math help!

[250r rider 88]

ok dont ask me why i cant seem to figure the equation out but i know the answers but i must show the equations to get the answers any help is appreciated:

Problem: A jar has $8.75 in nickels, quarters, dimes, there are 2 times as many dimes as there are nickels, and 10 more dimes than quarters. How many of each coin is in the jar?

Now without the equations i get 15 Q, 25 D, and 50 N which works out but i cant for the life of me remember how to set up the linear equation, hopefully someone smarter than me can help

thanks

[tgreenz400]

dude you messed up... their cant be 25 dimes cuz theirs supposed to be more dimes than nickels;)

[250r rider 88]

haha ur right i did ok scratch my answers, i still cant seem to get the equation been at this for over an hour

[tgreenz400]

well id help ya out but its been a while since ive dealt with that and i cant remember how to show equations for my life:p

[250r rider 88]

hey np thanks for trying im in the same boat been awhile since ive had to do these kind of problems and just plum forgot how they work hopefully someone on here can get me and answer

[tgreenz400]

Heres what i found

Complicated linear equations, such as the ones above, can be rewritten using the laws of elementary algebra into several simpler forms. In what follows, capital letters represent constants (unspecified but fixed numbers), while x and y are the variables.

General form

where A and B are not both equal to zero. The equation is usually written so that A ¡Ý 0, by convention. The graph of the equation is a straight line, and every straight line can be represented by an equation in the above form. If A is nonzero, then the x-intercept, that is the x-coordinate of the point where the graph crosses the x-axis (y is zero), is −C/A. If B is nonzero, then the y-intercept, that is the y-coordinate of the point where the graph crosses the y-axis (x is zero), is −C/B, and the slope of the line is −A/B.
Standard form

where, A and B are not both equal to zero and A, B, and C are integers whose greatest common factor is 1. As above, A ¡Ý 0. The standard form can be converted to the general form.
Slope-intercept form

where m is the slope of the line and b is the y-intercept, which is the y-coordinate of the point where the line crosses the y axis. This can be seen by letting x = 0, which immediately gives y = b.
Point-slope form

where m is the slope of the line and (x1,y1) is any point on the line. The point-slope and slope-intercept forms are easily interchangeable.
The point-slope form expresses the fact that the difference in the y coordinate between two points on a line (that is, y − y1) is proportional to the difference in the x coordinate (that is, x − x1). The proportionality constant is m (the slope of the line).
Occasionally point-slope form appears as the following:

However, in this form x = x1 does not satisfy the equation.
Intercept form

where E and F must be nonzero. The graph of the equation has x-intercept E and y-intercept F. The intercept form can be converted to the standard form by setting A = 1/E, B = 1/F and C = 1.
Two-point form

where p ¡Ù h. The graph passes through the points (h,k) and (p,q), and has slope m = (q−k) / (p−h).
The two-point form is easily had from the point-slope form by inserting an explicit calculation of the slope in place of m
Parametric form

and

Two simultaneous equations in terms of a variable parameter t, with slope m = V / T, x-intercept (VU−WT) / V and y-intercept (WT−VU) / T.
This can also be related to the two-point form, where T = p−h, U = h, V = q−k, and W = k:

and

In this case t varies from 0 at point (h,k) to 1 at point (p,q), with values of t between 0 and 1 providing interpolation and other values of t providing extrapolation.
Normal form

where ¦Õ is the angle of inclination of the normal and p is the length of the normal. The normal is defined to be the shortest segment between the line in question and the origin. Normal form can be derived from general form by dividing all of the coefficients by and if C > 0 multiply all coefficients by -1 to have the last constant negative. This form also called Hesse standard form, named after a German mathematician Ludwig Otto Hesse.
Special cases

This is a special case of the standard form where A = 0 and B = 1, or of the slope-intercept form where the slope M = 0. The graph is a horizontal line with y-intercept equal to F. There is no x-intercept, unless F = 0, in which case the graph of the line is the x-axis, and so every real number is an x-intercept.

This is a special case of the standard form where A = 1 and B = 0. The graph is a vertical line with x-intercept equal to E. The slope is undefined. There is no y-intercept, unless E = 0, in which case the graph of the line is the y-axis, and so every real number is a y-intercept.
and
In this case all variables and constants have canceled out, leaving a trivially true statement. The original equation, therefore, would be called an identity and one would not consider the graph (it would be the entire xy-plane). An example is 2x + 4y = 2(x + 2y). The two expressions on either side of the equal sign are always equal, no matter what values are used for x and y.
Note that if algebraic manipulation leads to a statement such as 1 = 0, then the original equation is called inconsistent, meaning it is untrue for any values of x and y. An example would be 3x + 2 = 3x − 5.

In addition, there may be more than two variables in the equation or several simultaneous equations. For more information see System of linear equations.


[edit] Connection with linear functions and operators
In all of the named forms above (assuming the graph is not a vertical line), the variable y is a function of x, and the graph of this function is the graph of the equation.

In the particular case that the line crosses through the origin, if the linear equation is written in the form y = f(x) then f has the properties:


and


where a is any scalar. A function which satisfies these properties is called a linear function, or more generally a linear map. This property makes linear equations particularly easy to solve and reason about.

Linear equations occur with great regularity in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state.


[edit] Linear equations in more than two variables
A linear equation can involve more than two variables. The general linear equation in n variables is:


In this form, a1, a2, ¡_, an are the coefficients, x1, x2, ¡_, xn are the variables, and b is the constant. When dealing with three or fewer variables, it is common to replace x1 with just x, x2 with y, and x3 with z, as appropriate.

[250r rider 88]

ok now i htink i have the correct answers: 15N 30D 20Q just still dont understand the equation part

[250r rider 88]

wow nice find but didnt make a bit of sense to me :ermm:

[tgreenz400]

haha Yea me either. well maybe you should try looking at your text book...if you dont have one im sure that theirs an online version for yours

[250r rider 88]

nope no text book for this class, its a MET lab and it has a manual which is basically just problems no help whatsoever:ermm:

[Ghost-Rider]

Just show how your converted the coins ?

[250r rider 88]

how?

[tgreenz400]

Originally posted by Ghost-Rider
Just show how your converted the coins ? please demenstrate (sp)...

[Ghost-Rider]

10 pennies = 1 dime, 10 dimes = $1 etc..

[rYan01400EX]

Hey man,

I got the solution.


The solution is:
15 Nickels
30 Dimes
20 Quarters

I got this from trial and error. As far as the equation goes, I've tried and tried and tried. But, I hope with the answer to the problem it'll push you in the right direction!

Originally posted by 250r rider 88
ok dont ask me why i cant seem to figure the equation out but i know the answers but i must show the equations to get the answers any help is appreciated:

Problem: A jar has $8.75 in nickels, quarters, dimes, there are 2 times as many dimes as there are nickels, and 10 more dimes than quarters. How many of each coin is in the jar?

Now without the equations i get 15 Q, 25 D, and 50 N which works out but i cant for the life of me remember how to set up the linear equation, hopefully someone smarter than me can help

thanks

I can remember doing something like this in Geometry...and let me tell you. It sucked.

It's right at the back of my head...Give me a bit to think about it, and I'm sure I'll remember.

[GPracer2500]

Here's the math. The tricky part is setting up the equation so you're only using one variable. You can start with this...

8.75 = .25Q + .10D + .05N

But then you need to make substitutions so you're only using one variable....

Twice as many dimes as nickels can be written as N = D/2.

And ten fewer dimes than quarters can be written as Q = D - 10.

Then you substitute for N and Q in the first equation and you get this....

8.75 = (.25)(D-10) + .10D + (.05)(D/2)

--and solve--

8.75 = .25D - 2.5 + .10D + .025D
11.25 = .375D
D = 30

Then take that answer for D and put it back into N = D/2 and Q = D-10 and you get....

D = 30
N = 15
Q = 20

$3.00 + $0.75 + $5.00 = $8.75

[rYan01400EX]

Man GP, you're good. I suck at math.

Originally posted by GPracer2500
Here's the math. The tricky part is setting up the equation so you're only using one variable. You can start with this...

8.75 = .25Q + .10D + .05N

But then you need to make substitutions so you're only using one variable....

Twice as many dimes as nickels can be written as N = D/2.

And ten fewer dimes than quarters can be written as Q = D - 10.

Then you substitute for N and Q in the first equation and you get this....

8.75 = (.25)(D-10) + .10D + (.05)(D/2)

--and solve--

8.75 = .25D - 2.5 + .10D + .025D
11.25 = .375D
D = 30

Then take that answer for D and put it back into N = D/2 and Q = D-10 and you get....

D = 30
N = 15
Q = 20

$3.00 + $0.75 + $5.00 = $8.75

God...I hated Geometry. In my opinion, Algebra 2 was easier.

[250r rider 88]

thanks for the help guys much appreciated

[Raptor68]

Originally posted by MrMan
God...I hated Geometry. In my opinion, Algebra 2 was easier.

That isn't Geometry...

[QuadRider55]

Originally posted by MrMan
God...I hated Geometry. In my opinion, Algebra 2 was easier.

im glad thats easy


agebra 1 is HARD!!!!

[Ace Sixx]

Originally posted by MrMan
God...I hated Geometry. In my opinion, Algebra 2 was easier.

*psst* (that *is* algebra)