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x + 7y = 33 (Eqn 1) 3x + 2y = 7 (Eqn 2) Multiply Eqn 1 by 3, now the equations are 3x + 21y = 99 (Eqn 3) 3x + 2y = 7 (Eqn 2) Subtract Eqn 2 from Eqn 4 and the result of subtraction is 19y = 92 Hence y = 92/19 Now substitute the value of y in Eqn 1 x + (7) x (92/19) = 33 x = 33 - 644/19 = -(17/19)
Eqn (A): => 2x + 5y = 16 Eqn (B): => 5x + 2y = -2 5*Eqn (A) - 2*Eqn (B): 21y = 84 => y = 4 Substituting for y in Eqn (a): x = -2
Eqn 1: x - 2y = 3 Eqn 2: 5x + 3y = 2 5 times Eqn 1: 5x - 10y = 15 Eqn 2 - 5*Eqn 1: 0x + 13y = -13 so that 13y = -13 or y = -1 Substitute in Eqn 1: x + 2*1 = 3 x + 2 = 3 x = 1 Solution: x = 1, y = -1
c - 5d = 2 2c + d = 4 Multiply eqn 1 by 2: 2c - 10d = 4 Subtract this from eqn 2: 0c + 11d = 0 which implies that d = 0 Then, by eqn 1, c = 2
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