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REAL NUMBERS
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By using elimination method,
5x + 7y = 17 ______(1)
3x - 2y = 7 _______(2)
Multiply, equation (1) by 3 and equation (2) by 5.
\bf\red{(5x+7y=17)}(5x+7y=17) × 3
\implies⟹ 15x + 21y = 51 _____(3)
\bf\red{(3x-2y=4)}(3x−2y=4) × 5
\implies⟹ 15x - 10y = 20 _____(4)
Subtract equation (4) from (3),
\cancel{15x}15x + 21y = 51
\cancel{-15x}−15x (+) - 10y = (-)20
_________________
31y = 31
\implies⟹ y = \dfrac{31}{31}3131
\implies⟹ y = 1
Now, put the value of y in eq (1),
5x + 7(1) = 17
\implies⟹ 5x + 7 = 17
\implies⟹ 5x = 17 - 7
\implies⟹ 5x = 10
\implies⟹ x = \dfrac{10}{5}510
\implies⟹ x = 2
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