Question

В
elimination method we obtain
3-64-b 0 —
- 39
- 6 = 0 -
2 2
2​

Answer 1

3x - 5y = 6 ____(i)

2x + 3y = 61 ____(ii)

Multiplying equation (i) by 2 and (ii) by 3 :

6x - 10y = 12 ___(iii)

6x + 9y = 183 ___(iv)

Subtracting equation (iv) from (iii) :

 6x - 10y  =  12

- 6x -  9y  = -183

--------------------------

-10y - 9y = 12 - 183

-19y = -171

y = 171 / 19

y = 9

Putting value of y in equation (iii) :

6x - 10y = 12

6x - 10(9) = 12

6x - 90 = 12

6x = 12 + 90

6x = 102

x = 17

Hence , x = 17 and y = 9 .

Answer 2

$\rm\huge\red{\underline{\underline{ Problem : }}}$

Solve this pair of linear equation by elimination method.

➡ 3x - 5y - 6 = 0 ; 2x + 3y - 61 = 0

$\rm\huge\red{\underline{\underline{ Solution : }}}$

★ Given that,

By using Elimination Method solve this pair of linear equation :

$\sf\:\implies 3x - 5y = 6 ..... (1)$

$\sf\:\implies 2x + 3y = 61 ..... (2)$

★ To find,

• Values of x & y.

★ Let,

Multiply equation (1) with 2 & equation (2) with 3.

We get,

$\sf\:\implies 6x - 10y = 12 ..... (3)$

$\sf\:\implies 6x + 9y = 183 ..... (4)$

• Subtract (3) & (4). We get,

$\sf\:\implies - 19y = - 171$

$\sf\large{\implies y = \frac{- 171}{-19}}$

$\sf\:\implies y = 9$

• Substitute value of y in (1)

$\sf\:\implies 3x - 5(9) = 6$

$\sf\:\implies 3x - 45 = 6$

$\sf\:\implies 3x = 6 + 45$

$\sf\:\implies 3x = 51$

$\sf\large{\implies x = \frac{51}{3}}$

$\sf\:\implies x = 17$

★ Verification,

Substitute the values of x & y in (1) to verify.

$\sf\:\implies 3x - 5y = > \green{LHS}$

• Substitute the values.

$\sf\:\implies 3(17) - 5(9)$

$\sf\:\implies 51 - 45$

$\sf\:\implies 6$

$\bf\blue{Since,\:LHS = RHS.\: Hence,\:it\:was\:verified.}$

$\underline{\boxed{\bf{\purple{ \therefore x = 17 ; y = 9}}}}\:\orange{\bigstar}$

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