Question

Solve the following by substitution method 2x+√3y=0 and√3x-√8y=10

Answer 1

$2x + \sqrt{3} y = 0 \\ 2x = - \sqrt{3} y \\ x = - \sqrt{3} y \div 2 \\ \\ \sqrt{3} x - \sqrt{8} y = 10 \\ \sqrt{3} ( - \sqrt{3} y \div 2) - \sqrt{8} y = 10 \\ (- 3y \div 2) - \sqrt{8} y = 10 \\ y(( - 3 \div 2) - \sqrt{8} ) = 10 \\ y( (- 3 - 2 \sqrt{8} ) \div 2) = 10 \\ y = (10 \times 2) \div ( - 3 - 2 \sqrt{4 \times 2} ) \\ y = - 20 \div(3 + 4 \sqrt{2} ) \\ \\ x = - \sqrt{3} y \div 2 \\ x = 10 \sqrt{3 } \div \: (3 + 4 \sqrt{2} )$

Step-by-step explanation: