Solve the system of equations by substitution.
x + y = 10
y = 9x
(Simplify your answer. Type an ordered pair.)
Substituting the value of y from the second equation into the first equation, we get:
x + 9x = 10
Combining like terms, we get:
10x = 10
Dividing both sides by 10, we get:
x = 1
Substituting this value of x into the second equation, we get:
y = 9(1)
y = 9
Therefore, the solution to the system of equations is (1, 9).
Solve the system using substitution. Check your answer.
y = -6x + 30
5y - x + 5 = 0
From the first equation, we can solve for y in terms of x:
y = -6x + 30
Substituting this expression for y into the second equation:
5(-6x + 30) - x + 5 = 0
Expanding the expression:
-30x + 150 - x + 5 = 0
Combining like terms:
-31x + 155 = 0
Solving for x:
-31x = -155
Dividing both sides by -31:
x = 5
Substituting this value of x back into the first equation to solve for y:
y = -6(5) + 30
y = -30 + 30
y = 0
Therefore, the solution to the system of equations is (5, 0).
To check our answer, we substitute these values of x and y into both equations:
For the first equation:
5 = -6(0) + 30
5 = 0 + 30
5 = 30
This is NOT true, so the solution (5, 0) is not valid for the first equation.
For the second equation:
5(0) - 5 + 5 = 0
0 - 5 + 5 = 0
0 = 0
This is true, so the solution (5, 0) is valid for the second equation.
Therefore, the solution (5, 0) is not valid for both equations simultaneously, and thus, it is not a solution to the system of equations.
Solve by the substitution method.
3x + 5y = -1
-5x + y = 39
We can start by solving the second equation for y:
-5x + y = 39
y = 5x + 39
Now we substitute this expression for y into the first equation:
3x + 5(5x + 39) = -1
Expanding the equation:
3x + 25x + 195 = -1
Combining like terms:
28x + 195 = -1
Subtracting 195 from both sides:
28x = -196
Dividing both sides by 28:
x = -7
Now we substitute this value of x back into the second equation to solve for y:
-5(-7) + y = 39
35 + y = 39
y = 39 - 35
y = 4
Therefore, the solution to the system of equations is (-7, 4).
To check our answer, we substitute these values of x and y into both equations:
For the first equation:
3(-7) + 5(4) = -1
-21 + 20 = -1
-1 = -1
This is true.
For the second equation:
-5(-7) + 4 = 39
35 + 4 = 39
39 = 39
This is true.
Therefore, the solution (-7, 4) is valid for both equations and is the correct solution to the system.
Solve the system using substitution. Check your answer.
9x - 3y = 1
3y = x - 1
(Simplify your answer. Type an ordered pair.)
We can start by solving the second equation for x:
3y = x - 1
x = 3y + 1
Now we substitute this expression for x into the first equation:
9(3y + 1) - 3y = 1
Expanding the equation:
27y + 9 - 3y = 1
Combining like terms:
24y + 9 = 1
Subtracting 9 from both sides:
24y = -8
Dividing both sides by 24:
y = -8/24
y = -1/3
Now we substitute this value of y back into the second equation to solve for x:
3(-1/3) = x - 1
-1 = x - 1
x = -1 + 1
x = 0
Therefore, the solution to the system of equations is (0, -1/3).
To check our answer, we substitute these values of x and y into both equations:
For the first equation:
9(0) - 3(-1/3) = 1
0 + 1 = 1
1 = 1
This is true.
For the second equation:
3(-1/3) = 0 - 1
-1 = -1
This is true.
Therefore, the solution (0, -1/3) is valid for both equations and is the correct solution to the system.
Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total of 11 people took the trip. She was able to purchase coach tickets for $220 and first class tickets for $1030. She used her total budget for airfare for the trip, which was $6470. How many first class tickets did she buy? How many coach tickets did she buy?
Let x be the number of coach tickets Sarah bought and y be the number of first class tickets Sarah bought.
According to the given information, there were a total of 11 people who took the trip, including Sarah. So we have the equation:
x + y + 1 = 11
x + y = 10
The cost of each coach ticket is $220, and the cost of each first class ticket is $1030. The total budget for airfare was $6470. So we have the equation:
220x + 1030y = 6470
We can solve this system of equations to find the values of x and y.
First, we can rewrite the first equation in terms of x:
x = 10 - y
Substituting this into the second equation:
220(10 - y) + 1030y = 6470
2200 - 220y + 1030y = 6470
810y = 4270
y = 4270/810
y ≈ 5.271
Since we cannot have a fraction of a person, we round y to the nearest whole number:
y ≈ 5
Substituting this value back into the first equation:
x + 5 = 10
x = 10 - 5
x = 5
Therefore, Sarah bought 5 first class tickets and 5 coach tickets.
To verify the solution, we can substitute these values into the second equation:
220(5) + 1030(5) = 1100 + 5150 = 6250
The total cost of the tickets is indeed $6470, which matches the given information.
So, Sarah bought 5 first class tickets and 5 coach tickets.