To solve this system of equations, we can use the method of elimination.
1. Multiply the first equation by 3 and the second equation by -3 to eliminate x:
-9x - 12y = -69
-9x + 15y = -15
2. Add the two equations together:
-12y + 15y = -69 - 15
3y = -84
y = -28
3. Substitute y = -28 back into one of the original equations:
-3x - 4(-28) = -23
-3x + 112 = -23
-3x = -135
x = 45
Therefore, the solution to the system of equations is (45, -28), which is not one of the answer choices provided.
Solve the system of two linear equations.
-3x - 4y = -23
3x - 5y = 5
A. (-5, -2)
B. (5, -2)
C. (-5, 2)
D. (5,2)
11 answers
Solve the system of two linear equations.
-3x - 4y = -23
3x - 5y = 5
A. (-5, -2)
B. (5, -2)
C. (-5, 2)
D. (5,2)
-3x - 4y = -23
3x - 5y = 5
A. (-5, -2)
B. (5, -2)
C. (-5, 2)
D. (5,2)
To solve the system of equations -3x - 4y = -23 and 3x - 5y = 5, we can use the method of elimination.
1. Add the two equations together to eliminate x:
(-3x - 4y) + (3x - 5y) = -23 + 5
-9y = -18
y = 2
2. Substitute y = 2 back into one of the equations:
-3x - 4(2) = -23
-3x - 8 = -23
-3x = -15
x = 5
Therefore, the solution to the system of equations is (5, 2), so the correct answer is option D. (5, 2).
1. Add the two equations together to eliminate x:
(-3x - 4y) + (3x - 5y) = -23 + 5
-9y = -18
y = 2
2. Substitute y = 2 back into one of the equations:
-3x - 4(2) = -23
-3x - 8 = -23
-3x = -15
x = 5
Therefore, the solution to the system of equations is (5, 2), so the correct answer is option D. (5, 2).
How many solutions does the system of equations have?
8x - 6y = 48
8x + 6y = 48
A. infinite
B. two
C. one
D. none
8x - 6y = 48
8x + 6y = 48
A. infinite
B. two
C. one
D. none
To determine the number of solutions the system of equations has, we can analyze the two equations:
1. 8x - 6y = 48
2. 8x + 6y = 48
Adding the two equations together, we get:
16x = 96
x = 6
Substitute x = 6 back into the first equation:
8(6) - 6y = 48
48 - 6y = 48
-6y = 0
y = 0
Therefore, the system has one unique solution, which is x = 6 and y = 0.
The answer is C. one.
1. 8x - 6y = 48
2. 8x + 6y = 48
Adding the two equations together, we get:
16x = 96
x = 6
Substitute x = 6 back into the first equation:
8(6) - 6y = 48
48 - 6y = 48
-6y = 0
y = 0
Therefore, the system has one unique solution, which is x = 6 and y = 0.
The answer is C. one.
Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30
A. (3,9)
B. (-21, -60
C. (-3, 6)
D. (-33,-9)
4x - 6y = -48
-2x + 4y = 30
A. (3,9)
B. (-21, -60
C. (-3, 6)
D. (-33,-9)
To solve the system of equations using elimination with multiplication, we can first multiply the second equation by 2 so that the coefficients of y in the two equations are opposites:
1. 4x - 6y = -48
2. -2x + 4y = 30 (multiply by 2)
-4x + 8y = 60
Now, we can add the two equations together to eliminate y:
4x - 6y + (-4x + 8y) = -48 + 60
2y = 12
y = 6
Substitute y = 6 back into the first equation and solve for x:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3
Therefore, the solution to the system of equations is x = -3, y = 6, which corresponds to answer choice C. (-3, 6).
1. 4x - 6y = -48
2. -2x + 4y = 30 (multiply by 2)
-4x + 8y = 60
Now, we can add the two equations together to eliminate y:
4x - 6y + (-4x + 8y) = -48 + 60
2y = 12
y = 6
Substitute y = 6 back into the first equation and solve for x:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3
Therefore, the solution to the system of equations is x = -3, y = 6, which corresponds to answer choice C. (-3, 6).
Solve the system of equations.
3x - 4y = 40
5x - 5y = 55
A. (4, -7)
B. (14, 3)
C. (18, 7)
D. (8, -3)
3x - 4y = 40
5x - 5y = 55
A. (4, -7)
B. (14, 3)
C. (18, 7)
D. (8, -3)
To solve the system of equations, we can use the method of substitution or elimination.
Let's use the method of substitution.
From the first equation: 3x - 4y = 40
Rearranging this equation to solve for x gives: x = (40 + 4y)/3
Substitute x = (40 + 4y)/3 into the second equation: 5x - 5y = 55
5((40 + 4y)/3) - 5y = 55
(200 + 20y)/3 - 5y = 55
(200 + 20y - 15y)/3 = 55
(200 + 5y)/3 = 55
200 + 5y = 165
5y = -35
y = -7
Now substitute y = -7 back into x = (40 + 4y)/3:
x = (40 + 4(-7))/3
x = (40 - 28)/3
x = 12/3
x = 4
Therefore, the solution to the system of equations is x = 4, y = -7, which corresponds to answer choice A. (4, -7).
Let's use the method of substitution.
From the first equation: 3x - 4y = 40
Rearranging this equation to solve for x gives: x = (40 + 4y)/3
Substitute x = (40 + 4y)/3 into the second equation: 5x - 5y = 55
5((40 + 4y)/3) - 5y = 55
(200 + 20y)/3 - 5y = 55
(200 + 20y - 15y)/3 = 55
(200 + 5y)/3 = 55
200 + 5y = 165
5y = -35
y = -7
Now substitute y = -7 back into x = (40 + 4y)/3:
x = (40 + 4(-7))/3
x = (40 - 28)/3
x = 12/3
x = 4
Therefore, the solution to the system of equations is x = 4, y = -7, which corresponds to answer choice A. (4, -7).
How many solutions does the system have?
12x - 15y = 18
4x - 5y = 6
A. two solutions
B. no solution
C. one solution
D. infinitely solutions
12x - 15y = 18
4x - 5y = 6
A. two solutions
B. no solution
C. one solution
D. infinitely solutions
To determine the number of solutions for the system of equations, we can use the method of elimination.
Given equations:
1. 12x - 15y = 18
2. 4x - 5y = 6
To start the elimination procedure by making the coefficients of x the same, we can multiply the second equation by 3:
2. 12x - 15y = 18
Subtracting equation 2 from equation 1:
12x - 15y = 18
- (12x - 15y = 18)
_______________
0 = 0
Since the equation 0 = 0 is always true, the lines representing these equations are coincident, which means they have infinitely many solutions. Therefore, the correct answer is D. infinitely many solutions.
Given equations:
1. 12x - 15y = 18
2. 4x - 5y = 6
To start the elimination procedure by making the coefficients of x the same, we can multiply the second equation by 3:
2. 12x - 15y = 18
Subtracting equation 2 from equation 1:
12x - 15y = 18
- (12x - 15y = 18)
_______________
0 = 0
Since the equation 0 = 0 is always true, the lines representing these equations are coincident, which means they have infinitely many solutions. Therefore, the correct answer is D. infinitely many solutions.