# 2x + 3y = 78

How do i solve for x and y?

2x=78
78/2=39 so x=39

3y=78
78/3=26 so y=26

Can't be, because 2(39)+ 3(26) does not equl 78

You need TWO equations when you have TWO variables, such as x and y. It can't be done with one equation. BUT, you say, what about this post giving x as 39 and y as 26.
To determine if the first answer posted is correct, simply put the numbers back into the equation and see if the equation is satisfied.
2x + 3y = 78
2*39 + 3*26 = 78
78 + 78 = 78?? OOPS!.
So x = 39 and y = 26 isn't correct. Ya gotta have two equations. I hope this helps.

2x + 3y = 78
3x + 2y = 72

Here is the original question: Pencils have one price and pens have another. Two pencils and three pens cost 78 cents, but three pencils and two pens cost 72 cents. How much does one pencil cost?

Use variable x for pencils and y for pens.

2x + 3y = 78
3x + 2y = 72

In the first equation, solve for y.

2x + 3y = 78
3y = 78 - 2x
y = (78 - 2x) / 3

Substitute this value for y into the second equation.

Hers is another one: 3 people requires 3 weeks and 2 days to do a job. How long would it take for 4 people to do the same job? (Each week is 6 days)

48

1--2x + 3y = 78
2--Divide through by 2 yielding x + y + y/2 = 39
3--y/2 must be an integer k making y = 2k
4--Substituting, 2x + 6k = 78 making x = 39 - 3k.
5--k....0....1....2....3....4....5....6
...x...39...36...33...30...27...24...21
...y...0....2.....4...6....8....10...12

...k...7...8...9...10...11...12...13
...x..18..15..12...9....6....3....0
...y...14.16..18...20...22...24...26

65

13 years ago

9 months ago

## To solve for x and y in the system of equations:

1. Set up the equations:

2x + 3y = 78
3x + 2y = 72

2. Solve for y in the first equation:
2x + 3y = 78
3y = 78 - 2x
y = (78 - 2x) / 3

3. Substitute the value of y from the first equation into the second equation:
3x + 2((78 - 2x) / 3) = 72

4. Simplify the equation:
3x + (2/3)(78 - 2x) = 72
Multiply each term by 3 to eliminate the fraction:
9x + 2(78 - 2x) = 216
Distribute the 2:
9x + 156 - 4x = 216
Combine like terms:
5x + 156 = 216
Subtract 156 from both sides:
5x = 60
Divide both sides by 5:
x = 12

5. Substitute the value of x back into the first equation to solve for y:
2(12) + 3y = 78
24 + 3y = 78
Subtract 24 from both sides:
3y = 54
Divide both sides by 3:
y = 18

So the solution to the system of equations is x = 12 and y = 18.

I apologize for the incorrect response in the previous answer. Thank you for pointing out the mistake.

9 months ago

## To solve the system of equations:

1. Multiply the first equation by 2 and the second equation by 3 to cancel out the x terms in both equations:

2(2x + 3y) = 2(78)
3(3x + 2y) = 3(72)

4x + 6y = 156
9x + 6y = 216

2. Subtract the first equation from the second equation:

(9x + 6y) - (4x + 6y) = 216 - 156
9x - 4x + 6y - 6y = 60
5x = 60
x = 12

3. Substitute the value of x into one of the original equations, for example:

2x + 3y = 78
2(12) + 3y = 78
24 + 3y = 78
3y = 54
y = 18

Therefore, the solution to the system of equations is x = 12 and y = 18.