Help Fast!! Solve equation

I hit a blank. Can you guys help me with these questions?

1. 5(2a-5)+4(1-a)=3 What is a?

2. 2m-3(4-m)=2(m+1)-17 What is m?

3.

4. y(y+3)-y(y-6)=3

5. Sam Lichinsky has 4 times more dimes than quarters in his coin collection. He has 65 half dollars. The face value of his half dollars is half the face value of the dimes and quarters combined. How many dimes does he have? How many quarters does he have?

6. A bus leaves the bus terminal with y passengers. When the bus comes to the first stop, 3/5 of the passengers get off and 4 passengers get off. Write an expression in y for the number of passengers on the bus when it leaves it's first stop.

At the second stop, 1/4 of the passengers leave the bus and 9 passengers get on. If the number of passengers on the bus now is only half that when the bus left the bus terminal, find the value of y.

7. ABC is an isosceles triangle with AB = AC

If angle B = 4/5x-14degrees and angle C = 1/2x+10degrees find the value of x.

8. Solve using the method of substitution.

y=2x-1
y-x=3

9. Solve using the method of elimination

4p+2q=10
4p-3q=25

10. 5 apple pies and 3 ice-cream sundaes cost $73. An apple pie and 2 ice-cream sundaes cost $28.6. Find the cost of an apple pie and the cost of an ice-cream sundae.

Thanks you guys! :) This forum is the best forum ever for math help.

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Originally Posted by Rocher

I hit a blank. Can you guys help me with these questions?

1. 5(2a-5)+4(1-a)=3 What is a?

2. 2m-3(4-m)=2(m+1)-17 What is m?

3.

4. y(y+3)-y(y-6)=3

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1. 5(2a-5)+4(1-a)=3
=> 10a - 25 + 4 - 4a = 3 ...............expanded the brackets, do you know how to do that?
=> 10a - 4a = 3 + 25 - 4 ...............get all the a's on one side and the numbers on the other
=> 6a = 24 .......................simplify
=> a = 4 ..........................divide both sides by 6

2. 2m-3(4-m)=2(m+1)-17
=> 2m - 12 + 3m = 2m + 2 - 17
=> 5m - 12 = 2m - 15
=> 5m - 2m = -15 + 12
=> 3m = -3
=> m = -1

3. Could you rewrite this question without LaTex?

4. y(y+3)-y(y-6)=3
=> y^2 + 3y - y^2 + 6y = 3
=> 9y = 3
=> y = 1/3

Is there anything you dont understand so far?

Quote:

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Originally Posted by Rocher

5. Sam Lichinsky has 4 times more dimes than quarters in his coin collection. He has 65 half dollars. The face value of his half dollars is half the face value of the dimes and quarters combined. How many dimes does he have? How many quarters does he have?

6. A bus leaves the bus terminal with y passengers. When the bus comes to the first stop, 3/5 of the passengers get off and 4 passengers get off. Write an expression in y for the number of passengers on the bus when it leaves it's first stop.

At the second stop, 1/4 of the passengers leave the bus and 9 passengers get on. If the number of passengers on the bus now is only half that when the bus left the bus terminal, find the value of y.

7. ABC is an isosceles triangle with AB = AC

If angle B = 4/5x-14degrees and angle C = 1/2x+10degrees find the value of x.

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5. Let x be the number of quaters Sam has. Then he has 4x dimes. He also has 65 half dollars = $32.50, which is (1/2)[ 4x(0.10) + x(0.25)].

So [4x(0.10) + x(0.25)]/2 = 32.50
=> 4x(0.10) + x(0.25) = 65 ...............multiply both sides by 2
=> x[4(0.10) + 0.25] = 65 ................factored out the x (do you know how to do that?)
=> x(0.65) = 65
=> x = 65/0.65
=> x = 100

So he has 100 quarters and 4(100) = 400 dimes.

100 quarters = $25 and 400 dimes = $40. So his dimes and quarters come up to $65, which is twice $32.50 as desired.

This problem is a little intricate, if you dont get something, say so.



6. We leave the terminal with y people. At the first stop, 3/5 of y people get off, and also 4 people get off (you sure this shouldn't be "get on"?). So, we are left with y - ((3/5)y + 4) passengers.

So we have [(2/5)y - 4] passengers when the bus leaves the first stop.

At the second stop (1/4)*[(2/5)y -4] people get off and 9 get on. So now we have (3/4)*[(2/5)y -4] + 9 passengers on the bus, and we are told that this is half the number of people that were on the bus at the terminal. so we get:

(3/4)*[(2/5)y -4] + 9 = (1/2)y
=> (3/10)y - 3 + 9 = (1/2)y
=> (3/10)y - (1/2)y = 3 - 9
=> (-1/5)y = -6
=> y = 30 ......................multiplied both sides by -5


7. ABC is an isosceles triangle with AB = AC

If angle B = 4/5x-14degrees and angle C = 1/2x+10degrees find the value of x.

can you clarify the angles. do you mean 4/(5x) and 1/(2x) or do you mean (4/5)x and (1/2)x?

Quote:

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Originally Posted by Rocher

8. Solve using the method of substitution.

y=2x-1
y-x=3

9. Solve using the method of elimination

4p+2q=10
4p-3q=25

10. 5 apple pies and 3 ice-cream sundaes cost $73. An apple pie and 2 ice-cream sundaes cost $28.6. Find the cost of an apple pie and the cost of an ice-cream sundae.

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8. Solve using the method of substitution.

y=2x-1 ................(1)
y-x=3 ..................(2)

From equation (1) we see that y = 2x - 1. Substitute 2x - 1 for y in equation (2):

=> (2x - 1) - x = 3
=> 2x - x -1 = 3
=> x = 3 + 1
=> x = 4

But y = 2x - 1
so y = 2(4) - 1
= 8 - 1
= 7

So x = 4, y = 7


9. Solve using the method of elimination

4p+2q=10 ......................(1)
4p-3q=25 ......................(2)

Subtract equation (1) from (2)

=> - 5q = 15 ...................(2) - (1)
=> q = -15/5 = -3

But 4p + 2q = 10
so 4p + 2(-3) = 10
=> 4p - 6 = 10
=> 4p = 10 + 6 = 16
=> p = 16/4
=> p = 4

So p = 4, q = -3


10. 5 apple pies and 3 ice-cream sundaes cost $73. An apple pie and 2 ice-cream sundaes cost $28.6. Find the cost of an apple pie and the cost of an ice-cream sundae.

Let the cost of an apple pie be x, and the cost of an ice-cream sundae be y, we have:

5x + 3y = 73 ..................(1)
x + 2y = 28.6 .................(2)

5x + 3y = 73 ..................(1)
5x + 10y = 143 ...............(3) = (2) multiplied by 5

=> 7y = 70 ....................(3) - (1)
=> y = 10

But 5x + 3y = 73
=> 5x + 3(10) = 73
=> 5x + 30 = 73
=> 5x = 73 - 30 = 43
=> x = 43/5 = 8.6

So an apple pie costs $8.60 and an ice-cream sundae costs $10

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Originally Posted by Jhevon

1. 5(2a-5)+4(1-a)=3
=> 10a - 25 + 4 - 4a = 3 ...............expanded the brackets, do you know how to do that?
=> 10a - 4a = 3 + 25 - 4 ...............get all the a's on one side and the numbers on the other
=> 6a = 24 .......................simplify
=> a = 4 ..........................divide both sides by 6

2. 2m-3(4-m)=2(m+1)-17
=> 2m - 12 + 3m = 2m + 2 - 17
=> 5m - 12 = 2m - 15
=> 5m - 2m = -15 + 12
=> 3m = -3
=> m = -1

3. Could you rewrite this question without LaTex?

4. y(y+3)-y(y-6)=3
=> y^2 + 3y - y^2 + 6y = 3
=> 9y = 3
=> y = 1/3

Is there anything you dont understand so far?

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1 How do you expand the brackets?

3
m .... m+2
--- = ---------
1 ......... 3
---
2

Quote:

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Originally Posted by Jhevon

5. Let x be the number of quaters Sam has. Then he has 4x dimes. He also has 65 half dollars = $32.50, which is (1/2)[ 4x(0.10) + x(0.25)].

So [4x(0.10) + x(0.25)]/2 = 32.50
=> 4x(0.10) + x(0.25) = 65 ...............multiply both sides by 2
=> x[4(0.10) + 0.25] = 65 ................factored out the x (do you know how to do that?)
=> x(0.65) = 65
=> x = 65/0.65
=> x = 100

So he has 100 quarters and 4(100) = 400 dimes.

100 quarters = $25 and 400 dimes = $40. So his dimes and quarters come up to $65, which is twice $32.50 as desired.

This problem is a little intricate, if you dont get something, say so.



6. We leave the terminal with y people. At the first stop, 3/5 of y people get off, and also 4 people get off (you sure this shouldn't be "get on"?). So, we are left with y - ((3/5)y + 4) passengers.

So we have [(2/5)y - 4] passengers when the bus leaves the first stop.

At the second stop (1/4)*[(2/5)y -4] people get off and 9 get on. So now we have (3/4)*[(2/5)y -4] + 9 passengers on the bus, and we are told that this is half the number of people that were on the bus at the terminal. so we get:

(3/4)*[(2/5)y -4] + 9 = (1/2)y
=> (3/10)y - 3 + 9 = (1/2)y
=> (3/10)y - (1/2)y = 3 - 9
=> (-1/5)y = -6
=> y = 30 ......................multiplied both sides by -5


7. ABC is an isosceles triangle with AB = AC

If angle B = 4/5x-14degrees and angle C = 1/2x+10degrees find the value of x.

can you clarify the angles. do you mean 4/(5x) and 1/(2x) or do you mean (4/5)x and (1/2)x?

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6

Oh sorry. You are right XD. It's get on. So 4 people get on.

7

It was actually fractions. Damn I wish I knew how to use that Latex thing. So it would be angle B: [(4 over 5)x-14] Angle C [(1 over 2)x+10)

Quote:

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Originally Posted by Jhevon

8. Solve using the method of substitution.

y=2x-1 ................(1)
y-x=3 ..................(2)

From equation (1) we see that y = 2x - 1. Substitute 2x - 1 for y in equation (2):

=> (2x - 1) - x = 3
=> 2x - x -1 = 3
=> x = 3 + 1
=> x = 4

But y = 2x - 1
so y = 2(4) - 1
= 8 - 1
= 7

So x = 4, y = 7


9. Solve using the method of elimination

4p+2q=10 ......................(1)
4p-3q=25 ......................(2)

Subtract equation (1) from (2)

=> - 5q = 15 ...................(2) - (1)
=> q = -15/5 = -3

But 4p + 2q = 10
so 4p + 2(-3) = 10
=> 4p - 6 = 10
=> 4p = 10 + 6 = 16
=> p = 16/4
=> p = 4

So p = 4, q = -3

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Can you help me solve two more? I thought I would get it if you showed me the steps for two of them but it turns out I was wrong >_<. I tried the elimination one first. I did this question :

3r-5s= -9
6r+2s= 18

So I followed what you did.

-7s= 27
s= -27(divided by)7. But then I got the answer as: 23.14285714. :(

Can you help me do that question as well as the one where you have to substitute:

2d-3c= 5
5d+2c = 22

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Also I have two other questions I hope somebody might be able to help me solve.

1. The least of three consecutive integers is divided by 10, the next is divided by 16, the greatest is divided by 26 and the sum of the quotients is 10.

2. In a Math contest, the teams gain 15 points for each correct answer and lose 8 for each incorrect answer. One team had answered 50 questions and got a score of 336. How many questions did the team answer correctly?

Quote:

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Originally Posted by Rocher

1 How do you expand the brackets?

3
m .... m+2
--- = ---------
1 ......... 3
---
2

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When you expand brackets, it means you take everything on the immediate outside of the bracket and multiply everything inside the brackets separately, make sure to use the correct signs. When the signs are the same and you multiply, the sign of the new number is plus, that is, a positive number multiplied by a positive number gives a positive number, and a negative number multiplied by a negative number also gives a positive number. When the signs are different, the new number is negative. So a positive number times a negative number gives a negative answer, and a negative number times a positive number gives a negative answer as well. So here goes:

We had 5(2a-5)+4(1-a)=3. Let's first do it piece by peice and then we'll do the whole thing. So the first piece is 5(2a - 5). So we take the 5 on the outside and multiply everything on the inside, so

5(2a - 5) = 5(2a) - 5(5) = 10a - 25.

The next piece is 4(1 - a), so we take the 4 on the outside and multiply everything on the inside. So

4(1 - a) = 4(1) - 4(a) = 4 - 4a. So now let's redo the problem.

5(2a-5)+4(1-a)=3
=> 10a - 25 + 4 - 4a = 3
=> 10a - 4a -25 + 4 = 3
=> 10a - 4a = 3 + 25 - 4 ...........i add 25 and subtract 4 to get rid of them on the left hand side.

=> 10a - 4a = 28 - 4
=> 6a = 24
=> a = 4 ....................divided by 6 on both sides.


I dont understand what you were trying to say here:

Quote:

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3
m .... m+2
--- = ---------
1 ......... 3
---
2

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Quote:

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Originally Posted by Rocher

6

Oh sorry. You are right XD. It's get on. So 4 people get on.

7

It was actually fractions. Damn I wish I knew how to use that Latex thing. So it would be angle B: [(4 over 5)x-14] Angle C [(1 over 2)x+10)

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Here is the corrected version of number 6:

We leave the terminal with y people. At the first stop, 3/5 of y people get off, and also 4 people get on. So, we are left with y - ((3/5)y) + 4 passengers.

So we have [(2/5)y + 4] passengers when the bus leaves the first stop.

At the second stop (1/4)*[(2/5)y + 4] people get off and 9 get on. So now we have (3/4)*[(2/5)y + 4] + 9 passengers on the bus, and we are told that this is half the number of people that were on the bus at the terminal. so we get:

(3/4)*[(2/5)y + 4] + 9 = (1/2)y
=> (3/10)y + 3 + 9 = (1/2)y
=> (3/10)y - (1/2)y = -3 - 9
=> (-1/5)y = -12
=> y = 60 ......................multiplied both sides by -5

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I dont understand what you were trying to say here:

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It's fractions. I had to put periods so it would be in its correct spot.

Quote:

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Originally Posted by Rocher

It's fractions. I had to put periods so it would be in its correct spot.

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Yes, but what question are you refering to?

Is this what you mean? m/(1/2) = (m + 2)/3

Quote:

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Originally Posted by Rocher

Can you help me solve two more? I thought I would get it if you showed me the steps for two of them but it turns out I was wrong >_<. I tried the elimination one first. I did this question :

3r-5s= -9
6r+2s= 18

So I followed what you did.

-7s= 27
s= -27(divided by)7. But then I got the answer as: 23.14285714. :(

Can you help me do that question as well as the one where you have to substitute:

2d-3c= 5
5d+2c = 22

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Also I have two other questions I hope somebody might be able to help me solve.

1. The least of three consecutive integers is divided by 10, the next is divided by 16, the greatest is divided by 26 and the sum of the quotients is 10.

2. In a Math contest, the teams gain 15 points for each correct answer and lose 8 for each incorrect answer. One team had answered 50 questions and got a score of 336. How many questions did the team answer correctly?

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Redo
3r-5s= -9
6r+2s= 18

and

2d-3c= 5
5d+2c = 22

and type out the steps you took, i'll tell you where you went wrong.

Here are the solutions to the other 2 questions.

1. The least of three consecutive integers is divided by 10, the next is divided by 16, the greatest is divided by 26 and the sum of the quotients is 10.

Consecutive integers means they are one after the other. For example, 1,2,3 or 7,8,9 are three consecutive intergers.

So if the first integer is x, the second is x + 1 and the third is x + 2.

Of course, the least integer will be the first one, so we have:

x/10 + (x + 1)/16 + (x + 2)/26 = 10

It's kind of hard coming up with the LCD for 10, 16, and 26, so let's just make the LCD the multiple of all of them = 10*16*26 = 4160.

x/10 + (x + 1)/16 + (x + 2)/26 = 10

416x + 260(x + 1) + 160(x + 2)
------------------------------ = 10
4160

Now multiply both sides by 4160 to cancel the denomenator:

=> 416x + 260(x + 1) + 160(x + 2) = 41600
=> 416x + 260x + 260 + 160x + 320 = 41600 ..........expanded the brackets
=> 836x + 580 = 41600
=> 836x = 41600 - 580 = 41020
=> x = 41020/836
=> x = 49.066

x is not an integer, which means you probably typed this question wrong as well.


2. In a Math contest, the teams gain 15 points for each correct answer and lose 8 for each incorrect answer. One team had answered 50 questions and got a score of 336. How many questions did the team answer correctly?

Let x be the number of questions they got correct. Since they answered 50 questions, 50 - x is the number of questions they got wrong. For each correct answer they got 15 points, so the points they got for answering x questions correctly is 15x. They lost 8 points for each question they got wrong, so the points they got for the questions they got wrong was -8(50 - x). Since their total points was 336, we have:

15x -8(50 - x) = 336
=> 15x - 400 + 8x = 336 ...............expanded the brackets.
=> 15x + 8x = 336 + 400 ...............added 400 to both sides
=> 23x = 736
=> x = 736/23 = 32

So the team got 32 questions correct.

Are you sure you're supposed to do
2d-3c= 5
5d+2c = 22
by substitution? I wouldn't do it that way becuase you'd have to work with fractions and stuff, elimination is an easier way to go. But hey, i fyour professor says substitution, we have no choice

Yeah the first question I did type out wrong. Sorry to make you go to the trouble of solving a wrong question. It was :

The least of three consecutive integers is divided by 10, the next is divided by 17, the greatest is divided by 26 and the sum of the quotients is 10.

Yeah, my teacher wants us to learn both elimination and substitution so that has to be substitution.

Quote:

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Originally Posted by Rocher

Yeah the first question I did type out wrong. Sorry to make you go to the trouble of solving a wrong question. It was :

The least of three consecutive integers is divided by 10, the next is divided by 17, the greatest is divided by 26 and the sum of the quotients is 10.

Yeah, my teacher wants us to learn both elimination and substitution so that has to be substitution.

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Did you do the two problems i asked you to? I want to see where you are going wrong so i can help you do them on your own