To solve systems of equations by graphing, just simplify the equations to be in slope intercept form (y = mx + b), and then graph them. Finally, find the intersection point... and you have your variable values. Easy... right?
|Algebra||Represent and solve equations and inequalities graphically|
|Systems of Equations||Graphing|
cryptic equations leading the way there.
Graph the equations, and they'll provide you with the coordinates to sugar overload.
Here's the scrap of paper your friend gave you.
We'll tackle the equations by changing them to slope-intercept form first...
Let's start with the top equation...
negative-three-x plus y equals six. You can do this one without sugar and caffeine
coursing through your veins.
Just add three-x to both sides. Doing that, we see that y equals three-x plus six.
The second one is slightly trickier. But if you can mix Coke and Pepsi until it tastes
like Dr. Pepper, this is nothing.
First, subtract x from both sides, giving us two-y equals negative x minus 2.
Then just divide all the terms by two.
We end up with y equals negative one-half x minus 1.
Now we just have to graph them.
We'll do the first equation in blue.
The y-intercept is 6, so we can plot a point at zero-six, which is six up the y-axis.
Because we know slope is rise over run, for every one we run or move to the right along
the x axis, we rise, or move three up the y-axis.
Reversing this, we move three spaces down the y-axis for every one we move left along
The blue line will intersect the x-axis at negative-two, zero.
We'll do the second equation in red.
The y-intercept is negative 1, so we can plot a point at zero, negative-one on the y-axis.
For every one we run, or move right along the x-axis, we'll move one-half down.
Flipping that, we'll move 1/2 up for every one we move left along the x-axis.
This line, too, intersects the x-axis at negative-two, zero.
So that's where the Black's Market is.
At (-2, 0).
You may have to take the subway there, but we're pretty sure you'll be able to run back
home on a pure sugar rush.