Our equations from the problem statement are....
(1/3)x+(1/2)y=0..........Eq1
(1/2)x+(1/5)y=11/5.....Eq2
Recommendation: Whenever you have an equation(s) with
fractions in the terms, it is a good idea to write an equivalent
equations with integers only. This greatly reduces the possibility
for making an error in calculations.
So let's rewrite the equations multiplying each by the least
common multiple (LCM). For Eq1, LCM is 6. For Eq2, LCM
is 10...
2x+3y=0........Eq1a, multiply each term by 6
5x+2y=22.......Eq2a, multiply each term by 10
Now our strategy is to eliminate the "y" term by again multiplying
each equation by a factor to get "y" to cancel...
4x+6y=0........Eq1b, multiply each term by 2
-15x-6y=-66.....Eq2b, multiply each term by (-3)
------------------
-11x =-66.....add each term vertically
∴ x=6........divide both sides by (-11)
Substitute 6 for "x" into any of the two ORIGINAL equations
Eq1 or Eq2 and solve for "y." Let's use Eq1...
(1/3)(6)+(1/2)y=0........substitute 6 for "y"
(6/3)+(1/2)y=0.......simplify
2+(1/2)y=0
(1/2)y=-2......subtract 2, both sides
∴ y=-4......multiply both sides by 2
So our values are x=6 and y=-4. Now substitute these values
into the ORIGINAL equations Eq1, Eq2 to check for truth...
(1/3)(6)+(1/2)(-4)=0..........Eq1, substitute values for "x and y"
(6/3)+(-4/2)=0..........simplify
2-2=0
0=0...........true, √check
(1/2)(6)+(1/5)(-4)=11/5.....Eq2, substitute values for "x and y"
(6/2)+(-4/5)=11/5.....simplify
(30/10)-(8/10)=22/10
22/10=22/10
11/5=11/5.....true, √check
So the values of x=6 and y=-4 are correct. Always check your
solutions and work!