[amp_mcq option1=”infinitely many” option2=”two distinct solutions” option3=”unique” option4=”none” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{C}}$, two distinct solutions.<\/p>\n
To solve a system of linear equations, we can use the elimination method. In this method, we add or subtract the equations in such a way that one of the variables cancels out. We can then solve for the remaining variable and substitute it back into one of the original equations to solve for the other variable.<\/p>\n
In this case, we can eliminate $x$ by adding the first and second equations together. This gives us the equation $6y = 3$. Solving for $y$, we get $y = \\frac{1}{2}$.<\/p>\n
We can then substitute $y = \\frac{1}{2}$ into the third equation to solve for $x$. This gives us the equation $x + \\frac{3}{2} = 3$. Solving for $x$, we get $x = \\frac{1}{2}$.<\/p>\n
Therefore, the system of equations has two distinct solutions, $(x, y) = \\left(\\frac{1}{2}, \\frac{1}{2}\\right)$ and $(x, y) = \\left(-\\frac{1}{2}, -\\frac{1}{2}\\right)$.<\/p>\n
Option A is incorrect because the system of equations has two distinct solutions, not infinitely many solutions.<\/p>\n
Option B is incorrect because the system of equations has two distinct solutions, not one solution.<\/p>\n
Option D is incorrect because the system of equations has two distinct solutions, not no solutions.<\/p>\n","protected":false},"excerpt":{"rendered":"
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