There are two ways to solve for x and y in a pair of equations.  One way is to combine (add) the two equations together in order to eliminate one variable. In this case, to solve for y, you would have to multiply the top equation by -7 in order to negate the y variables. We will choose the second way, which is to solve for one of the variables then plug in that value into the other equation to solve for the other variable. If you look at both equations, solving for y on the top equation is easiest, so let's do that:

now just add 3x to both sides to solve for y:

now plug this value into the other (bottom) equation to solve for x:

now subtract 14 to both sides to isolate x:

now just divide both sides by 28 to solve for x:

now just plug in the value of x into either equation to solve for y.

Let's pick the top equation:

now just add 3 to both sides to solve for y:

You can check your answers by plugging them into either equation, and since they are equations, if both sides are equal, then you have the correct answers.

Multiply the first equation by two.

Add both equations to cancel out the y-variable.

Divide by three on both sides.

The answer is:  

This question requires setting up two equations and then solving the system of two equations. One equation represents the total number of tickets sold, and the other equation represents the total revenue. Both equations will use the same two variables,  and , for the number of children and adults who were admitted, respectively. 

The first equation is the easier of the two. The sum of the number of children and the number of adults admitted to the fair will equal 2300.

The second equation will incorporate the price charged to children and adults

Children's tickets cost $4 and adults' tickets cost $8. The total amount of money raised was $13,688.

To solve the system of equations, you can eliminate one of the variables. One way to do this is via the elimination method. To eliminate 

Now add the two equations together to eliminate a variable.

--------------------------------

Solve for .

Substitute  into either of the original equations and solve for .

There were 1178 children and 1122 adults admitted to the festival. Finally, subtract the results.

1) This is a relatively standard quadratic equation. List and add factors of 18.

1 + 18 = 19

2 + 9 = 11

3 + 6 = 9

2) Pull out common factors of each pair, "x" from the first and "6" from the second.

3) Factor again, pulling out "(x+3)" from both terms.

4) Set each term equal to zero and solve.

x + 3 = 0, x = –3

x + 6 = 0, x = –6

1) After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

2) Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

1) Before we can figure out when Billy will be 1.5 times Johnny's age, we have to figure out their current ages. So let's define our variables in terms of the first part of the question.

B = Billy's age and J = Johnny's age

It's easier to solve if we put one variable in terms of the other. If Billy were just twice as old as Johnny, we could write his age as B = 2J.

But Billy is one less than twice as old as Johnny, so B = 2J – 1

2) We know that the two boys' ages multiply together to make ninety-one.

B * J = J(2J – 1) = 91

3) Now we have our factored quadratic. We just need to multiply it out and set everything equal to zero to begin.

4) Now we need to factor back out. We start by multiplying the first coefficient by the final term and listing off the factors.

2 * –91 = –182

1 + –182 = –181

2 + –91 = –89

7 + –26 = –19

13 + –14 = –1

5) Split up the middle term in so that factoring by grouping is possible.

6) Factor by grouping, pulling out "2J" from the first set of terms and "13" from the second.

7) Factor out the "(J-7)" from both terms.

8) Set both parentheses equal to zero and solve.

2J + 13 = 0, J = –13/2

J – 7 = 0, J = 7

Clearly, only of the two solutions works, since Johnny's age has to be positive. Johnny is 7, therefore Billy is 2(7) – 1=13. But we're not done yet!

9) We need to figure out at what point Billy will 1.5 times Johnny's age. Guess and check would be a fairly efficient way to do this problem, but setting up an equation would be even faster. First, though, we need to figure out what our variable is. We know Billy's and Johnny's current ages; we just need to figure out their future ages. One variable is always better than two, so instead of using two different variables to represent their respective future ages, we'll use one variable to represent the number of years we have to add to each of their current ages in order to make Billy 1.5 times older than Johnny. Let's call that variable "x."

1.5(J + x) = B + x

We know the values of J and B, so we can go ahead and fill those in.

1.5(7 + x) = 13 + x

10) Then we solve for x algebraically, with inverse order of operations.

10.5 + 1.5x = 13 + x

0.5x = 2.5

x = 5

J = 7 + 5 = 12

B = 13 + 5 = 18

This requires the use of the quadratic formula.  Recall that:

 for .

For this problem, . 

So,

.

.

Therefore, the two solutions are:

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

First:

Inside:

Outside:

Last:

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum  and  product ; these numbers are .

and

or

The solution set is .

Solve by factoring. We need to find two factors that multiply to eight and add to six.

One of these factors must equal zero in order for the equation to be true.

First, we must factor out any common factors between the two terms.  Both 3 and 12 share a factor of 3, so we can "take" 3 out, like this:

.

Inside the parentheses, it becomes clear that this is a difference of squares problem (a special factor), which can be solved with the equation

.

Thus, . 

Now, we can set each factor to 0, and solve for :

and

.