Pretty simple math question?

Question

ok so I have a test tommorow that I'm kinda ****** for.
anyways what's going to be on the test is solving equations in many differnt ways

the problems will be ones like

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

on top of eachother and I have to solve them using
graphing
substitution
addition
muiltiplication and addition
finding out how many solutions the equation has
whether the graphs of these systems of equations have 0,1 or infinitely points of intersection

if you could help by giving examples that would be great to.
thanks

Answer 1

Best Answer - Chosen by Asker

haha, we are doing this stuff right now.
umm, so...for graphing:
you take both, and make it so that it is in the form y equals.
using the example u have:
2x-y=5 would be y=2x-5
3x+2y=18 would be y=(18-3x)/2
you can just punch it into your graphing calculator...and find the intersect, lol. just remember to use your brackets...or else your graph will be wacked.

lemme get some examples from my homework i just finished, haha.

so...you have 3 options...this is a summary of what i have in my notes:

a linear system is: y1 = m1 x + b1
y2 = m2 x + b2

if m1 does not equal m2, there is one solution, and it is a consistant system - or one intersection. ex. 2 lines crossing each other once

if m1 does equal m2, but b1 does not equal b2, there are no solutions, and is an inconsistant system - or there is no intersection. ex. two parallel lines.

if m1 does equal m2 and b1 equals b2, there are infinate solutions, and it is consistant - or, it is the same line.

so for example:
x-2 = -3 goes to y=(-x-3)/-2 or y= x/2 + 3/2
2x-4y = -6 goes to y= (-6-2x)/-4 or y= 1/2 + 3/2
m1 = m2, and b1 = b2, so there are infinate solutions, and it is consistant.

now, if you want to solve by substitution...some key points are:
- it works best when you can find a variable by itself
- you need to isolate the variable if it is not by itself

example:
2x+y = 4 goes to y=4-2x
3x+2y = -3 goes to 3x+2(4-2x) = -3 *since y=4-2x, you substitute y for (4-2x)*
3x +8 -4x = -3
8-x = -3
-x = -5
x = 5
then substitute x in the equation:
y = 4-2x
y = 4-2(5)
y = 4-10
y = -6
so, your answer is, that there is one intersection, and the point is (5,-6)

lets do one that has two different intersections:
y=x squared
y=x-2
so, if they intersect, then at some point, both y's would equal each other.
so, you make them equal to each other, lol.
x squared = x-2 *then put them all on one side*
x squared -x +2 = 0 *then factor*
(x-3)(x-2)=0
so x = 3,2
then substitute x in the equation for both x's:
y=x+6
y=3+6
y=9

y=x+6
y=2+6
y=8
so, your answer is two different intersections....and they are (3,9) and (2,8)

I hope this helps!!!!

Source(s):

I'm in pure math 20...(grade 11)

Asker's Rating:

Asker's Comment:

thanks a lot

Answer 2

Substitution!
2X- Y=5
3X-2Y=18

2x-y-5=0.........(1)
3x-2y-18=0.....(2)

(1)x -2 -4x+2y+10
3x-2y -18

-x = -8
x=8

(1) 2(8)-y-5
16-y-5=0
-y=5-16
-y=-9
y=9