Let's write both equations in the standard form of [ y = mx + b ],
and then see what we can tell about their graphs.
First equation: x + 3y = 9
Subtract 'x' from each side: 3y = -x + 9
Divide each side by 3: y = -1/3 x + 3
This line crosses the y-axis at y=3, and it has a slope of -1/3 .
Second equation: 3x - y = 7
Subtract 3x from each side: -y = -3x + 7
Multiply each side by -1 : y = 3x - 7
This line crosses the y-axis at y=-7, and it has a slope of 3 .
-- The two lines have different slopes, so they're not parallel.
They must intersect somewhere.
-- They're not the same line, so they can't 'intersect' everywhere.
-- They have slopes of -1/3 and 3 . Their slopes are negative reciprocals,
so the lines are perpendicular.
All of this says that the two equations can't have no solution, and they can't have
infinitely many solutions. They must have one and only one solution.
I guess that means that it's our job to find it now.
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For each equation, the "mx + b" form is equal to 'y' . Since these two things are
equal to the same thing, they must be equal to each other, and we can write:
-1/3 x + 3 = 3x - 7
Multiply each side by 3 : -x + 9 = 9x - 21
Add 'x' to each side: 9 = 10x - 21
Add 21 to each side: 30 = 10x
Divide each side by 10 : 3 = x
The intersection/solution is some place where x=3 .
Let's put that back into the first equation:
x + 3y = 9
3 + 3y = 9
Subtract 3 from each side: 3y = 6
Divide each side by 3 : y = 2
And there's your solution: x = 3
y = 2
On the graph, the two lines intersect at the point (3, 2) .
We used the first equation to get part of the solution, so we can't use
the same equation to check the solution. We'll put our solution into the
second equation, and see whether it checks there:
3x - y = 7
3(3) - (2) = 7
9 - 2 = 7
7 = 7 yay !
The two equations have one and only one solution,
and it is definitely x = 3, y = 2 .
