Let's solve each question step by step:
Question 1:
Solve the following system of equations: \[ x + 3y = -4 \] \[ x + 5y = -6 \]
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Subtract the first equation from the second: \[ (x + 5y) - (x + 3y) = -6 - (-4) \] \[ 2y = -2 \] \[ y = -1 \]
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Substitute \( y = -1 \) into the first equation: \[ x + 3(-1) = -4 \] \[ x - 3 = -4 \] \[ x = -1 \]
Answer: (−1, −1)
Question 2:
Two families attended a baseball game. Let \( p \) be the price of a bag of popcorn and \( c \) be the price of a souvenir cup.
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From the first family: \[ 5p + 6c = 65 \] (Equation 1)
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From the second family: \[ 10p + 4c = 90 \] (Equation 2)
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Solve Equation 1 for \( c \): \[ 6c = 65 - 5p \] \[ c = \frac{65 - 5p}{6} \]
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Substitute into Equation 2: \[ 10p + 4\left(\frac{65 - 5p}{6}\right) = 90 \] \[ 10p + \frac{260 - 20p}{6} = 90 \] Multiplying everything by 6 to clear the fraction: \[ 60p + 260 - 20p = 540 \] \[ 40p = 280 \] \[ p = 7 \]
Answer: $7
Question 3:
Let \( x \) be Sam's age and \( y \) be Harry's age. You are given:
- \( x = 5y \) (Equation 1)
- \( x = 10 + 3y \) (Equation 2)
Set the two equations equal: \[ 5y = 10 + 3y \] \[ 2y = 10 \] \[ y = 5 \]
Now substitute \( y \) back to find \( x \): \[ x = 5(5) = 25 \]
Answer: Sam is 25 years old, and Harry is 5 years old.
Question 4:
Solve the following system of equations: \[ -2x + 3y = 5 \] \[ y = 3x + 4 \]
Substituting \( y \) from the second into the first: \[ -2x + 3(3x + 4) = 5 \] \[ -2x + 9x + 12 = 5 \] \[ 7x + 12 = 5 \] \[ 7x = -7 \] \[ x = -1 \]
Substituting back to find \( y \): \[ y = 3(-1) + 4 = 1 \]
Answer: (−1, 1)
Question 5:
Modeling with the given equations: \[ 4x = 7 + 3y \] \[ x - 3y = -5 \]
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From the second equation: \[ x = 3y - 5 \]
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Substitute into the first: \[ 4(3y - 5) = 7 + 3y \] \[ 12y - 20 = 7 + 3y \] \[ 9y = 27 \] \[ y = 3 \]
Substituting \( y \) back to get \( x \): \[ x = 3(3) - 5 = 4 \]
Answer: City A was 4° C, and City B was 3° C.
Question 6:
Using substitution:
- Substitute \( y = -2x + 1 \) into \( 4x + 2y = -1 \): \[ 4x + 2(-2x + 1) = -1 \] \[ 4x - 4x + 2 = -1 \] This simplifies to \( 2 = -1 \), which is a contradiction.
Answer: No solution
Question 7:
Express \( x \) from the second equation: From \( -x - 2y = -10 \), we have: \[ -x = -10 + 2y \Rightarrow x = 10 - 2y \]
Now, substitute into the first equation: \[ 3(10 - 2y) + y = 8 \]
Answer: 2y - 10
Question 8:
Solve the following system:
- Substitute \( x = 60 + 4y \) into \( 7x + 12y = 500 \): \[ 7(60 + 4y) + 12y = 500 \] \[ 420 + 28y + 12y = 500 \] \[ 40y = 80 \] \[ y = 2 \]
Now substitute back: \[ x = 60 + 4(2) = 68 \]
Answer: (68, 2)
Question 9:
Let \( n \) be the number of notebooks and \( t \) be the number of t-shirts.
- \( n + t = 70 \) (Equation 1)
- \( 5n + 11t = 500 \) (Equation 2)
Substituting from Equation 1 into Equation 2: \[ 5n + 11(70 - n) = 500 \] \[ 5n + 770 - 11n = 500 \] \[ -6n = -270 \] \[ n = 45 \]
Answer: 45
Question 10:
Let \( x \) be the number of children and \( y \) the number of adults.
- \( x + y = 320 \) (Equation 1)
- \( 7x + 10y = 2900 \) (Equation 2)
This set of equations models the scenario correctly.
Answer: 7x + 10y = 2900; x + y = 320
If you have any further questions or need additional assistance, feel free to ask!
9 answers