Question

which statement about the following system is correct?
$4x - 2y = 10$
$y = -2x + 5$

• the system is inconsistent because it has no solution.
• the system is consistent because it has no solution.
• the system is inconsistent because the two equations actually represent the same line.
• the system is consistent because it has a solution.

Explanation:

Step1: Simplify the first equation

Start with the first equation \(4x - 2y = 10\). Let's solve for \(y\) to see its slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept).
Subtract \(4x\) from both sides: \(-2y=-4x + 10\).
Divide every term by \(- 2\): \(y = 2x-5\)? Wait, no, wait. Wait, let's do it again. \(4x-2y = 10\), subtract \(4x\): \(-2y=-4x + 10\), divide by \(-2\): \(y = 2x-5\)? Wait, no, the second equation is \(y=-2x + 5\). Wait, no, I made a mistake. Let's re - solve the first equation:

Starting with \(4x-2y = 10\).
We want to solve for \(y\), so first, isolate the term with \(y\): \(-2y=10 - 4x\).
Then divide both sides by \(-2\): \(y=\frac{4x - 10}{2}=2x - 5\)? Wait, no, that's wrong. Wait, \(4x-2y = 10\), let's factor out a 2 from the left - hand side: \(2(2x - y)=10\), then divide both sides by 2: \(2x - y = 5\), then \(-y=5 - 2x\), so \(y = 2x-5\)? But the second equation is \(y=-2x + 5\). Wait, no, maybe I misread the second equation. Wait, the second equation is \(y=-2x + 5\). Wait, let's substitute the second equation into the first equation.

Substitute \(y=-2x + 5\) into \(4x-2y = 10\).

Step2: Substitute and solve

Substitute \(y=-2x + 5\) into \(4x-2y = 10\):
\(4x-2(-2x + 5)=10\)
Expand the left - hand side: \(4x + 4x-10 = 10\)
Combine like terms: \(8x-10 = 10\)
Add 10 to both sides: \(8x=20\)
Divide both sides by 8: \(x=\frac{20}{8}=\frac{5}{2}\)
Then substitute \(x = \frac{5}{2}\) into the second equation \(y=-2x + 5\):
\(y=-2\times\frac{5}{2}+5=-5 + 5=0\)

Wait, but wait, earlier when I solved the first equation, I made a mistake. Let's re - solve the first equation correctly. \(4x-2y = 10\), divide both sides by 2: \(2x - y=5\), so \(y = 2x - 5\)? No, that's incorrect. Wait, \(2x-y = 5\) can be rewritten as \(y=2x - 5\), but the second equation is \(y=-2x + 5\). Wait, no, there is a miscalculation. Wait, let's do the substitution again.

Wait, the first equation: \(4x-2y = 10\), let's solve for \(y\):

\(4x-2y=10\)

Subtract \(4x\) from both sides: \(-2y=10 - 4x\)

Divide both sides by \(-2\): \(y = 2x-5\)? No, \(y=\frac{4x - 10}{-2}=-2x + 5\). Oh! I see my mistake. \(\frac{10-4x}{-2}=\frac{- (4x - 10)}{-2}=\frac{4x - 10}{2}=2x - 5\)? No, \(\frac{10-4x}{-2}=\frac{-(4x - 10)}{-2}=\frac{4x - 10}{2}=2x - 5\) is wrong. Wait, \(\frac{10-4x}{-2}=\frac{10}{-2}-\frac{4x}{-2}=-5 + 2x=2x - 5\)? But if we factor out a \(-2\) from the numerator: \(\frac{-2(2x - 5)}{-2}=2x - 5\). But the second equation is \(y=-2x + 5\). Wait, no, I think I messed up the sign. Let's start over.

First equation: \(4x-2y = 10\)

Let's solve for \(y\):

\(-2y=10 - 4x\)

Multiply both sides by \(-1\): \(2y = 4x-10\)

Divide by 2: \(y = 2x-5\)

Second equation: \(y=-2x + 5\)

Wait, now substitute \(y=-2x + 5\) into \(y = 2x-5\):

\(-2x + 5=2x-5\)

Add \(2x\) to both sides: \(5 = 4x-5\)

Add 5 to both sides: \(10 = 4x\)

\(x=\frac{10}{4}=\frac{5}{2}\)

Then \(y=-2\times\frac{5}{2}+5=-5 + 5 = 0\)

Wait, but if we rewrite the first equation correctly:

Wait, \(4x-2y = 10\) can be rewritten as \(2y=4x - 10\), so \(y = 2x-5\)? No, that's not right. Wait, no, \(4x-2y = 10\) is equivalent to \(2x - y=5\), and the second equation is \(y=-2x + 5\), which can be rewritten as \(2x + y=5\).

Now we have the system:

\(\begin{cases}2x - y=5\\2x + y=5\end{cases}\)

If we add the two equations together: \((2x - y)+(2x + y)=5 + 5\)

\(4x=10\), so \(x=\frac{10}{4}=\frac{5}{2}\)

Substitute \(x = \frac{5}{2}\) into \(2x + y=5\): \(2\times\frac{5}{2}+y=5\), \(5 + y=5\), so \(y = 0\)

So the system has a solution \((x=\frac{5}{2},y = 0)\). A consistent system is a system of equations that has at least one solution. An inconsistent system has no solutions.

Now let's analyze the options:

- Option 1: "The system is inconsistent because it has no solution." False, because we found a solution.
- Option 2: "The system is consistent because it has no solution." False, a consistent system has at least one solution, an inconsistent system has no solutions.
- Option 3: "The system is inconsistent because the two equations actually represent the same line." False, the two equations \(4x - 2y = 10\) (or \(2x - y=5\)) and \(y=-2x + 5\) (or \(2x + y=5\)) have different slopes (the slope of \(2x - y=5\) is 2, the slope of \(2x + y=5\) is - 2), so they are not the same line. Also, inconsistent systems with two lines either are parallel (no solution) or the same line (infinite solutions). But we have a solution, so it's consistent.
- Option 4: "The system is consistent because it has a solution." True, because we found that \(x = \frac{5}{2}\) and \(y = 0\) is a solution, and a consistent system is defined as a system with at least one solution.

Answer:

The system is consistent because it has a solution.