Thread: please help me solve the following (show the work done):

5

seriously.....these are the last 3 problems on my take home test for my college algebra class.....i'm not comprehending it at all.....google and youtube haven't helped and neither has the notes from class or the practice test with answers and work

If you don't understand how to do the work.. why should you earn the marks on a test for it.

Originally Posted by Muscle_Girl

If you don't understand how to do the work.. why should you earn the marks on a test for it.

it's either that or the military is going to pay for it again......i only have this one math class for my degree......all my other classes i'm totally fine in......also if i'm not understanding it (along with the majority of my class) then maybe it's not me and it's the instructor?......is the school going to reimburse me for not passing then?

It's a simple 2 equations two unknowns. The substitution method takes longer. You basically solve for one in terms of the other and then substitute. Addition method is a quick short cut but there are ruels. There are internet sites that can help you.

Originally Posted by Glycomann

It's a simple 2 equations two unknowns. The substitution method takes longer. You basically solve for one in terms of the other and then substitute. Addition method is a quick short cut but there are ruels. There are internet sites that can help you.

ya, i've been viewing youtube and other math site all weekend.....i've figured out the 1st and 3rd problem, but i'm not seeing anything that will help me show my work/check my work.....i'm assuming i just plug the answer in and redo the original problem?.......i also think that for the addition method i need to find common denominators and then use recipricals to isolate and solve for x or y first and then go back and do it again for the other one? I have til wednesday nights class to figure it out

Dude this shit is easy if you don't figure it out now you'll forever fall behind in harder math classes and when you start physics and other science courses. I thought I would never use trig and somehow the stars have aligned so that I have to use it often in my job to find optimum back support heights for tilting solar arrays.....

Trig is used to make large amounts of meth

Originally Posted by maniclion

Dude this shit is easy if you don't figure it out now you'll forever fall behind in harder math classes and when you start physics and other science courses. I thought I would never use trig and somehow the stars have aligned so that I have to use it often in my job to find optimum back support heights for tilting solar arrays.....

i don't have any of those other classes....i'm only doing a 2 year associates in radiography

Originally Posted by irish_2003

i don't have any of those other classes....i'm only doing a 2 year associates in radiography

Don't ever short yourself man, ManicLion's Strategies For Being Successful While Living On The Fringe....if you are given an opportunity to learn from someone take it cause later on you'll have to do it yourself.....

i will show you later tonight or in the morning.

Can it wait until tomorrow ?

L1: -9x + 6y=2
L2: 3x - 2y=10


sub: -9x = -6y+2---> x=(6y+2)/9
L2: 3x-2y=10---->3(6y-2)/9 - 2y = 10----->18y-6-2y=90---->16y=96, y=6

i think.....looks right. its been 10 yrs since ive done something like this.

if its really important PM me with an email address and i will write it out in more detail and scan it along with the rest of the problems. it will be tomorrow morning though, i have class myself

Originally Posted by ROID

L1: -9x + 6y=2
L2: 3x - 2y=10


sub: -9x = -6y+2---> x=(6y+2)/9
L2: 3x-2y=10---->3(6y-2)/9 - 2y = 10----->18y-6-2y=90---->16y=96, y=6

i think.....looks right. its been 10 yrs since ive done something like this.

if its really important PM me with an email address and i will write it out in more detail and scan it along with the rest of the problems. it will be tomorrow morning though, i have class myself

thanks......i got the answers done.....it was a take home test since we missed a week of class due to instructor being sick......i just now have to plug the answers into the original problems and show me checking my work....it's due wed nights class

Addition method prob 2

L1: 3x - 2y=4
L2: -5x + y= -3

You need to match the coefficients for one of the variables. I chose the x variable. To do so, I multiplied L1 by 5 and L2 by 3. This results in:

L1':15x - 10y = 20
L2'-15x + 3y = -9

You can now add the equations together.

0x + (-7y) = 11
-7y =11

Now solve for y:
-7y = 11
y = -11/7

You now plug the solution for y into either equation and solve
3x - 2y = 4
3x - 2(-11/7) = 4
3x + 22/7 = 4
3x = 28/7 - 22/7
3x =6/7
x = 2/7

Your solutions are:
x = 2/7
y = -11/7

to check, plug the values into an equation:

3x - 2y = 4
3(2/7) - 2(-11/7) = 4

6/7 - (-22/7) 4

6/7 + 22/7 = 4

28/7 = 4
4=4

I don't believe their is a solution to the first problem.

Originally Posted by ROID

L1: -9x + 6y=2
L2: 3x - 2y=10


sub: -9x = -6y+2---> x=(6y+2)/9
L2: 3x-2y=10---->3(6y-2)/9 - 2y = 10----->18y-6-2y=90---->16y=96, y=6

-9x = -6y + 2

Solving for x, x = (6y-2)/9

Plugging it into L2, 3[(6y-2)/9] - 2y = 10
Here's the issue: you cannot multiply both side by 9 without taking into account the "-2y" on the left side.

(18y-6)/9 - 2y = 10
[(18y-6)/9 - 2y]9 = 10(9) This is the step where all factors on both sides of the equation need to be multiplied by 9.
18y - 6 - 18 y = 90
6 = 90
6 =/= 90

The variables annihilate each other and you are left with no solution.


Is this not correct? Irish, what answer did you arrive to?

by any method:

L1: 3x - 2y=6
L2: 4x + 3y=8

I chose the addition method. I multiplied L1 by -4 and L2 by 3 in order to eliminate x. This gives me -12x and 12x and they cancel out to 0 which turns the equation into a 1 variable equation.

L1':-4(3x - 2y) = 6(-4)
L1':-12x + 8y = -24

L2' 3(4x+3y)=8
L2':12x + 9y =24

Now I have two new equations and can eliminate x by adding the equations together. Solving gives me the solution for y.
L1': -12x + 8y = -24
L2': 12x + 9y = 24
72y = 0
y = 0

Solving for x is done by substituting the y into either equation:
3x - 2y = 6
3x - 2(0) = 6
3x = 6
x = 2

Solutions are x = 2 and y = 0

Checking the answer:

3x - 2y = 6
3(2) - 2(0) = 6
6=6

4x + 3y = 8
4(2) + 3(0) = 8
8 = 8

by the substitution method:

L1: -9x + 6y=2
L2: 3x - 2y=10

Originally Posted by Marat

I don't believe their is a solution to the first problem.

Indeed, there is none. Their slopes are equal, and their intercepts are not - thus, they are parallel.

The slope-intercept form y = mx + b illustrates this nicely:


L1 can be re-written:
6y = 9x + 2
y = (9/6)x + 2/6
y = 1.5x + 1/3
(Slope m = 1.5; y-intercept b = 1/3)
L2 can be re-written:
-2y = -3x + 10
y = 1.5x - 5
(Slope m = 1.5; y-intercept b = -5)

Note: had their intercepts also been equal, L1 and L2 would have been coincident lines - that is to say, both equations would have described the same line.

Originally Posted by irish_2003

i don't have any of those other classes....i'm only doing a 2 year associates in radiography

Dude I did a 1 1/2 year associates and had to take Trig and Precalc plus Physics if they'd had your requirements I could have skipped since I took college algebra in HS

I took AP Calc in H.S., for college credit. I'm not even that great in math. My brother got all the math genes.

Originally Posted by MDR

I took AP Calc in H.S., for college credit. I'm not even that great in math. My brother got all the math genes.

I was supposed to but my dyslexia and a teacher who didn't give a shit caused me to have to repeat Algebra....she would fly through shit and assumed that since everyone was smart in the class we got it so she would leave and let us do our homework....all I would see is a bunch of numbers mixed with letters jumbled on a page and my brain would shutdown from chaos..the next year I retook it with a new teacher and she had worked with dyslexic students when she first started so she showed me some techniques to help my brain make sense of what I was seeing...since then I graduated top of all classes from my sophomore to SR HS yrs thru the military and college....

once I got to algerbra 1 it was all downhill from there

Originally Posted by maniclion

I was supposed to but my dyslexia and a teacher who didn't give a shit caused me to have to repeat Algebra....she would fly through shit and assumed that since everyone was smart in the class we got it so she would leave and let us do our homework....all I would see is a bunch of numbers mixed with letters jumbled on a page and my brain would shutdown from chaos..the next year I retook it with a new teacher and she had worked with dyslexic students when she first started so she showed me some techniques to help my brain make sense of what I was seeing...since then I graduated top of all classes from my sophomore to SR HS yrs thru the military and college....

Sometimes a decent teacher makes all the difference. In my case I had a good teacher and my best friend was great in math. Both showed a lot of patience when going over things with me.

Originally Posted by Built

Indeed, there is none. Their slopes are equal, and their intercepts are not - thus, they are parallel.

The slope-intercept form y = mx + b illustrates this nicely:


L1 can be re-written:
6y = 9x + 2
y = (9/6)x + 2/6
y = 1.5x + 1/3
(Slope m = 1.5; y-intercept b = 1/3)
L2 can be re-written:
-2y = -3x + 10
y = 1.5x - 5
(Slope m = 1.5; y-intercept b = -5)

Note: had their intercepts also been equal, L1 and L2 would have been coincident lines - that is to say, both equations would have described the same line.

Originally Posted by danzik17

I'm embarrassed to say what level of math i'm in.

i wonder if that was intentional or an accident

Originally Posted by maniclion

Dude I did a 1 1/2 year associates and had to take Trig and Precalc plus Physics if they'd had your requirements I could have skipped since I took college algebra in HS

it's only a requirement for me because i haven't had a math class in the last 3 years (school/program requirement).....i took algebra in 8th grade in 1988 lol!

Originally Posted by MDR

Sometimes a decent teacher makes all the difference. In my case I had a good teacher and my best friend was great in math. Both showed a lot of patience when going over things with me.

i won't be embarassed if i have to take it again....i firmly believe i DO in fact have a poor teacher of the subject......i know SHE knows the material, but she's not a good teacher of it.....(also we only meet 1 day a week for 3 hours so you forget it as quick as you learn it)