Solve the following pair of simultaneous equations: $3\sqrt {2}x-5\sqrt {3}y+\sqrt {5}=0$ and $2\sqrt {3}x+7\sqrt {2}y-2\sqrt {5}=0$
- $x=\cfrac { 18\sqrt { 1 } +6\sqrt { 4 } }{ 56 } ;\quad y=\cfrac { 12\sqrt { 6 } +2\sqrt { 2 } }{ 52 } $
- $x=\cfrac { 11\sqrt { 15 } +7\sqrt { 16 } }{ 109 } ;\quad y=\cfrac { 1\sqrt { 13 } -7\sqrt { 11 } }{ 2 } $
- $x=\cfrac { 3\sqrt { 12 } -5\sqrt { 14 } }{ 23 } ;\quad y=\cfrac { 8\sqrt { 13 } +6\sqrt { 11 } }{ 56 } $
- $x=\cfrac { 10\sqrt { 15 } -7\sqrt { 10 } }{ 72 } ;\quad y=\cfrac { 2\sqrt { 15 } +6\sqrt { 10 } }{ 72 } $
We can solve this system of linear equations using the elimination method. Multiplying the first equation by 2*sqrt(3) and the second by 3*sqrt(2) allows us to eliminate the x-term, yielding y = (2*sqrt(15) + 6*sqrt(10)) / 72. Similarly, eliminating the y-term by multiplying the equations by 7*sqrt(2) and 5*sqrt(3) respectively gives x = (10*sqrt(15) - 7*sqrt(10)) / 72.
Rewrite the simultaneous equations as 3 times the square root of 2 times x minus 5 times the square root of 3 times y equals negative the square root of 5, and 2 times the square root of 3 times x plus 7 times the square root of 2 times y equals 2 times the square root of 5. By Cramer's rule, x equals D divided by D1, where D evaluates to 56 times the square root of 6. The numerator D1 evaluates to 10 times the square root of 30 plus 42 times the square root of 10, so x simplifies to the fraction (10 times the square root of 15 minus 7 times the square root of 10) divided by 72. Similarly finding y gives the fraction (2 times the square root of 15 plus 6 times the square root of 10) divided by 72.