Solve System 657886

1. **State the problem:** Solve the system of linear equations: $$\frac{x}{3} + \frac{y}{2} = 2$$ $$\frac{x}{3} - \frac{y}{2} = 0$$ 2. **Recall the method:** We can solve this system by adding or subtracting the equations to eliminate one variable. 3. **Add the two equations:** $$\left(\frac{x}{3} + \frac{y}{2}\right) + \left(\frac{x}{3} - \frac{y}{2}\right) = 2 + 0$$ Simplify the left side: $$\frac{x}{3} + \frac{x}{3} + \frac{y}{2} - \frac{y}{2} = \frac{2x}{3} + 0 = \frac{2x}{3}$$ So: $$\frac{2x}{3} = 2$$ 4. **Solve for $x$:** Multiply both sides by 3: $$3 \times \frac{2x}{3} = 3 \times 2$$ $$2x = 6$$ Divide both sides by 2: $$\cancel{2}x = \frac{6}{\cancel{2}}$$ $$x = 3$$ 5. **Substitute $x=3$ into one of the original equations to find $y$:** Use the second equation: $$\frac{x}{3} - \frac{y}{2} = 0$$ Substitute $x=3$: $$\frac{3}{3} - \frac{y}{2} = 0$$ Simplify: $$1 - \frac{y}{2} = 0$$ 6. **Solve for $y$:** $$- \frac{y}{2} = -1$$ Multiply both sides by $-2$: $$\cancel{-2} \times - \frac{y}{\cancel{2}} = -2 \times -1$$ $$y = 2$$ 7. **Final solution:** $$x = 3, \quad y = 2$$