To use the system of linear equations to find the price of the journal and cookbook, we use substitution to solve for one variable in terms of the other. We can use equation 2 to solve for j in terms of c:
j = 2c + 3
Then we substitute this expression for j in equation 1:
2c + 3 + c = 54
Simplifying and solving for c:
3c = 51
c = 17
Now that we know the price of the cookbook, we can use equation 2 to solve for the price of the journal:
j = 2c + 3 = 2(17) + 3 = 37
Therefore, the price of the journal is $37 and the price of the cookbook is $17.
At Barnes and Noble, Sylvia purchased a journal and a cookbook that cost a total of $54, not including tax. If the price of the journal, j, is $3 more than 2 times the price of the cookbook, c, which system of linear equations could be used to determine the price of each item?
1. The equation for the total cost
j + c = 54
2. The equation for the price of the journal
2c + 3 = j
3. You can choose any method to solve this system: graphing, substitution, or elimination. Choose a method and solve for the price of the journal and the cookbook.
The price of the journal is
The price of the cookbook is
7 answers
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The point (5, −3)
is a solution to which system of equations?
a. y=x−8
2x+y=7
b. y=x+2
y=x+5
c. y=-1/2x+6
y=3x-1
d. y=2/3x+6
3y+6x-18=0
is a solution to which system of equations?
a. y=x−8
2x+y=7
b. y=x+2
y=x+5
c. y=-1/2x+6
y=3x-1
d. y=2/3x+6
3y+6x-18=0
To test whether the point (5, -3) is a solution to a system of equations, we substitute the values of x and y into each equation and check if both equations are satisfied.
a. y = x - 8 2x + y = 7
When x = 5 and y = -3,
y = x - 8 becomes -3 = 5 - 8, which is true
2x + y = 7 becomes 2(5) + (-3) = 7, which is also true
Therefore, (5, -3) is a solution to system a
b. y = x + 2 y = x + 5
When x = 5 and y = -3,
y = x + 2 becomes -3 = 5 + 2, which is false
y = x + 5 becomes -3 = 5 + 5, which is also false
Therefore, (5, -3) is not a solution to system b
c. y = -1/2x + 6 y = 3x - 1
When x = 5 and y = -3,
y = -1/2x + 6 becomes -3 = -1/2(5) + 6, which is false
y = 3x - 1 becomes -3 = 3(5) - 1, which is also false
Therefore, (5, -3) is not a solution to system c
d. y = 2/3x + 6 3y + 6x - 18 = 0
When x = 5 and y = -3,
y = 2/3x + 6 becomes -3 = 2/3(5) + 6, which is false
3y + 6x - 18 = 0 becomes 3(-3) + 6(5) - 18 = 0, which is true
Therefore, (5, -3) is not a solution to system d
Therefore, the answer is (a) y=x−8 and 2x+y=7.
a. y = x - 8 2x + y = 7
When x = 5 and y = -3,
y = x - 8 becomes -3 = 5 - 8, which is true
2x + y = 7 becomes 2(5) + (-3) = 7, which is also true
Therefore, (5, -3) is a solution to system a
b. y = x + 2 y = x + 5
When x = 5 and y = -3,
y = x + 2 becomes -3 = 5 + 2, which is false
y = x + 5 becomes -3 = 5 + 5, which is also false
Therefore, (5, -3) is not a solution to system b
c. y = -1/2x + 6 y = 3x - 1
When x = 5 and y = -3,
y = -1/2x + 6 becomes -3 = -1/2(5) + 6, which is false
y = 3x - 1 becomes -3 = 3(5) - 1, which is also false
Therefore, (5, -3) is not a solution to system c
d. y = 2/3x + 6 3y + 6x - 18 = 0
When x = 5 and y = -3,
y = 2/3x + 6 becomes -3 = 2/3(5) + 6, which is false
3y + 6x - 18 = 0 becomes 3(-3) + 6(5) - 18 = 0, which is true
Therefore, (5, -3) is not a solution to system d
Therefore, the answer is (a) y=x−8 and 2x+y=7.
5 answers