substitution method and linear combination method

Well, if
6b+2t=232 and
13b+4t=487
the difference of 255$ is in 7 bushes and 2 trees.
Compare that to the price of six trees and two bushes, and you've found the price for a single bush.

Conclusion: The problem is too simple to use complex math like combi...linear...stution-whatzit.

Linear combination simply means setting up the equation in the form: 13 bushes + 4 trees = $487

Substitution means solve one equation for one variable and substitute that answer into the second equation.

"Linear combination simply means setting up the equation in the form: 6 bushes + 4 trees = $487"

That's the linear part. The combination part is when you combine them to eliminate a variable. In this case, the easiest way is to multiply the first equation by 2 to get:
12b+4t=464
13b+4t=487
then subtract the first from the second to get:
1b+0t=23 or b=23

Or you could use cramer's rule. It finds where the two lines meet, and that's what you're looking for. It works like this with two equations in the form of ax+by=c
first you find the matrices which represent those equations, so you take
a[size=7]1[/size]x+b[size=7]1[/size]=c[size=7]1[/size]
a[size=7]2[/size]x+b[size=7]2[/size]=c[size=7]2[/size]
and get something like
|a[size=7]1[/size] b[size=7]1[/size]||x|_|c[size=7]1[/size]|
|a[size=7]2[/size] b[size=7]2[/size]||y|=|c[size=7]2[/size]|
then from there, you can take two different routes. You could find the inverse of the first matrice, but that would be a bit complex and probably not what you're looking for. Instead, you could use Cramer's Rule. To do this, you go
__|c[size=7]1[/size] b[size=7]1[/size]|
x=|c[size=7]2[/size] b[size=7]2[/size]|
__|a[size=7]1[/size] b[size=7]1[/size]|
__|a[size=7]2[/size] b[size=7]2[/size]|

__|a[size=7]1[/size] c[size=7]1[/size]|
y=|a[size=7]2[/size] c[size=7]2[/size]|
__|a[size=7]1[/size] b[size=7]1[/size]|
__|a[size=7]2[/size] b[size=7]2[/size]|
To evaluate those funky-lookin box things in the form of
__|a[size=7]1[/size] b[size=7]1[/size]|
__|a[size=7]2[/size] b[size=7]2[/size]|
you take a[size=7]1[/size]b[size=7]2[/size]-a[size=7]2[/size]b[size=7]1[/size]
so that's how you do it. applying it to this problem, we go
6b+2t=232
13b+4t=487
__|6_ 232|
x=|13 487|
__|6 _2|
__|13 4|

__|232_2|
y=|487 4|
__|6 _2|
__|13 4|
giving x= -354/-2 = 177 and y=-46/-2 = 23
Keep in mind that this is a lot more work than it's worth at this point. With only those 2 terms, it takes a while, but it's worth knowing for solving equations with higher numbers of variables, and of course, it changes a little bit with every extra variable. Man, if anyone wants to quote that to check out how much html that was, be my guest. And also, those little white lines are meant to be spaces as i haven't learned how to do invisitext yet and the spaces wouldn't stack up at all.