Answer :
To determine the cost of one sundae, [tex]\( s \)[/tex], and one cone, [tex]\( c \)[/tex], we can set up and solve a system of linear equations based on the information provided.
### Step-by-Step Solution:
1. Identify the given information and formulate the equations:
- Mark buys 4 sundaes and 3 cones for a total of [tex]$26. - This can be written as: \[ 4s + 3c = 26 \] - Brian buys 5 sundaes and 2 cones for a total of $[/tex]29.
- This can be written as:
[tex]\[ 5s + 2c = 29 \][/tex]
2. Write the system of linear equations:
[tex]\[ \begin{cases} 4s + 3c = 26 \\ 5s + 2c = 29 \end{cases} \][/tex]
3. Solve the system of equations using the method of substitution or elimination.
Here, I will outline the steps using the elimination method:
- Multiply the first equation by 2 and the second equation by 3 to make the coefficients of [tex]\(c\)[/tex] the same:
[tex]\[ \begin{cases} 2 \times (4s + 3c) = 2 \times 26 \implies 8s + 6c = 52 \\ 3 \times (5s + 2c) = 3 \times 29 \implies 15s + 6c = 87 \end{cases} \][/tex]
- Subtract the first modified equation from the second modified equation to eliminate [tex]\(c\)[/tex]:
[tex]\[ (15s + 6c) - (8s + 6c) = 87 - 52 \\ 7s = 35 \][/tex]
- Solve for [tex]\(s\)[/tex]:
[tex]\[ s = \frac{35}{7} = 5 \][/tex]
4. Substitute [tex]\(s = 5\)[/tex] back into one of the original equations to solve for [tex]\(c\)[/tex]:
Using the first equation [tex]\(4s + 3c = 26\)[/tex]:
[tex]\[ 4(5) + 3c = 26 \\ 20 + 3c = 26 \\ 3c = 26 - 20 \\ 3c = 6 \\ c = \frac{6}{3} = 2 \][/tex]
### Conclusion:
The cost of one sundae ([tex]\(s\)[/tex]) is [tex]$\$[/tex]5[tex]$ and the cost of one cone (\(c\)) is $[/tex]\[tex]$2$[/tex].
### Step-by-Step Solution:
1. Identify the given information and formulate the equations:
- Mark buys 4 sundaes and 3 cones for a total of [tex]$26. - This can be written as: \[ 4s + 3c = 26 \] - Brian buys 5 sundaes and 2 cones for a total of $[/tex]29.
- This can be written as:
[tex]\[ 5s + 2c = 29 \][/tex]
2. Write the system of linear equations:
[tex]\[ \begin{cases} 4s + 3c = 26 \\ 5s + 2c = 29 \end{cases} \][/tex]
3. Solve the system of equations using the method of substitution or elimination.
Here, I will outline the steps using the elimination method:
- Multiply the first equation by 2 and the second equation by 3 to make the coefficients of [tex]\(c\)[/tex] the same:
[tex]\[ \begin{cases} 2 \times (4s + 3c) = 2 \times 26 \implies 8s + 6c = 52 \\ 3 \times (5s + 2c) = 3 \times 29 \implies 15s + 6c = 87 \end{cases} \][/tex]
- Subtract the first modified equation from the second modified equation to eliminate [tex]\(c\)[/tex]:
[tex]\[ (15s + 6c) - (8s + 6c) = 87 - 52 \\ 7s = 35 \][/tex]
- Solve for [tex]\(s\)[/tex]:
[tex]\[ s = \frac{35}{7} = 5 \][/tex]
4. Substitute [tex]\(s = 5\)[/tex] back into one of the original equations to solve for [tex]\(c\)[/tex]:
Using the first equation [tex]\(4s + 3c = 26\)[/tex]:
[tex]\[ 4(5) + 3c = 26 \\ 20 + 3c = 26 \\ 3c = 26 - 20 \\ 3c = 6 \\ c = \frac{6}{3} = 2 \][/tex]
### Conclusion:
The cost of one sundae ([tex]\(s\)[/tex]) is [tex]$\$[/tex]5[tex]$ and the cost of one cone (\(c\)) is $[/tex]\[tex]$2$[/tex].
