There were 510 people at a play. The admission price was $3 for adults and $1 for children. The admission receipts were $970. How many adults and how many children attended

adults --- a

children --- 150 - a

3a + 1(150-a) = 970

solve for a, quite easy

To solve this problem, we can use a system of equations to represent the given information.

Let's assume that the number of adults who attended the play is represented by 'a', and the number of children who attended is represented by 'c'.

From the given information, we know that there were 510 people at the play. Therefore, we can write the equation:

a + c = 510 ...(Equation 1)

We also know that the admission price for adults was $3 and for children was $1. The total revenue from the admission receipts was $970. We can write this as another equation:

3a + 1c = 970 ...(Equation 2)

Now we have a system of two equations with two variables. We can solve this system to find the values of 'a' and 'c', which will give us the number of adults and the number of children who attended the play.

There are multiple methods to solve such a system of equations, such as substitution, elimination, or graphing. Let's solve this system using the method of substitution.

First, solve Equation 1 for 'a', in terms of 'c':
a = 510 - c

Now substitute this value of 'a' into Equation 2:

3(510 - c) + c = 970

Expanding and simplifying:

1530 - 3c + c = 970

Combine like terms:

-2c = 970 - 1530

-2c = -560

Divide both sides by -2 to solve for 'c':

c = -560 / -2
c = 280

Now substitute the value of 'c' back into Equation 1 to find the value of 'a':

a + 280 = 510

Subtract 280 from both sides:

a = 510 - 280
a = 230

So, there were 230 adults and 280 children who attended the play.