Algebra help?

Question

I am having trouble trying to figure out these 4 problems. I have tried to work them out myself, but it is not going to well for me.

For all of the equations / stands for division, not fraction.

x/3 + y/4 = 5
x + 3y/8 = 12

1/s + 3/t = 7/9
1/s – 1/(2t) = 0

3x + 5y - 30 = 0
x - y - 2 = 0

Jim has 18 coins consisting dimes and quarters that total $3.45. Find how many coins of each kind does she have? (8 points)

Answer 1

ah - simultaneous equations!

My favourite!

(1) x/3 + y/4 = 5
(2) x + 3y/8 = 12

if you multiply (1) X 12
you get
4x + 3y = 60

if you multiply (2) X 8 you get
8x + 3y = 96
if you subtract (1)X12 from (2)X8

8x + 3y = 96
4x + 3y = 60
-----------------
4x = 36
x = 9

Then you put the x in one of the equations and get y


(1) 1/s + 3/t = 7/9
(2) 1/s - 1/(2t) = 0

if you manipulate (2) you get 1/s = 1/(2t)
so you can substitute this into (1)

1/(2t) + 3/t = 7/9

1/(2t) + 6/(2t) = 7/9
7/(2t) = 7/9
t = 9/2

Substitute this back into (1) to find s


(1) 3x + 5y - 30 = 0
(2) x - y - 2 = 0

you could use elimination or substitution for this one, but I think I'd use substitution and manipulate (2) so as to get x = y+2

then put this into equation (1)
3(y+2) + 5y -30 = 0
so 3y + 6 + 5y = 30
8y = 24
y = 3

substitute this back into one of the equations to find x


Jim has 18 coins
let's say there are d dimes and q quarters

(1) d + q = 18 because he has 18 coins in total

(2) 10d + 25q = 345 because they total $3.45 in value

d = 18-q if you manipulate equation (1)

substitute this into (2)

10(18-q) + 25q = 345
180 - 10q +25q = 345
180 + 15q = 345
15q = 345 - 180
q = 165/15
q = 11

Now we can find the number of dimes he has

Answer 2

I can help you with the first three. Use substitution and get either Y or X by itself like in the last equation:

3x+5y-30=0
x-y-2=0 (move the -y and -2 to the other side of the equation, which gives you X= y+2)

Now plug (Y+ 2) in for X in the above equation
3(Y+2)+ 5Y Now you are able to solve just for Y. Distribute the 3 (3 x Y + 3 x 2) now u get this-- (3Y+6+5Y-30) combine like terms. (8Y-24=0). Solve for Y.
8Y= 24 (divide by 8)
Y= 3
Place the value you have for Y into the equation and solve for X.
3X+5(3)-30= 0 (distribute 5) (5 x 3)
3X + 15 - 30= 0 (combine like terms and move to other side making negatives positive)
3X= 15 (divide by 3)
X= 5

Got it

Answer 3

OK, let's start with the first one. These are fairly simple.

What I think is best for these annoying fraction problems where you have a lot of division going on is to just multiply through by the denominators. So for example let's take the first problem:

x/3 + y/4 = 5
x + 3y/8 = 12
Let's start with the first equation. One of the denominators is 3, so multiply through by 3:
x + 3y/4 = 15
Now the only denominator you have left is 4, so multiply through by that:
4x + 3y = 60

Great, now you have a much-less complicated-looking equation. Work on the second equation in that first problem by multiplying through by 8, which is one of the denominators:

8x + 3y = 72

OK. Now what are the equations you have? Let's write 'em down:

4x + 3y = 60
8x + 3y = 72

Now what do you do? Well, you want to add the equations in such a way that one of the variables (x or y) disappears. How can you do that? Well, look at the equation: if you add them now, 3y + 3y makes 6y, but if you SUBTRACT them, 3y - 3y makes... ZERO! which means you'd have gotten rid of the y term! So subtract the equations:

-4x = -12
Divide through by -4:
x = 3

There you go. Now all you have to do is sub that value back into one of your upper equations, let's choose the first one:

3/3 + y/4 = 5
1 + y/4 = 5
Subtract 1 from both sides.
y/4 = 4
Multiply through by your denominator, 4:
y = 16

There you go! x = 3 and y = 16!


OK, what about the second one? Well, here you just have to notice something: that 1/s shows up in both the equations. That means if you can get it by itself in one equation, you can substitute that value into the other equation. So let's take the second equation:

1/s - 1/(2t) = 0
Add 1/(2t) to both sides.
1/s = 1/(2t)

Well, now that you know 1/s is the same thing as 1/(2t), you can put 1/(2t) in for 1/s in the FIRST equation!

1/(2t) + 3/t = 7/9
OK, now just multiply through by the denominators like I showed you. Multiply through by 2t in this case:
1 + 6t/t = 14t/9
7 = 14t/9
Then multiply through by 9:
63 = 14t
So t = 4.5, or 9/2. Then you can sub this value back into the second equation to get a value for s:

1/s - 1/9 = 0
1/s = 1/9
s = 9

Now let's tackle the third equation. What you want to do in this case, where there are no evil fractions or anything, is just eliminate one of the variables - that means when you add the equations, one of the variables should add to 0. Right now, if you look at the equations, you have 3x and x, which would sum to 4x, and you also have 5y and -y, which would sum to 4y. So that doesn't work. But what if you multiply the second equation by 5? Then you'd have 5y and -5y, which would sum to 0, which is what you want. So let's do that.

x - y - 2 = 0
Multiply through by 5:
5x - 5y - 10 = 0
Now sum them!

3x + 5y - 30 = 0
5x - 5y - 10 = 0
Add each column....
8x - 40 = 0
Now add 40 to both sides, then divide through by 8.
8x = 40
x = 5

There's your answer! Sub it back into the second equation to get a value for y:

5 - y - 2 = 0
3 - y = 0
y = 3

Hope I helped!

Answer 4

1) 4x + 3y = 60 { Multiplied whole thing by 12 }
8x + 3y = 96 { Multiplied whole thing by 8 }
-------------------
4x = 36 { Subtracted first from second, and y's canceled out }
x = 9
y = 8

2) 1/s + 3/t = 7/9
1/s - 1/(2t) = 0
-----------------------
3/t + 1/(2t) = 7/9 { Subtracted second from first }
54 + 9 = 14t { Multiplied by 18t }
63 = 14t
t = 3.5
s = 7

3) 3x + 5y = 30
5x - 5y = 10 { Multiplied by 5 }
------------------
8x = 40
x = 5
y = 3

4) d + q = 18
10d + 25q = 345

10d + 25q = 345
10d + 10q = 180
-------------------------
15q = 165
q = 11
d = 7

Answer 5

For increasing cubic binomials the final formula is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D