To solve the system of equations:
6x - 3y = -6 (1)
9x - 3y = -18 (2)
We can start by manipulating one of the equations to make the coefficients of y the same. We can multiply the first equation by -3 to make the coefficient of y -3 in both equations. This gives us:
-18x + 9y = 18 (3)
9x - 3y = -18 (2)
Now we can subtract equation (3) from equation (2):
9x - 18x - 3y - 9y = -18 - 18
-9x - 12y = -36
Now we can solve for x in terms of y:
-12y = -36
y = 3
Now substitute y=3 into equation (1):
6x - 3(3) = -6
6x - 9 = -6
6x = 3
x = 3/6
x = 1/2
Therefore, the solution to the system of equations is: (1/2, 3)
Solve the system of equations.
6x−3y=−6
9x−3y=−18(1 point)
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11 answers
Solve the system of equations.
5x–4y=–44
–x+3y=22 (1 point)
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5x–4y=–44
–x+3y=22 (1 point)
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To solve the system of equations:
5x - 4y = -44 (1)
-x + 3y = 22 (2)
First, let's multiply equation (2) by 5 to make the coefficients of x the same:
-5x + 15y = 110 (3)
5x - 4y = -44 (1)
Now, add equations (1) and (3) together to eliminate x:
-4y + 15y = -44 + 110
11y = 66
y = 6
Now, substitute y = 6 into equation (2) to solve for x:
-x + 3(6) = 22
-x + 18 = 22
-x = 22 - 18
-x = 4
x = -4
Therefore, the solution to the system of equations is: (-4, 6)
5x - 4y = -44 (1)
-x + 3y = 22 (2)
First, let's multiply equation (2) by 5 to make the coefficients of x the same:
-5x + 15y = 110 (3)
5x - 4y = -44 (1)
Now, add equations (1) and (3) together to eliminate x:
-4y + 15y = -44 + 110
11y = 66
y = 6
Now, substitute y = 6 into equation (2) to solve for x:
-x + 3(6) = 22
-x + 18 = 22
-x = 22 - 18
-x = 4
x = -4
Therefore, the solution to the system of equations is: (-4, 6)
Solve the system of equations.
−6x+2y=−32
4x+3y=−9 (1 point)
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−6x+2y=−32
4x+3y=−9 (1 point)
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To solve the system of equations:
-6x + 2y = -32 (1)
4x + 3y = -9 (2)
First, let's multiply equation (1) by 2 to make the coefficients of y the same:
-12x + 4y = -64 (3)
4x + 3y = -9 (2)
Now, we can add equations (3) and (2) together to eliminate y:
-12x + 4y + 4x + 3y = -64 - 9
-8x + 7y = -73
Let's solve for y in terms of x:
7y = 8x - 73
y = (8/7)x - 73/7
Now, let's substitute y = (8/7)x - 73/7 into either equation (1) or (2) to solve for x:
-6x + 2((8/7)x - 73/7) = -32
-6x + (16/7)x - 146/7 = -32
Multiply by 7 to resolve fractions:
-42x + 16x - 146 = -224
-26x = -78
x = -78 / -26
x = 3
Now, substitute x = 3 back into y = (8/7)x - 73/7:
y = (8/7)(3) - 73/7
y = 24/7 - 73/7
y = -49/7
Therefore, the solution to the system of equations is: (3, -49/7)
-6x + 2y = -32 (1)
4x + 3y = -9 (2)
First, let's multiply equation (1) by 2 to make the coefficients of y the same:
-12x + 4y = -64 (3)
4x + 3y = -9 (2)
Now, we can add equations (3) and (2) together to eliminate y:
-12x + 4y + 4x + 3y = -64 - 9
-8x + 7y = -73
Let's solve for y in terms of x:
7y = 8x - 73
y = (8/7)x - 73/7
Now, let's substitute y = (8/7)x - 73/7 into either equation (1) or (2) to solve for x:
-6x + 2((8/7)x - 73/7) = -32
-6x + (16/7)x - 146/7 = -32
Multiply by 7 to resolve fractions:
-42x + 16x - 146 = -224
-26x = -78
x = -78 / -26
x = 3
Now, substitute x = 3 back into y = (8/7)x - 73/7:
y = (8/7)(3) - 73/7
y = 24/7 - 73/7
y = -49/7
Therefore, the solution to the system of equations is: (3, -49/7)
The Math Club sold slices of pizza, p, at a profit of $3 each and cans of soda, s, at $2 each to raise money for a trip. They sold 200 items and made a profit of $550. Write the pair of linear equations that model this situation.(1 point)
Equation to represent the total number of items sold:
= 200
Equation to represent the total profit:
= 550
Equation to represent the total number of items sold:
= 200
Equation to represent the total profit:
= 550
Let's denote the number of slices of pizza sold as "p" and the number of cans of soda sold as "s".
1. The equation to represent the total number of items sold:
p + s = 200
2. The equation to represent the total profit:
3p + 2s = 550
Therefore, the pair of linear equations that model this situation is:
p + s = 200
3p + 2s = 550
1. The equation to represent the total number of items sold:
p + s = 200
2. The equation to represent the total profit:
3p + 2s = 550
Therefore, the pair of linear equations that model this situation is:
p + s = 200
3p + 2s = 550
The Lakewood baseball team is selling T-shirts for a fundraiser. The shirts cost $100 for the printing design and setup, plus $10 per shirt. The team is going to sell the shirts for $15 each. How many shirts do they need to sell to break even?(1 point)
shirts
shirts
Let's denote the number of shirts as x.
The total cost for printing the design and setup is $100 and the cost per shirt is $10. Therefore, the total cost is given by:
Total cost = $100 + $10x
The selling price of each shirt is $15. The revenue from selling x shirts is given by:
Total revenue = $15x
For the team to break even, the total revenue should equal the total cost:
Total revenue = Total cost
$15x = $100 + $10x
Now, we can solve for x:
$15x = $100 + $10x
$15x - $10x = $100
$5x = $100
x = $100 / $5
x = 20
Therefore, the team needs to sell 20 shirts to break even.
The total cost for printing the design and setup is $100 and the cost per shirt is $10. Therefore, the total cost is given by:
Total cost = $100 + $10x
The selling price of each shirt is $15. The revenue from selling x shirts is given by:
Total revenue = $15x
For the team to break even, the total revenue should equal the total cost:
Total revenue = Total cost
$15x = $100 + $10x
Now, we can solve for x:
$15x = $100 + $10x
$15x - $10x = $100
$5x = $100
x = $100 / $5
x = 20
Therefore, the team needs to sell 20 shirts to break even.
The Kesling Middle School PTA is planning a carnival to raise money for the school’s art department. They estimate that the event will be very popular and that they will have 500 people attend. They plan to charge adults $10 and children $5 for admission. The PTA wants to earn $3,500 from admission charges. How many adults and how many children need to attend for the PTA to reach their goal of $3,500?(1 point)
adults;
children
adults;
children
Let's denote the number of adults attending as "a" and the number of children attending as "c".
The total number of people attending is given as 500:
a + c = 500
The admission charge for adults is $10 and for children is $5. The total revenue from admission charges for adults and children should be $3,500:
10a + 5c = 3500
To solve the system of equations, we can use the substitution method. Rearrange the first equation to get a = 500 - c and substitute it into the second equation:
10(500 - c) + 5c = 3500
5000 - 10c + 5c = 3500
-5c = -1500
c = 300
Now substitute c = 300 back into a + c = 500:
a + 300 = 500
a = 200
Therefore, the PTA needs 200 adults and 300 children to reach their goal of $3,500.
The total number of people attending is given as 500:
a + c = 500
The admission charge for adults is $10 and for children is $5. The total revenue from admission charges for adults and children should be $3,500:
10a + 5c = 3500
To solve the system of equations, we can use the substitution method. Rearrange the first equation to get a = 500 - c and substitute it into the second equation:
10(500 - c) + 5c = 3500
5000 - 10c + 5c = 3500
-5c = -1500
c = 300
Now substitute c = 300 back into a + c = 500:
a + 300 = 500
a = 200
Therefore, the PTA needs 200 adults and 300 children to reach their goal of $3,500.