Solve the system using any method?!

[flora33]

Here is yet another one. I can use any method substitution or addition, I've tried both and I'm not sure if I'm getting the right answer...

x = 34y + 7
4x - 3y = 29

I assume the substitution method makes most sense, because the first equation is already solved for x:

4(34y + 7) - 3y = 29
136y + 28 - 3y = 29
133y + 28 = 29
133y + 28 - 28 = 29 - 28
133y = 1
133y/133 and 1/133
y = 1/133??

If I check:
4(34(1/133) + 7) - 3(1/133) = 29
4(265/33) - 3(1/133) = 29
1060/33 - 3/133 = 29
32.09865573 = 29

Wow, ok, as long as I did all the calculations correct- would this be considered an inconsistent system???

 

[tkhunny]

1) x = 34y + 7
2) 4x - 3y = 29
3) 4(34y + 7) - 3y = 29
4) y = 1/133

3) is excellent.
4) is exactly correct.

When you "checked" 4) in 3), you should have achieved 0 = 0. You just solved a single-variable equation "3)" and there should be no confusion on the result. This has little to do with the system in its entirety. The fact that you managed something inconsistent means only that your arithmetic was faulty. Try that again and be VERY careful. Note: Please don't switch to decimals. Learn to deal with the fractions.

If you wish to solve the system, substitute 4) into 1), then check both values in both equations. Note: Please don't switch to decimals. Learn to deal with the fractions.

 

[flora33]

Re: Solve the system using any method?! EDITED!!

1) x = 34y + 7
2) 4x - 3y = 29
3) 4(34y + 7) - 3y = 29
4) y = 1/133

3) is excellent.
4) is exactly correct.

When you "checked" 4) in 3), you should have achieved 0 = 0. You just solved a single-variable equation "3)" and there should be no confusion on the result. This has little to do with the system in its entirety. The fact that you managed something inconsistent means only that your arithmetic was faulty. Try that again and be VERY careful. Note: Please don't switch to decimals. Learn to deal with the fractions.

If you wish to solve the system, substitute 4) into 1), then check both values in both equations. Note: Please don't switch to decimals. Learn to deal with the fractions.

Wow, I feel really dumb. I just looked back at the problem- and I solved for the WRONG problem! It should have been:
x =3/4y + 7
4x - 3y = 29

4(3/4y + 7) - 3y = 29
3y + 28 - 3y = 29
28 = 29

Ok, so would THIS equation be considered an inconsistent system since the y terms drop out? According to my textbook "the systems are independent and the system is inconsistent. Since the system has no solution, the graphs of the equation in the system will be parallel."

 

[tkhunny]

Put the second equation in slope-intercept form and PROVE that the two lines are parallel.

 

[flora33]

Put the second equation in slope-intercept form and PROVE that the two lines are parallel.

Original problems:
x =3/4y + 7
4x - 3y = 29

I'm not quite sure how to prove that the two lines are paralell, but I will put the second equation in slope-intercept form:
4x - 3y = 29 -4x
3y = -4x + 29
3y/3 = -4x/3 + 29/3
y = -4/3x + 29/3
m = -4/3
Would I do the first equation as well?
x =3/4y + 7
3/4y = -1x + 7
3/4y/3/4 = -1x/3/4 + 7/3/4
y = -1/3x +7/3
m = -1/3

So, since the slope of each equation is not the same, these lines are not paralell and they are also not perpindicular because they are not negative recriprocals...

 

[tkhunny]

That does present a problem. Your original conclusion says they are parallel and now you are saying they are not. There must be something wrong.

Try that second equation again. This time, be more careful.

x =3/4y + 7

-(3/4)y = -x + 7 <== You missed a sign, there.

 

[flora33]

That does present a problem. Your original conclusion says they are parallel and now you are saying they are not. There must be something wrong.

Try that second equation again. This time, be more careful.

x =3/4y + 7

-(3/4)y = -x + 7 <== You missed a sign, there.

Ok, I see:
x =3/4y + 7
-3/4y = -1x + 7
-3/4y/-3/4 = -1x/-3/4 + 7/-3/4
y = -4/3x +-28/3
m = -4/3

I know I really have to watch those signs!

 

[Deleted member 4993]

Put the second equation in slope-intercept form and PROVE that the two lines are parallel.

Original problems:
x =3/4y + 7
4x - 3y = 29

I'm not quite sure how to prove that the two lines are paralell, but I will put the second equation in slope-intercept form:
4x - 3y = 29 -4x <<<< Where did this come from?
3y = -4x + 29<<<< Does not follow from above
3y/3 = -4x/3 + 29/3
y = -4/3x + 29/3
m = -4/3<<<< Follows from above - but incorrect. correct answer m = +4/3 (I put the extra + sign to emphasize)


Would I do the first equation as well?
x =3/4y + 7
3/4y = -1x + 7<<<< Does not follow from above
3/4y/3/4 = -1x/3/4 + 7/3/4
y = -1/3x +7/3<<<< Does not follow from above
m = -1/3

Last time I told you, I would have given you 90-95% of the points. This time, however, you would get ~10% from me. You are making careless mistakes (most dangerous kind) - not only with "signs" - but algebraic manipulations also.

So, since the slope of each equation is not the same, these lines are not paralell and they are also not perpindicular because they are not negative recriprocals...

 

[Deleted member 4993]

That does present a problem. Your original conclusion says they are parallel and now you are saying they are not. There must be something wrong.

Try that second equation again. This time, be more careful.

x =3/4y + 7

-(3/4)y = -x + 7 <== You missed a sign, there.

Ok, I see:
x =3/4y + 7
-3/4y = -1x + 7
-3/4y/-3/4 = -1x/-3/4 + 7/-3/4
y = -4/3x +-28/3<<<< sign mistake again. (negative)/(negative) is a (positive)
m = -4/3 <<<<< correct answer m = +4/3

I know I really have to watch those signs!