Question

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REAL NUMBERS

Answer 1

Answer:

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