Math Help!? Algebra 2 world problem?

Question

Four times the number of black marbles exceeded three times the number of yellow marbles by 17. Also, 6 times the number of black marbles was 2 less than 10 times the number of yellow marbles. How many of each were there?

Answer 1

let b=number of black marbles
let y=number of yellow marbles

4b=3y+17
6b=10y-2

multiply the first equation by 3 and the second by 2

12b=9y+51
12b=20y-4

since both are equal to 12b they are equal to eachother

9y+51=20y-4
55=11y
y=5

plug that into one of your original equations
6b=10y-2
6b=50-2
6b=48
b=8

5 yellow marbles
8 black marbles

Answer 2

Let x be the number of black marbles.
Let y be the number of yellow marbles.

Now convert the information into equations:
4x = 3y + 17
6x = 10y - 2

Now it's a straightforward problem of solving a system of equations. I'd start by tripling the first equation and doubling the second:
12x = 9y + 51
12x = 20y - 4

From this, you get 9y + 51 = 20y - 4
Solve that to get y = 5
Use that y-value in any of the above equations, then grind through some simple algebra to figure out that x = 8.

Correct answer: There were 8 black marbles and 5 yellow marbles.

Hope that helps!

Answer 3

4b = 3y + 17
6b = 10y - 2

Multiply the top by three and the bottom by two...

12b = 9y + 51
12b = 20y - 4

Since both right-hand sides equal 12b, they are equal:

9y + 51 = 20y - 4
11y = 55
y = 5

Knowing the number of yellow marbles is 5, you can use either of the two original equations to solve for the number of black:

4b = 3y + 17
4b = 3*5 + 17
4b = 32
b = 8

You have 5 yellow and 8 black marbles.

Answer 4

There are 5 yellow and 8 black marbles. Your question creates two equations. You have to write out both and then substitute one into the other to solve. Here goes:

4b = 3y+17 and 6b+2 = 10y

Now, you have to get the whole thing down to a single variable because you can't solve for both. To do that, isolate one variable (it doesn't matter which one) and then stick it into the other formula. For example:

b = (3y+17)/4

Now you can substitute this value of "b" into your other equation:

6((3y+17)/4)+2 = 10y

(18y+102)/4 +2 = 10y.... multiply both sides by 4...

18y+102 +8 = 40y

110 = 22y

y = 110/22.... or, y = 5

Now, just for fun, we can solve it the other way! By solving the other equation and demonstrating that you still get the same answer....

You know that 4b = 3y+17 and 6b+2 = 10y.... so,

6b = 10y -2

b = (10y-2)/6

4((10y-2)/6) = 3y+17

(40y-8)/6 = 3y +17...... multiply both sides by 6 to get

40y-8 = 18y + 102

22y = 110

y = 110/22...... or, y=5


Now that you have y's value, you can drop it into either equation to get b. For example:
4b = 3y+17

4b = 3(5)+17

4b = 15+17

4b = 32...... so, b = 8


In both examples I solved for y first but I could have just as easily solved either equation for b first and done the math to get to the same point.

Answer 5

In such style of issues you ought to locate out what proportion unknowns are there. provide a popularity to each and each unknown, e.g. x,y,z, Write the observe problem in mathematical notations utilising +,-,*, and /. for 2 unknowns try to be waiting to style 2 equations. clean up those equations to locate the values of the unknown. enable some coffee well worth eighty cents be x pounds enable some coffee well worth 50 cents be y pounds 80x+50y=70(x+y) x-y=10 or x=y+10 or eighty(y+10) = 70(y+10+y) or 80y+800 = 140y + seven hundred or 60y = a hundred or y = 5/3 pound or a million pound and 32/3 oz. and x = 11pound and 32/3 oz. enable the facets of triangle be x,y,z thus 3 equations are mandatory to clean up for 3 unknown. For next problem additionally there are 3 unknowns s,m and l. 3 equations are mandatory to clean up this problem

Answer 6

I'll get you started, you can substitute and finish


4b - 17 = 3y
6b +2 = 10y

Answer 7

1)4B-3Y=17
2)6B=10Y-2
12B-9Y=51
12B-20Y=-4
11Y=55
Y=5
4B-3X5=17
4B=32
B=8