How many solutions does the following system of equations have? Choose among the following three answers: 0, 1, inf (You should be able to see this at a glance. No calculations necessary!) [math: x+2y+3z=42x+4y+6z=8−x−2y−3z=−4]\begin{eqnarray} x + 2y +3z &=& 4\ 2x + 4y +6z &=& 8\ -x - 2y -3z &=& -4\ \end{eqnarray}

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The given system actually consists of three equations that are all multiples of each other, so they represent the same single linear constraint.

Option 0: '0 solutions'. This would occur if the equations contradicted each other. Here, there is no contradiction; the third equation is just the negative of the first, and the second is twice the first. Therefore, there are infinitely many solutions rather than none.

Option 1: '1 solution'. A single unique solution would require the three equations to be independent and intersect at a single point. Since all three equations describe the same plane (they are scalar multiples of one another), there are infinitely many points satisfying them, not a unique one.

Option inf: 'infinitely many solutions'. This is consistent with the fact that all three equations reduce to the same constraint x + 2y + 3z = 4. In three variables, a single independent linear equation describes a plane with infinitely many points, so the solution set is infinite.

In summary, the first and second options describe scenarios (inconsistency or a unique solution) that do not occur here because the equations are not independent; they are dependent and define the same plane, leading to an infinite number of solutions.

MTH1030 -1035 - S1 2025 MTH1030/5 Week 2 lesson quiz: Solving systems of equations

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