Multiplying the second equation by 2 and subtracting it from the first gives the equation:

5y + 2z = 20

Multiplying the first equation by 3, multiplying the third equation by 2, and then adding those equations gives:

13y + 16z = 106

Now we have a system of two equations and two unknowns. 

Multiply the equation 5y + 2z = 20 by 8, and then subtract this from 13y + 16z = 106

This yields –27y = –54, so y = 2. Then using the equation 5y + 2z = 20, z = 5. Then using any of the original three equations, x = –3.

The point of intersection is (-3, 2, 5), and the sum of these coordinates is 4.

We start by setting up a system of equations. The price of one ticket and one candy bar is $5, so t+c=5. The price of two tickets and one candy bar is $8.75, so 2t+c=8.75. We can use the first equation to find out that c=5-t. We then substitute that value into the second equation, giving us 2t+(5-t)=8.75. This simplifies to 2t-t+5=8.75, so t+5=8.75, and finally t=3.75. We use the final value for t in the first equation, so 3.75+c=5. We solve for c, and get c=1.25.

One can describe the ages of Jacob, Sarah, and Caroline with the letters J, S, and C, respectively. From the information in the problem, J = S + 3, and C = 2S. Since C = 28, S = 28/2 = 14, and J = 14 + 3 = 17. Jacob is 17 years old.

Solving the second equation for b and substituting b = 10 – a into the equation ab = 24 gives us

a(10 – a) = 10a – a² = 24

which can be set up and solved as the following quadratic equation:

a1 – 10a +24 = 0

(a – 6) (a – 4) = 0

a = 6, a = 4

The general formula for money problems is V₁ x N₁ + V₂ x N₂ = $Value where V is the value of the coin and N is the number of coins for each separate type of coin involved.  $Value is the total value of all the money when counted.  With this problem N = number of nickels and

Q = N + 2 is the number of quarters.  Substituting into the general formula we get

0.25(N + 2) + 0.05N = 1.40.  Solving for N yields N = 3, therefore Q = 5.  So the answer to the question is actually N + Q = 3 + 5 = 8 total coins.

This problem is probably best solved by developing a series of equations.  To relate the second and third years we can set up the equation 15 = 2x where x is the amount of money made in the second year, which is two times less than the amount of money made in the third.  By dividing each side by two we see that the business made 7.5 million dollars in the second year.  To relate the first and second years we can set up the equation x – 5 = 7.5 where x is the amount of money made in the first year.  By adding 5 to both sides of the equation we are able to see that the business made 12.5 million dollars in the first year.  

Sally works 40 regular hours and 5 overtime hours. Her regular hourly pay becomes 40 * 9.50 = $380.00, and her overtime pay becomes 5 * 9.50 * 1.5 = $71.25, so her weekly gross (before taxes) pay is $451.25

The commission she gets for selling seven cars is $500 * 7 = $3,500 and added to the salary of $1,000 yields $4,500 for the month.

Using algebra, we can write the system of equations J = (1/2)B and J + 3 = B, where J is John's age in 2010 and B is Bob's age in 2010. 

We plug in and get that J = 3 and B = 6

So 3 years later in 2013, we know Bob is 9 years old.

Substitute the second equation into the first to find that y = –5. Then plug y = –5 into either equation to find x = 15.