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Sagot :
Certainly! Let's solve this problem step by step by setting up and solving a system of linear equations.
First, define our variables:
- [tex]\( s \)[/tex]: the price of one can of soup (in dollars)
- [tex]\( c \)[/tex]: the price of one bag of chips (in dollars)
Given the information in the problem, we can set up the following two equations based on the given costs:
1. Two cans of soup and three bags of chips cost [tex]$10.30: \[ 2s + 3c = 10.30 \] 2. Four cans of soup and two bags of chips cost $[/tex]14.20:
[tex]\[ 4s + 2c = 14.20 \][/tex]
Now, let's line up these two equations for clarity:
[tex]\[ \begin{cases} 2s + 3c = 10.30 \\ 4s + 2c = 14.20 \end{cases} \][/tex]
To solve this system of equations, we can use the method of substitution or elimination. Here, we'll use elimination to solve the system.
Step 1: Align the equations in standard form and prepare to eliminate one of the variables by making the coefficients of one of the variables (either [tex]\( s \)[/tex] or [tex]\( c \)[/tex]) the same in both equations.
Let's multiply the first equation by 2 to align the coefficients of [tex]\( s \)[/tex] in both equations:
[tex]\[ 2 \times (2s + 3c) = 2 \times 10.30 \][/tex]
[tex]\[ 4s + 6c = 20.60 \tag{1} \][/tex]
We now have:
[tex]\[ 4s + 6c = 20.60 \tag{1} \][/tex]
[tex]\[ 4s + 2c = 14.20 \tag{2} \][/tex]
Step 2: Subtract the second equation from the first equation to eliminate [tex]\( s \)[/tex]:
[tex]\[ (4s + 6c) - (4s + 2c) = 20.60 - 14.20 \][/tex]
[tex]\[ 4c = 6.40 \][/tex]
Step 3: Solve for [tex]\( c \)[/tex] by dividing both sides of the equation by 4:
[tex]\[ c = \frac{6.40}{4} \][/tex]
[tex]\[ c = 1.60 \][/tex]
So, the price of one bag of chips, [tex]\( c \)[/tex], is [tex]$1.60. Step 4: Substitute \( c = 1.60 \) back into one of the original equations to solve for \( s \). Using the first equation: \[ 2s + 3c = 10.30 \] \[ 2s + 3(1.60) = 10.30 \] \[ 2s + 4.80 = 10.30 \] Step 5: Solve for \( s \) by isolating \( s \) on one side of the equation: \[ 2s = 10.30 - 4.80 \] \[ 2s = 5.50 \] \[ s = \frac{5.50}{2} \] \[ s = 2.75 \] So, the price of one can of soup, \( s \), is $[/tex]2.75.
To summarize:
- The price of one can of soup ([tex]\( s \)[/tex]) is [tex]$2.75. - The price of one bag of chips (\( c \)) is $[/tex]1.60.
First, define our variables:
- [tex]\( s \)[/tex]: the price of one can of soup (in dollars)
- [tex]\( c \)[/tex]: the price of one bag of chips (in dollars)
Given the information in the problem, we can set up the following two equations based on the given costs:
1. Two cans of soup and three bags of chips cost [tex]$10.30: \[ 2s + 3c = 10.30 \] 2. Four cans of soup and two bags of chips cost $[/tex]14.20:
[tex]\[ 4s + 2c = 14.20 \][/tex]
Now, let's line up these two equations for clarity:
[tex]\[ \begin{cases} 2s + 3c = 10.30 \\ 4s + 2c = 14.20 \end{cases} \][/tex]
To solve this system of equations, we can use the method of substitution or elimination. Here, we'll use elimination to solve the system.
Step 1: Align the equations in standard form and prepare to eliminate one of the variables by making the coefficients of one of the variables (either [tex]\( s \)[/tex] or [tex]\( c \)[/tex]) the same in both equations.
Let's multiply the first equation by 2 to align the coefficients of [tex]\( s \)[/tex] in both equations:
[tex]\[ 2 \times (2s + 3c) = 2 \times 10.30 \][/tex]
[tex]\[ 4s + 6c = 20.60 \tag{1} \][/tex]
We now have:
[tex]\[ 4s + 6c = 20.60 \tag{1} \][/tex]
[tex]\[ 4s + 2c = 14.20 \tag{2} \][/tex]
Step 2: Subtract the second equation from the first equation to eliminate [tex]\( s \)[/tex]:
[tex]\[ (4s + 6c) - (4s + 2c) = 20.60 - 14.20 \][/tex]
[tex]\[ 4c = 6.40 \][/tex]
Step 3: Solve for [tex]\( c \)[/tex] by dividing both sides of the equation by 4:
[tex]\[ c = \frac{6.40}{4} \][/tex]
[tex]\[ c = 1.60 \][/tex]
So, the price of one bag of chips, [tex]\( c \)[/tex], is [tex]$1.60. Step 4: Substitute \( c = 1.60 \) back into one of the original equations to solve for \( s \). Using the first equation: \[ 2s + 3c = 10.30 \] \[ 2s + 3(1.60) = 10.30 \] \[ 2s + 4.80 = 10.30 \] Step 5: Solve for \( s \) by isolating \( s \) on one side of the equation: \[ 2s = 10.30 - 4.80 \] \[ 2s = 5.50 \] \[ s = \frac{5.50}{2} \] \[ s = 2.75 \] So, the price of one can of soup, \( s \), is $[/tex]2.75.
To summarize:
- The price of one can of soup ([tex]\( s \)[/tex]) is [tex]$2.75. - The price of one bag of chips (\( c \)) is $[/tex]1.60.
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