x - 2y = 6

⇒3x = 6 + 2y

⇒x = (6+2y)/3                  (iii)

Bow put the value of x on (i) eq

⇒2x + 3y = 17

⇒2(6 + 2y)/3 + 3y = 17

By taking LCM , we get

⇒2(6 + 2y) + 9y = 17×3

⇒12 + 4y + 9y = 51

⇒12 + 13y = 51

⇒13y = 51-12

⇒13y = 39

⇒y = 39/13

⇒y = 3

Put the value of y on (iii)equation

⇒x = (6+2y)/3                  (iii)

⇒x = (6+ 2×3)/3

⇒x = (6+ 6)/3

⇒x = 12/3

⇒x = 4

Now

⇒x - y

⇒4 - 3

⇒1

Answer is 1

⇒So 1 is Neither Prime Nor Composite Number

Answer 2

$\bf Given \begin {cases} & \sf 2x + 3y = 17 \qquad \longrightarrow \:Eq.1 \:\: \\\\ & \sf 3x - 2y = 6 \qquad \longrightarrow \:Eq.2 \:\: \end{cases}$

Exigency To Find : The nature of the value of x - y .

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⠀⠀⠀⠀⠀Firstly we have to find value of x & y using Substitution method to find value of x - y :

⠀⠀⠀⠀⠀Given :

$\begin {cases} & \sf 2x + 3y = 17 \qquad \longrightarrow \:Eq.1 \:\: \\\\ & \sf 3x - 2y = 6 \qquad \longrightarrow \:Eq.2 \:\: \end{cases}$

⠀⠀⠀⠀⠀From Eq.1 :

$\qquad :\implies \sf 2x + 3y = 17 \qquad \longrightarrow \:Eq.1 \:$

$\qquad :\implies \sf 2x + 3y = 17 \:$

$\qquad :\implies \sf 2x = 17 - 3y \:$

$\qquad :\implies \bf x = \dfrac{ 17 - 3y }{2} \qquad \:\longrightarrow \: Eq.3 \: \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Eq.3\:in\:Eq.2 \::}}$

$\qquad :\implies\sf 3x - 2y = 6 \qquad \longrightarrow \:Eq.2 \:\:$

$\qquad :\implies\sf 3x - 2y = 6 \qquad \:\:$

$\qquad :\implies\sf 3 \dfrac{ (17 - 3y) }{2} - 2y = 6 \qquad \:\:$

$\qquad :\implies\sf 3 (17 - 3y) - 4y = 6 \times 2 \qquad \:\:$

$\qquad :\implies\sf 3 (17 - 3y) - 4y = 12 \qquad \:\:$

$\qquad :\implies\sf 51 - 9y - 4y = 12 \qquad \:\:$

$\qquad :\implies\sf 51 - 13y = 12 \qquad \:\:$

$\qquad :\implies\sf - 13y = 12 - 51 \qquad \:\:$

$\qquad :\implies\sf - 13y = -39 \qquad \:\:$

$\qquad :\implies\sf y = \dfrac{-39}{-13} \qquad \:\:$

$\qquad :\implies\sf y = \cancel {\dfrac{-39}{-13}} \qquad \:\:$

$\qquad :\implies \frak{\underline{\purple{\:y = 3 \:}} }\bigstar$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value \:of\:y\:in\:Eq.2 \::}}$

$\qquad :\implies\sf 3x - 2y = 6 \qquad \longrightarrow \:Eq.2 \:\:$

$\qquad :\implies\sf 3x - 2y = 6 \qquad \:\:$

$\qquad :\implies\sf 3x - 2(3) = 6 \qquad \:\:$

$\qquad :\implies\sf 3x - 6 = 6 \qquad \:\:$

$\qquad :\implies\sf 3x = 6 + 6 \qquad \:\:$

$\qquad :\implies\sf 3x = 12 \qquad \:\:$

$\qquad :\implies\sf x = \dfrac{12}{3} \qquad \:\:$

$\qquad :\implies \frak{\underline{\purple{\:x = 4\:\: }} }\bigstar$

⠀⠀⠀⠀⠀Now Finding the Value of x - y :

Here,

• x = 4
• y = 3

Therefore,

$\qquad :\implies\sf x - y \qquad \:\:$

$\qquad :\implies\sf 4 - 3 \qquad \:\:$

$\qquad :\implies \frak{\underline{\purple{\: 1\:\: }} }\bigstar$

Therefore,

• 1 is neither prime nor composite number .

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