0

# For a show adult tickets are \$8.75 and children's tickets are \$3.50. 460 tickets were sold for a total of 3143 how many adults tickets were sold?

child ticket = 3.50

460 tickets we're sold totaling 3143

how many adults tickets were sold

### 7 Answers by Expert Tutors

Erik S. | Math is the best subject in which I have ever tutoredMath is the best subject in which I have...
0

Now, the main unknown here is "the # of adult tickets."  Another thing that's unknown is the "the # of children's tickets."  So, we will assign 2 variables as follows:

Let a = # of adult tickets

and c = # of children's tickets.

The next step is to set up the equations out of the unknowns that follow the word problem.  Here they are as follows:

a adult tickets + c children's tickets = 460 tickets altogether, so

a + c = 460.

Now, since we know how much all the tickets cost altogether, \$3143, here comes the next equation:

((\$8.75/adult ticket) * a adult tickets) + ((\$3.50/children's ticket) * c children's tickets) = \$3143, so

8.75a + 3.50c = 3143 since the units, adult tickets and children's tickets, cancel each other out.

Now, since we are looking for the # of adult tickets, we can express the simple equation, a + c = 460, as c in terms of a:

a + c  = 460

-a       = -a         (i.e. -a on each side of the equation to eliminate a on the left side.)

c = 460 - a

Now that we have the equation c in terms of a, we can plug it into the other equation, 8.75a + 3.50c = 3143.  Here it is as follows:

8.75a + 3.50 * (460-a) = 3143                              substitute method

8.75a + [(3.50 * 460) - (3.50 * a)] = 3143             distributive property

8.75a + (1610 - 3.50a) = 3143       I multiplied the two terms in parentheses inside the brackets.

8.75a + 1610 - 3.50a = 3143          Everything's out of the parentheses.

5.25a + 1610 = 3143                     I combined the like terms above.

- 1610 = -1610                    I am eliminating the constant, 1610, on the left side of the equation.

5.25a            = 1533

5.25a/5.25    = 1533/5.25

a = 292

So, now I have found the number of adult tickets, 292.  Therefore, 292 adult tickets were sold.

Now let's check.  Now that we know the value of a of 292, we can plug it into our equations, a + c = 460 and 8.75a + 3.50c = 3143:

292 + c = 460

(8.75 * 292) + 3.50c = 3143

Now, we need to find the value of c in the first equation so that we can plug it into the second equation:

292 + c = 460

-292     = -292

c =  168

(8.75 * 292) + (3.50 * 168) = 3143

2555 + 588 = 3143

3143 = 3143   check!

You're a life saver
James W. | Michigan Grad: Computer Engineer with Mathematics MinorMichigan Grad: Computer Engineer with Ma...
5.0 5.0 (5 lesson ratings) (5)
0

this is an algebra system of two equations problem. you can read about it here.
http://en.wikipedia.org/wiki/System_of_linear_equations

let a = (adult tickets sold) and c = (childrens tickets sold)

so if 460 tickets total were sold, that means that the sum of the adult and childrens tickets were 460. translated into algebra, it is:

460 = a + c

if there was a total of 3143 dollars, and each adult ticket cost 8.75, each childs ticket cost 3.50, that translates into this equation:

8.75 * a + 3.50 * c = 3143

sub out c and solve for a, and remember order of operations (multiply before you add or subtract)

Kevin S. |
5.0 5.0 (4 lesson ratings) (4)
0

Let x = the number of adult tickets. Since 460 tickets were sold in all, then the number of children's tickets would be 460 - x.

We then take the number of each ticket, multiply by the price of each ticket, and that gives us total revenue of \$3143:

8.75x + 3.50 (460 - x) = 3143

8.75x + 1610 - 3.50x = 3143

5.25 x = 1533

Matthew S. | Statistics, Algebra, Math, Computer Programming TutorStatistics, Algebra, Math, Computer Prog...
4.9 4.9 (22 lesson ratings) (22)
0
Let's use A to represent the number of adult tickets sold, and C for the number of children's tickets sold. We know the cost of each ticket, and the total amount of sales. We also know how many tickets were sold. Let's write the known information in equation form. A + C = 460 <-- Total number of tickets sold, adult and children combined 8.75A + 3.50C = 3143 <-- Multiply adult tickets times the cost of an adult ticket, and repeat for children's tickets. The sum of both is the total sales amount. Solving for A in the first equation is easy, just subtract C from both sides: A + C - C = 460 - C A = 460 - C Substitute this into the second equation, and we'll now have one equation and one variable: 8.75(460-C) + 3.50C = 3143 Multiply out: 8.75*460 - 8.75C + 3.50C = 3143 <-- 8.75*460=4025 4025 - 8.75C + 3.50C = 3143 Next add our 2 C terms together: 4025 - 5.25C = 3143 Add 5.25C to both sides: 4025 - 5.25C + 5.25C = 3143 + 5.25C 4025 = 3143 + 5.25C Subtract 3143 from both sides: 4025 - 3143 = 3143 - 5.25C - 3143 882 = 5.25C Divide through by 5.25: 882/5.25 = 5.25C/5.25 168 = C Recall that A = 460 - C, so A = 460 - 168 = 292. So 292 adult tickets and 168 child tickets were sold.
Asok B. | Ph. D in Biochemistry, willing to teach Chemistry, Biochemistry, MathPh. D in Biochemistry, willing to teach ...
4.7 4.7 (18 lesson ratings) (18)
0

If x and y are the number of adult and child tickets then 8.75 x + 3.5 y= 3143....... (1)

and, x + y = 460.......(2)

Multiplying  both sides of the equation with 3.5 we get, 3.5 x + 3.5 y = 3.5 x 460 = 1610.....(3)

Subtracting (3) from (1) we have,

8.75 x-3.5 x = 3143-1610 = 1533

5.25 x= 1533

x= 1533/5.25 = 292

y= 460-292= 168

So the number of adult and child tickets sold were 292 and 168 respectively.

Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)
0

Assign variables to each parameter:

x = # of adult tickets sold

y = # of children tickets sold

Since we are given that the total # of tickets sold is 460, then we can arrive at the following equation:

x + y = 460

We are given that the adult tickets were sold for \$8.75 per ticket and the children tickets were sold for \$3.50 per ticket. With this, we can get the total amount made from the adult tickets sold by multiplying the # of adult tickets sold (x) by the cost of the adult tickets (\$8.75). Likewise, the total amount made from the children tickets sold is determined by multiplying the # of children tickets sold (y) by the cost of each children ticket (\$3.50). Adding these 2 quantities yield the total amount made from the total tickets sold, which were are given to be \$3143. That is,

8.75x + 3.50y = 3143

This generates a system of linear equations, which contain the two equations we found above:

8.75x + 3.50y = 3143

x    +    y     = 460

Since the question only asks you to find how many adult tickets were sold, we only need to solve for x. We can do so by either one of two methods, those being the elimination method and the substitution method. I find the elimination method to be simpler, so I will solve for the # of adult tickets sold (x) by eliminating the other variable which is the # of children tickets sold (y). To eliminate y, multiply the second equation above by -3.50 then add this equation to the first equation.

-3.50(x + y = 460)   ==>   -3.50·x  +  -3.50·y = -3.50·460

==>     -3.50x - 3.50y = -1610

8.75x + 3.50y = 3143

+   -3.50x - 3.50y = -1610

________________________

8.75x - 3.50x + 3.50y - 3.50y = 3143 - 1610

5.25x = 1533

5.25x/5.25 = 1533/5.25

x = 292

Thus, 292 adult tickets were sold.

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
0

x = # of adult tickets sold out

460-x = #of children's tickets sold out

8.75x + 3.50(460-x) = 3143

Solve for x,