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Drag each number to the correct location in the equations. Each number can be used more than once, but not all numbers will be used.

Mark is treating his family to ice cream. He buys 4 sundaes and 3 cones for a total of [tex]\[tex]$26[/tex]. Brian also buys ice cream for his family. His total is [tex]\$[/tex]29[/tex] for purchasing 2 cones and 5 sundaes. Determine the system of equations that can be used to find the cost of one sundae, [tex]s[/tex], and the cost of one cone, [tex]c[/tex].

Numbers to use:
4, 26, 5, 7, 3, 29, 2, 55

Equations:
Mark: __ [tex]s[/tex] + __ [tex]c[/tex] = __
Brian: __ [tex]s[/tex] + __ [tex]c[/tex] = __

Drag the numbers to fill in the blanks.

Sagot :

GinnyAnswer
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].
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