To determine how many solutions the given system of equations has, we need to analyze the equations:
- The first equation is \(y = -2x + 12\).
- The second equation is \(4y + x - 20 = 0\).
We can rewrite the second equation in terms of \(y\):
\[ 4y = -x + 20 \]
\[ y = -\frac{1}{4}x + 5 \]
Now, we have the two equations:
- \(y = -2x + 12\) (a line with a slope of -2)
- \(y = -\frac{1}{4}x + 5\) (a line with a slope of -\(\frac{1}{4}\))
Step 1: Compare Slopes
The slopes of the two lines are different (-2 vs -\(\frac{1}{4}\)). When two lines have different slopes, they will intersect at exactly one point.
Step 2: Finding the Intersection Point
To find the exact intersection, set the equations equal to each other:
\[ -2x + 12 = -\frac{1}{4}x + 5 \]
To eliminate the fraction, multiply the entire equation by 4:
\[ -8x + 48 = -x + 20 \]
Rearranging gives:
\[ -8x + x = 20 - 48 \]
\[ -7x = -28 \]
\[ x = 4 \]
Now, substitute \(x = 4\) back into one of the original equations to find \(y\):
Using \(y = -2x + 12\):
\[ y = -2(4) + 12 = -8 + 12 = 4 \]
Thus, the intersection point is \((4, 4)\).
Conclusion
Since we found one intersection point, the two equations have one solution.
Answer: One solution