Question

nc math 1—released items
20 this is a paper/pencil copy of an online technology enhanced item.
three systems of equations are shown in the table below.
place (click and drag) the choice that describes the number of solutions of each system into the appropriate column in the table below.
$2x + 2y = 15$ $2x + 2y = 0$ $2x + 3y = 12$
$4x + 3y = 27$ $4x + 4y = 16$ $2x + 3y = 10$
one solution no solution infinitely many solutions

Explanation:

Step1: Analyze the first system: \(2x + 2y = 15\) and \(4x + 3y = 27\)

We can use the elimination method. Multiply the first equation by 2: \(4x + 4y = 30\). Now subtract the second equation (\(4x + 3y = 27\)) from this new equation: \((4x + 4y)-(4x + 3y)=30 - 27\), which simplifies to \(y = 3\). Substitute \(y = 3\) into \(2x + 2y = 15\): \(2x+6 = 15\), so \(2x = 9\) and \(x=\frac{9}{2}\). So this system has one solution.

Step2: Analyze the second system: \(2x + 2y = 0\) and \(4x + 4y = 16\)

Simplify the first equation: \(x + y = 0\) (divide by 2). Simplify the second equation: \(x + y = 4\) (divide by 4). But \(x + y\) can't be both 0 and 4. So this system has no solution.

Step3: Analyze the third system: \(2x + 3y = 12\) and \(2x + 3y = 10\)

Subtract the second equation from the first: \((2x + 3y)-(2x + 3y)=12 - 10\), which gives \(0 = 2\), a contradiction. Wait, no, wait: Wait, the two equations are \(2x + 3y = 12\) and \(2x + 3y = 10\). Since the left - hand sides are the same but the right - hand sides are different, there is no solution? Wait, no, I made a mistake. Wait, no, the third system: Wait, the first equation of the third system is \(2x + 3y = 12\), the second is \(2x + 3y = 10\). So if we subtract them, \(0=2\), which is impossible. Wait, but that's not right. Wait, no, maybe I misread. Wait, the original problem: the third system is \(2x + 3y = 12\) and \(2x + 3y = 10\)? Wait, no, that can't be. Wait, no, maybe it's a typo? Wait, no, the user's image: the third system is \(2x + 3y = 12\) and \(2x + 3y = 10\)? Wait, no, that would mean no solution, but that seems odd. Wait, no, maybe I misread. Wait, no, let's check again. Wait, the first system: \(2x + 2y = 15\), \(4x + 3y = 27\) (one solution). Second system: \(2x + 2y = 0\), \(4x + 4y = 16\) (no solution, because \(2x + 2y = 0\) implies \(4x + 4y = 0\), but the second equation is \(4x + 4y = 16\), so parallel lines). Third system: \(2x + 3y = 12\) and \(2x + 3y = 10\) (no solution? Wait, no, that can't be. Wait, maybe it's \(2x+3y = 12\) and \(2x + 3y = 10\)? No, that would be inconsistent. Wait, but maybe I made a mistake. Wait, no, the key is to check the slopes and intercepts.

For a system of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\):

- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} eq\frac{c_1}{c_2}\), no solution (parallel lines).
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), infinitely many solutions (same line).
- If \(\frac{a_1}{a_2} eq\frac{b_1}{b_2}\), one solution (intersecting lines).

First system: \(a_1 = 2\), \(b_1 = 2\), \(c_1 = 15\); \(a_2 = 4\), \(b_2 = 3\), \(c_2 = 27\). \(\frac{2}{4}=\frac{1}{2}\), \(\frac{2}{3} eq\frac{1}{2}\), so \(\frac{a_1}{a_2} eq\frac{b_1}{b_2}\), one solution.

Second system: \(a_1 = 2\), \(b_1 = 2\), \(c_1 = 0\); \(a_2 = 4\), \(b_2 = 4\), \(c_2 = 16\). \(\frac{2}{4}=\frac{2}{4}=\frac{1}{2}\), \(\frac{0}{16}=0 eq\frac{1}{2}\), so \(\frac{a_1}{a_2}=\frac{b_1}{b_2} eq\frac{c_1}{c_2}\), no solution.

Third system: \(a_1 = 2\), \(b_1 = 3\), \(c_1 = 12\); \(a_2 = 2\), \(b_2 = 3\), \(c_2 = 10\). \(\frac{2}{2}=1\), \(\frac{3}{3}=1\), \(\frac{12}{10}=\frac{6}{5} eq1\), so \(\frac{a_1}{a_2}=\frac{b_1}{b_2} eq\frac{c_1}{c_2}\), no solution? Wait, no, that's what I thought earlier. But that contradicts my first analysis. Wait, no, in the first analysis of the third system, I think I misread the equations. Wait, the third system is \(2x + 3y = 12\) and \(2x + 3y = 10\)? Then \(\frac{2}{2}=\frac{3}{3} eq\frac{12}{10}\), so no solution. But that can't be. Wait, maybe the third system is \(2x + 3y = 12\) and \(2x+3y = 12\)? No, the user's image says \(2x + 3y = 10\). Wait, maybe it's a mistake in the problem, but according to the given equations:

Wait, let's re - check the second system: \(2x + 2y = 0\) (equation 1), \(4x + 4y = 16\) (equation 2). Divide equation 1 by 2: \(x + y = 0\). Divide equation 2 by 4: \(x + y = 4\). So \(\frac{2}{4}=\frac{2}{4}\), \(\frac{0}{16}=0 eq\frac{1}{2}\) (wait, \(\frac{2}{4}=\frac{2}{4}=\frac{1}{2}\), and \(\frac{0}{16}=0 eq\frac{1}{2}\)), so no solution.

Third system: \(2x + 3y = 12\) and \(2x + 3y = 10\). \(\frac{2}{2}=\frac{3}{3}\), \(\frac{12}{10}=\frac{6}{5} eq1\), so no solution? But that would mean two systems with no solution, which is odd. Wait, maybe I misread the third system. Wait, the user's image: the third system is \(2x + 3y = 12\) and \(2x + 3y = 10\)? Or is it \(2x+3y = 12\) and \(2x - 3y = 10\)? No, the image shows \(2x + 3y = 10\).

Wait, maybe there's a mistake in my initial analysis. Let's re - do the first system:

First system: \(2x + 2y = 15\) (Equation 1), \(4x + 3y = 27\) (Equation 2). Multiply Equation 1 by 2: \(4x + 4y = 30\) (Equation 3). Subtract Equation 2 from Equation 3: \((4x + 4y)-(4x + 3y)=30 - 27\Rightarrow y = 3\). Then substitute \(y = 3\) into Equation 1: \(2x+6 = 15\Rightarrow 2x = 9\Rightarrow x = 4.5\). So one solution.

Second system: \(2x + 2y = 0\) (Equation A), \(4x + 4y = 16\) (Equation B). Equation A can be written as \(x + y = 0\), Equation B as \(x + y = 4\). These are parallel lines (same slope, different y - intercepts), so no solution.

Third system: \(2x + 3y = 12\) (Equation C), \(2x + 3y = 10\) (Equation D). Equation C and Equation D have the same left - hand side but different right - hand sides. So they are parallel lines (same slope, different y - intercepts), so no solution? But that would mean two systems with no solution, which is strange. Wait, maybe the third system is \(2x+3y = 12\) and \(2x + 3y = 12\)? No, the image says 10.

Wait, maybe the original problem has a typo, but based on the given equations:

- System 1: \(2x + 2y = 15\), \(4x + 3y = 27\): one solution.
- System 2: \(2x + 2y = 0\), \(4x + 4y = 16\): no solution.
- System 3: \(2x + 3y = 12\), \(2x + 3y = 10\): no solution.

But that can't be right. Wait, maybe the third system is \(2x+3y = 12\) and \(2x + 3y = 12\)? No, the image shows 10. Alternatively, maybe I made a mistake in the second system. Let's check the second system again. \(2x + 2y = 0\) implies \(y=-x\). Substitute into \(4x + 4y = 16\): \(4x+4(-x)=16\Rightarrow 4x - 4x = 16\Rightarrow 0 = 16\), which is false. So no solution.

Third system: \(2x + 3y = 12\) and \(2x + 3y = 10\). Substitute \(y=\frac{12 - 2x}{3}\) into the second equation: \(2x+3\times\frac{12 - 2x}{3}=10\Rightarrow 2x + 12 - 2x = 10\Rightarrow 12 = 10\), which is false. So no solution.

But the problem says "three systems" and three categories: one solution, no solution, infinitely many solutions. So maybe I misread the third system. Wait, maybe the third system is \(2x + 3y = 12\) and \(4x + 6y = 24\)? Then it would have infinitely many solutions. But according to the image, it's \(2x + 3y = 10\).

Assuming the problem is as given:

- \(2x + 2y = 15\), \(4x + 3y = 27\): one solution.
- \(2x + 2y = 0\), \(4x + 4y = 16\): no solution.
- \(2x + 3y = 12\), \(2x + 3y = 10\): no solution.

But this seems odd. Maybe there's a mistake in the problem statement or my reading. However, based on the given equations:

Answer:

• \(2x + 2y = 15\), \(4x + 3y = 27\): one solution.
• \(2x + 2y = 0\), \(4x + 4y = 16\): no solution.
• \(2x + 3y = 12\), \(2x + 3y = 10\): no solution.

(If we assume that the third system was supposed to be \(2x + 3y = 12\) and \(2x + 3y = 12\), then it would have infinitely many solutions. But based on the given image, the third system has \(2x + 3y = 10\), so the above is the answer.)