Question

determine how many solutions usin 9) $2y = -8 - x$ $x - 4 - 2y = 0$

Explanation:

Step1: Rewrite equations in standard form

First equation: \(2y = -8 - x\) can be rewritten as \(x + 2y = -8\) (by adding \(x\) to both sides).
Second equation: \(x - 4 - 2y = 0\) can be rewritten as \(x - 2y = 4\) (by adding \(2y\) and 4 to both sides).

Step2: Analyze the system of equations

We have the system:
\[ \begin{cases} x + 2y = -8 \\ x - 2y = 4 \end{cases} \]
Let's check the slopes (or the coefficients of \(x\) and \(y\)). For the first equation, the coefficients of \(x\) and \(y\) are \(1\) and \(2\). For the second equation, the coefficients of \(x\) and \(y\) are \(1\) and \(-2\). The ratios of the coefficients of \(x\) (\(\frac{1}{1} = 1\)) and the ratios of the coefficients of \(y\) (\(\frac{2}{-2} = -1\)) are not equal, so the lines are not parallel and not coincident. Also, we can solve the system by elimination or substitution. Let's add the two equations:
\((x + 2y) + (x - 2y) = -8 + 4\)
\(2x = -4\)
\(x = -2\)
Then substitute \(x = -2\) into the first equation: \(-2 + 2y = -8\), \(2y = -6\), \(y = -3\). Wait, but let's check the original equations. Wait, maybe I made a mistake in rewriting? Wait, no, let's check the original second equation: \(x - 4 - 2y = 0\) => \(x - 2y = 4\). First equation: \(2y = -8 - x\) => \(x + 2y = -8\). Now, if we write both equations in slope-intercept form (\(y = mx + b\)):
First equation: \(2y = -x -8\) => \(y = -\frac{1}{2}x - 4\)
Second equation: \(x - 2y = 4\) => \(-2y = -x + 4\) => \(y = \frac{1}{2}x - 2\)
The slopes are \(-\frac{1}{2}\) and \(\frac{1}{2}\), which are not equal and not negative reciprocals (so not perpendicular, but still intersecting). Wait, but when we solved, we got a solution. But wait, let's check with the original equations. Wait, maybe I messed up the sign. Wait, original second equation: \(x - 4 - 2y = 0\) => \(x - 2y = 4\). Original first equation: \(2y = -8 - x\) => \(x + 2y = -8\). Now, let's add the two equations: \(x + 2y + x - 2y = -8 + 4\) => \(2x = -4\) => \(x = -2\). Then substitute into first equation: \(2y = -8 - (-2) = -6\) => \(y = -3\). Now check in second equation: \(-2 - 4 - 2*(-3) = -6 + 6 = 0\), which works. Wait, but that means there is one solution? Wait, no, wait, maybe I made a mistake. Wait, no, the two lines have different slopes, so they intersect at one point. Wait, but let's check the coefficients again. Wait, the first equation: \(x + 2y = -8\), second: \(x - 2y = 4\). The determinant of the coefficient matrix \(\begin{pmatrix}1 & 2 \\ 1 & -2\end{pmatrix}\) is \((1)(-2) - (2)(1) = -2 - 2 = -4 eq 0\), so the system has a unique solution. Wait, but wait, maybe the problem was to check for parallel or coincident lines. Wait, no, the slopes are \(-\frac{1}{2}\) and \(\frac{1}{2}\), which are not equal, so the lines intersect at one point. Wait, but let's re-express the original equations. Wait, first equation: \(2y = -8 - x\) => \(y = -\frac{1}{2}x - 4\). Second equation: \(x - 4 - 2y = 0\) => \(2y = x - 4\) => \(y = \frac{1}{2}x - 2\). So the slopes are \(-\frac{1}{2}\) and \(\frac{1}{2}\), which are not equal, so the lines are not parallel, hence they intersect at exactly one point. Wait, but earlier when I solved, I got a solution. So the system has one solution? Wait, but maybe I made a mistake in the problem statement. Wait, the original problem says "Determine how many solutions usin" (probably "using" some method, like comparing slopes or solving). Wait, but according to the equations, when we solve, we get a unique solution. Wait, but let's check again. Wait, first equation: \(2y = -8 - x\) => \(x + 2y = -8\). Second equation: \(x - 4 - 2y = 0\) => \(x - 2y = 4\). Now, if we subtract the second equation from the first: \((x + 2y) - (x - 2y) = -8 - 4\) => \(4y = -12\) => \(y = -3\). Then \(x = 4 + 2y = 4 + 2*(-3) = -2\). So the solution is \(x = -2\), \(y = -3\), which satisfies both equations. So the system has one solution. Wait, but maybe the problem was supposed to have no solution? Wait, no, the calculations show a solution. Wait, maybe I misread the equations. Let me check again. First equation: \(2y = -8 - x\). Second equation: \(x - 4 - 2y = 0\). So substituting \(2y\) from first equation into second: \(x - 4 - (-8 - x) = 0\) => \(x - 4 + 8 + x = 0\) => \(2x + 4 = 0\) => \(2x = -4\) => \(x = -2\). Then \(2y = -8 - (-2) = -6\) => \(y = -3\). So yes, one solution.

Answer:

The system of equations has one solution.