QUESTION IMAGE
Question
drag each tile to the correct box.
arrange the systems of equations in order from least to greatest based on the number of solutions for each system.
$-5x + y = 10$
$-25x + 5y = 50$
$y = 6x - 2$
$y = 6x - 4$
$3x - 7y = 9$
$-4x + 5y = 1$
boxes for dragging
Step1: Analyze first system
The first system is:
Notice that the second equation is \(5\times(-5x + y)=5\times10\), so it's a multiple of the first equation. This means the two lines are coincident (same line), so there are infinitely many solutions.
Step2: Analyze second system
The second system is:
These are two parallel lines (same slope \(m = 6\), different y - intercepts). Parallel lines never intersect, so the number of solutions is \(0\).
Step3: Analyze third system
The third system is:
The slopes of the two lines: For \(3x-7y = 9\), we can rewrite it as \(y=\frac{3}{7}x-\frac{9}{7}\), slope \(m_1=\frac{3}{7}\). For \(- 4x + 5y=1\), rewrite as \(y=\frac{4}{5}x+\frac{1}{5}\), slope \(m_2=\frac{4}{5}\). Since \(m_1
eq m_2\), the two lines intersect at one point, so there is \(1\) solution.
Now, we order the systems from least to greatest number of solutions. The second system has \(0\) solutions, the third system has \(1\) solution, and the first system has infinitely many solutions. So the order is: \(y = 6x - 2\) and \(y = 6x - 4\) (system with \(0\) solutions), \(3x - 7y = 9\) and \(-4x + 5y = 1\) (system with \(1\) solution), \(-5x + y = 10\) and \(-25x + 5y = 50\) (system with infinitely many solutions).
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\(y = 6x - 2\), \(y = 6x - 4\) ; \(3x - 7y = 9\), \(-4x + 5y = 1\) ; \(-5x + y = 10\), \(-25x + 5y = 50\)