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Intersection is not always in the form of y = mx + b, this is a linear equation involving one variable (the x, or in our case the number of jumps). We could also have a linear equation with 2 variables, we could also have a mixture of linear and non-linear equations (such as an exponential equation and a linear equation and finding their point of intersection).
Honestly the only shorcut is improving upon math, otherwise you'll be limited by trying to remember certain solutions to equations and when to apply them, but you'll rarely know when NOT to apply them, or which ones to apply for a problem that's a bit more complex :(
A bit more explanation for this one, the Kangaroos jump at a linear rate (their veolcity/jump movement never changes, it's always v1 or v2). Therefore we have something along the lines of:
y = velocity * jumps + startLocation
We are already given velocity AND start location for each kangaroo. let's abbreviate using the actual equation variables (x1,x2,v1,v2), we get two separate equations for y (the total distance travelled after J jumps)
y = v1 * J + x1
y = v2 * J + x2
We want to set these equal to each other, this would give us a distance that they both share when their J variable's are equal to one another:
v1 * J + x1 = v2 * J + x2
And using our algebra we get the EQ described above in the other posters solution:
v1 * J - v2 * J + x1 - x2 = 0
Which can be written as: J(v1-v2) = x2 - x1 or (x2-x1) / (v1-v2) = J
So given 2 velocity values and 2 starting positions,
like 4 = v1 and 2 = v2 and 0 = x1 and 4 = x2, we can evaluate:
(4 - 0) / (4-2) = J
4 / 2 = J
J = 2 (so at 2 jumps we should intersect)
Showing the values below of the jumps this is right (the far left value is the starting, arrows point to the next jump location):
0 -> 4 -> 8
4 -> 6 -> 8
And then of course, as the above states, we need to make sure the jump total is a whole number, otherwise we're intersecting in the air. Which is fine, but this problem doesn't really allow that. After all, given any 2 kangaroos if the "BACK" kangaroo is moving FASTER than the one in front of it, they will eventually collide. It's like 2 cars and the one in back is moving faster, if it doesn't slow down it will hit the one in front of it!
Number Line Jumps
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Intersection is not always in the form of y = mx + b, this is a linear equation involving one variable (the x, or in our case the number of jumps). We could also have a linear equation with 2 variables, we could also have a mixture of linear and non-linear equations (such as an exponential equation and a linear equation and finding their point of intersection).
Honestly the only shorcut is improving upon math, otherwise you'll be limited by trying to remember certain solutions to equations and when to apply them, but you'll rarely know when NOT to apply them, or which ones to apply for a problem that's a bit more complex :(
A bit more explanation for this one, the Kangaroos jump at a linear rate (their veolcity/jump movement never changes, it's always v1 or v2). Therefore we have something along the lines of:
y = velocity * jumps + startLocation
We are already given velocity AND start location for each kangaroo. let's abbreviate using the actual equation variables (x1,x2,v1,v2), we get two separate equations for y (the total distance travelled after J jumps)
We want to set these equal to each other, this would give us a distance that they both share when their J variable's are equal to one another:
v1 * J + x1 = v2 * J + x2
And using our algebra we get the EQ described above in the other posters solution:
v1 * J - v2 * J + x1 - x2 = 0 Which can be written as: J(v1-v2) = x2 - x1 or (x2-x1) / (v1-v2) = J
So given 2 velocity values and 2 starting positions, like 4 = v1 and 2 = v2 and 0 = x1 and 4 = x2, we can evaluate:
Showing the values below of the jumps this is right (the far left value is the starting, arrows point to the next jump location):
And then of course, as the above states, we need to make sure the jump total is a whole number, otherwise we're intersecting in the air. Which is fine, but this problem doesn't really allow that. After all, given any 2 kangaroos if the "BACK" kangaroo is moving FASTER than the one in front of it, they will eventually collide. It's like 2 cars and the one in back is moving faster, if it doesn't slow down it will hit the one in front of it!