Question

$(3x + 5y - 2x { }^{2} - 3y {}^{2} )$
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Answer 1

Answer:

2x+3y=7 -----(1)

(1)3x - 5y = 1 -----(2)

2)equation (1) x 5 = 10x + 15y = 35 =

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2put the value of 'x' in equation (1) we get,

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2put the value of 'x' in equation (1) we get,=> 3y = 7 - 2x

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2put the value of 'x' in equation (1) we get,=> 3y = 7 - 2x=> 3y = 7-4

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2put the value of 'x' in equation (1) we get,=> 3y = 7 - 2x=> 3y = 7-4=> y = 3=3

2)equation (1) x 5 = 10x + 15y = 35 =equation (2) х 3 = 9x - 15y = 3add these two equation => 19x = 38=> x = 38 = 19=> x = 2put the value of 'x' in equation (1) we get,=> 3y = 7 - 2x=> 3y = 7-4=> y = 3=3=> y=1

Answer 2

Answer:

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