Question

$2x + 9y = 5 \\ 6x + 25y = 16 \\ elimination \: method$
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Answer 1

Solution:

2x + 9y =5            ......(i)

6x + 25y =16        ......(ii)

By Elimination Method

Multiplying equation (i) with 3

3(2x + 9y =5)

6x + 27y = 15        ......(iii)

From (i) and (iii)

6x + 25y = 16

6x + 27y = 15

-      -         -

      -2y  =  1

          y = -1/2

Putting y = -1/2 in equation (i) {can put in any of the 3 }

2x + 9 × $\frac{-1}{2}$ = 5

2x +  $\frac{-9}{2}$ = 5

2x = 5 + $\frac{9}{2}$

2x = $\frac{19}{2}$

Therefore,

x = $\frac{19}{4}$

Answer 2

Answer:-

$\bf{x = \dfrac{19}{4}}$

$\bf{y = \dfrac{-1}{2}}$

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• Given:-

$2x + 9y = 5 \: \: \: \: \: \: \longrightarrow\bf[eqn.1]$

$6x + 25y= 16 \: \: \: \: \: \: \longrightarrow\bf[eqn.2]$

• Method:-

$Elimination \: Method$

• Solution:-

Multiplying [eqn.1] with 3 :-

$6x + 27y = 15 \: \: \: \: \: \: \longrightarrow\bf[eqn.3]$

Now, Subtracting [eqn.2] from [eqn.3]:-

$\implies{6x + 27y -(6x + 25y) = 15 - (16)}$

$\implies{6x + 27y - 6x - 25y = -1}$

$\implies{ 2y = -1}$

$\implies\bf\boxed{y ={\dfrac{-1}{2}}}$

Substituting the above value of $\bf{y=\dfrac{-1}{2}}$ in [eqn.1]:-

$\implies{2x + 9y = 5}$

$\implies{2x + 9 \bigg(\dfrac{-1}{2}\bigg) = 5}$

$\implies{2x - \dfrac{9}{2} = 5}$

$\implies{2x = 5 + \dfrac{9}{2}}$

$\implies{2x = \dfrac{19}{2}}$

$\implies\bf\boxed{x = \dfrac{19}{4}}$

Hence,

• $\bf{x = \dfrac{19}{4}}$

• $\bf{y = \dfrac{-1}{2}}$

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