Question

23x+17y= 6 & 39x-19y= 58 by elemination method

Answer 1

Step-by-step explanation:

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Answer 2

☯GIVEN :

• $\bf{23x +17y = 6}$ and $\bf{39x-19y = 58}$

☯TO DO :

• Solve by Elimination Method.

➲SOLUTION :

▪Multiply the first equation by 19:

$\implies{ \sf{19(23x + 17y) = 6 \times 19}}$

$\implies{ \bf{437x+323y = 114 (i)}}$

▪Multiply the second equation by 17 :

$\implies{ \sf {17(39x -19y) = 58 \times 17}}$

$\implies{ \bf {663x-323y = 986 - (ii)}}$

★Adding both equations:

$:\implies \sf(437x + 323y) +(663x -323y) = 114+986$

$:\implies \sf 437x + 323y +663x -323y = 1100$

$:\implies \sf437x + 663x + \cancel{323y } - \cancel{323y }= 1100$

$:\implies \sf 437x+663x = 1100$

$:\implies \sf 1100x = 1100$

$:\implies \sf x = \cancel{\dfrac{1100}{1100}}$

$:\implies \huge{ \underline{\boxed{ \blue{\bf {x = 1 }}}}}$

✞Substituting the obtained value of x in the first equation :

$:\implies \sf {23x +17y = 6}$

$:\implies \sf 23\times 1+ 17y = 6$

$:\implies \sf 23+17y = 6$

$:\implies \sf 17y = 6-23$

$:\implies \sf 17y = -17$

$:\implies \sf y = \cancel{\dfrac{-17}{17}}$

$:\implies\huge{ \underline{\boxed{ \purple{ \bf {y = -1 }}}}}$

$\huge{\pink{\therefore}}$ The value of x and y is 1 and -1 respectively.