Solve by substitution method: root3x + root2y = 5;
3 ×root3x – 2× root2y = 5
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Explanation:
√3x + √2y = 0 _ _ _ (1)
√5x + √7y = 0 _ _ _ (2)
In (1)→ y = -- √3x\√2
Put in eq. (2)
√5x + √7[-- √3x\√2 ] =0
√5x + [ -- √21x\√2 ] = 0
√5x -- √21x = 0*√2 => 0
-- √16x = 0
x=0
so , y=0 (on putting x=0 in eq. 1o
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