Write a system of equations to describe the situation below, solve using any method, and fill in the blanks. $\newline$ Castroville Photography Studio is taking graduation portraits for students at local schools. Eighth graders from Castroville Elementary School ordered $75$ basic portrait packages and $89$ deluxe portrait packages, for a total of $\$13,204$ . The seniors at Weston High ordered $52$ basic portrait packages and $98$ deluxe portrait packages, for a total of $\$12,276$ . How much does each type of package cost? $\newline$ A basic package costs $\$$ _, and a deluxe package costs $\$$ _.

View step-by-step help

Home

Math Problems

Algebra 1

Solve a system of equations using any method: word problems

Full solution

Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

\newline

Castroville Photography Studio is taking graduation portraits for students at local schools. Eighth graders from Castroville Elementary School ordered

75

basic portrait packages and

89

deluxe portrait packages, for a total of

\$13,204

. The seniors at Weston High ordered

52

basic portrait packages and

98

deluxe portrait packages, for a total of

\$12,276

. How much does each type of package cost?

\newline

A basic package costs

\$

_____, and a deluxe package costs

\$

_____.

Set Up Equations: Let's denote the cost of a basic package as $b$ and the cost of a deluxe package as $d$ . We need to set up two equations based on the information given. $\newline$ Eighth graders ordered $75$ basic and $89$ deluxe packages for a total of $\$13,204$ . This can be represented by the equation: $\newline$ $75b + 89d = 13,204$ $\newline$ Seniors ordered $52$ basic and $98$ deluxe packages for a total of $\$12,276$ . This can be represented by the equation: $\newline$ $52b + 98d = 12,276$
Elimination Method: We now have a system of equations: $\newline$ $75b + 89d = 13,204$ $\newline$ $52b + 98d = 12,276$ $\newline$ We can use either substitution or elimination to solve this system. Let's use the elimination method to solve for one of the variables.
Multiply Equations: To eliminate one of the variables, we can multiply the first equation by $52$ and the second equation by $75$ to make the coefficients of $b$ the same in both equations. $\newline$ $(75b + 89d) \times 52 = 13,204 \times 52$ $\newline$ $(52b + 98d) \times 75 = 12,276 \times 75$
Subtract Equations: After multiplying, we get the new system of equations: $\newline$ $3900b + 4628d = 686,608$ $\newline$ $3900b + 7350d = 920,700$ $\newline$ Now we can subtract the first equation from the second to eliminate 'b'. $\newline$ $(3900b + 7350d) - (3900b + 4628d) = 920,700 - 686,608$
Solve for d: Subtracting the equations gives us: $\newline$ $3900b + 7350d - 3900b - 4628d = 920,700 - 686,608$ $\newline$ $2722d = 234,092$ $\newline$ Now we can solve for 'd' by dividing both sides by $2722$ . $\newline$ $d = \frac{234,092}{2722}$ $\newline$ $d = 86$
Substitute and Solve for b: Now that we have the value of 'd', we can substitute it back into one of the original equations to solve for 'b'. Let's use the first equation: $\newline$ $75b + 89d = 13,204$ $\newline$ $75b + 89(86) = 13,204$ $\newline$ $75b + 7654 = 13,204$ $\newline$ Now we can solve for 'b' by subtracting $7654$ from both sides. $\newline$ $75b = 13,204 - 7654$ $\newline$ $75b = 5550$ $\newline$ $b = \frac{5550}{75}$ $\newline$ $b = 74$