Question

quick question, will fan and medal

Answer 1

what are the three ways to solve a system of equations? and how can i apply each one?

Answer 2

http://mathforum.org/library/drmath/view/61608.html

Answer 3

thank you @cj49

Answer 4

np..

Answer 5

@cj49 how can i "apply" these?

Answer 6

1st one is the substitution method

Answer 7

2x - y = 1
3x - 2y = 5

Answer 8

well i know the methods now but when it says i need to know how to "apply" them, what is it asking?

Answer 9

3x + 2y = 12
6x + 5y = 26

To make elimination work in a situation like this, you need to
multiply one of the equations by a constant factor so that you end up
in a situation you like better, i.e., with matching coefficients. In
this case, we can multiply the first equation by 2 to get

6x + 4y = 24
6x + 5y = 26

Answer 10

Inspection is the easiest method, and it's always a good idea to check
to see if you can use it before jumping in with another method. When
can you use inspection? It's easiest when one of the variables has
the same coefficient in both equations, e.g.,

3x + 2y = 12
3x + 5y = 18

To use inspection, you reason this way: "In moving from the first
equation to the second, all that changes is that we're adding 3y. On
the right side, we add 6. So it must be the case that 3y is the same
as 6, which means that y must equal 2."

In fact, all you’re really doing here is using elimination, but not
bothering to do the formal addition or subtraction of equations. If
we write the equations in the opposite order,

3x + 5y = 18
3x + 2y = 12

we can subtract the left sides and the right sides to get

(3x+5y) - (3x+2y) = (18) - (12)

3x - 3x + 5y - 2y = 6

3y = 6

Answer 11

application of graphing method
http://www.regentsprep.org/regents/math/algebra/ae3/grsys.htm