+9466 We can get both of these equations into slope-intercept form, which is....
y = mx + b , where " m " is the slope and " b " is the y-intercept .
First equation:
y = 1/3 x - 2 This equation is in slope-intercept form. We can see that m = 1/3 = the slope.
Second equation:
-2x + 6y = 18 Add 2x to both sides of this equation.
6y = 2x + 18 Divide through by 6 .
y = 2/6 x + 3
y = 1/3 x + 3 Now we can see that m = 1/3 = the slope.
The slopes are both = 1/3 , and the y-intercepts are different.
This means that the lines are parallel, but they are not the same.
+2441 Here's another method to solve this problem:
Not all authorities agree on the standard form of a line; some believe it to be \(Ax+By=C\) and others believe it to be \(y=mx+b\). However, for the purposes of this problem, I will utilize the first form, \(Ax+By=C\).
For a system, however, we have this form:
1. \(A_1x+B_1y=C_1\)
2. \(A_2x+B_2y=C_2\)
Let's convert both equations into that form:
1. \(y=\frac{1}{3}x-2\)
| \(y=\frac{1}{3}x-2\) | Subtract y to both sides |
| \(\frac{1}{3}x-2-y=0\) | Add 2 to both sides |
| \(\frac{1}{3}x-y=2\) | |
2. \(-2x+6y=18\)
Luckily for us, this equation is already in its desired form, so we can move on.
Ok, let's look at the equations side by side again:
1. \(\frac{1}{3}x-y=2\)
2. \(-2x+6y=18\)
Now, look at the ratio of the coefficients. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), then the system has no solution. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\), then the system has infinitely many solutions. If \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\), then there is exactly one solution.
Let's try to see if this works.
Let's check to see if \(\frac{\frac{1}{3}}{-2}=\frac{-1}{6}\neq\frac{2}{18}\)
First, let's simplify \(\frac{\frac{1}{3}}{-2}\)
| \(\frac{\frac{1}{3}}{-2}\) | Apply the rule that \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\) |
| \(\frac{1}{3*-2}\) | |
| \(\frac{1}{-6}\) | |
| \(\frac{-1}{6}\) | |
Ok, let's check the condition again: Does \(\frac{-1}{6}=\frac{-1}{6}\neq\frac{2}{18}\)? Yes, 2/18 simplifies to 1/9, which is not equal to -1/6.
As I stated above, when a system of equations has the relationship of \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\) then there is no solution. Therefore, this system has no solution. If there is no solution to this system, then the lines must be parallel because otherwise the lines would intersect and would have a solution.
Despite all of this work that I have shown, Hecticlar's method appears to be easier to follow and is probably more efficient because this method cannot tell if the lines are perpendicular...
