How many solutions does the system of equations have? 8x−6y=48 8x+6y=48 (1 point) Responses two two none none infinite infinite one one
9 answers
The system of equations has infinite solutions.
Solve the system of two equations using elimination with multiplication. 4x−6y=−48 −2x+4y=30 (1 point) Responses (−21,−6) left parenthesis negative 21 comma negative 6 right parenthesis (−33,−9) left parenthesis negative 33 comma negative 9 right parenthesis (3, 9) left parenthesis 3 comma 9 right parenthesis (−3, 6) left parenthesis negative 3 comma 6 right parenthesis
To solve the system of equations using elimination with multiplication, we will eliminate one variable by multiplying one or both equations by appropriate numbers to create coefficients that will cancel each other out.
First, let's multiply the first equation by 2 and the second equation by 3 to create coefficients that will cancel out the x terms:
(2)(4x - 6y) = (2)(-48) [Multiply the first equation by 2]
(3)(-2x + 4y) = (3)(30) [Multiply the second equation by 3]
Simplifying these equations, we have:
8x - 12y = -96
-6x + 12y = 90
Now, if we add the two equations together, the y terms will cancel out:
(8x - 12y) + (-6x + 12y) = -96 + 90
Simplifying further:
8x - 12y - 6x + 12y = -6
2x = -6
x = -6/2
x = -3
Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation:
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -48 + 12
-6y = -36
y = -36 / -6
y = 6
Therefore, the solution to the system of equations is (-3, 6).
The correct response is (−3, 6) left parenthesis negative 3 comma 6 right parenthesis.
First, let's multiply the first equation by 2 and the second equation by 3 to create coefficients that will cancel out the x terms:
(2)(4x - 6y) = (2)(-48) [Multiply the first equation by 2]
(3)(-2x + 4y) = (3)(30) [Multiply the second equation by 3]
Simplifying these equations, we have:
8x - 12y = -96
-6x + 12y = 90
Now, if we add the two equations together, the y terms will cancel out:
(8x - 12y) + (-6x + 12y) = -96 + 90
Simplifying further:
8x - 12y - 6x + 12y = -6
2x = -6
x = -6/2
x = -3
Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation:
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -48 + 12
-6y = -36
y = -36 / -6
y = 6
Therefore, the solution to the system of equations is (-3, 6).
The correct response is (−3, 6) left parenthesis negative 3 comma 6 right parenthesis.
Solve the system of equations. 3x−4y=40 5x−5y=55 (1 point) Responses (18, 7) left parenthesis 18 comma 7 right parenthesis (4,−7) left parenthesis 4 comma negative 7 right parenthesis (8,−3) left parenthesis 8 comma negative 3 right parenthesis (14, 3)
To solve the system of equations:
3x - 4y = 40 (Equation 1)
5x - 5y = 55 (Equation 2)
We can solve this system of equations by using the method of substitution or elimination. Let's use the method of substitution.
From Equation 1, we can isolate x:
3x = 4y + 40
x = (4y + 40)/3
Now, substitute this expression for x into Equation 2:
5((4y + 40)/3) - 5y = 55
(20y + 200)/3 - 5y = 55
(20y + 200) - 15y = 165
20y - 15y = 165 - 200
5y = -35
y = -35/5
y = -7
Now substitute the value of y back into Equation 1 to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 40 - 28
3x = 12
x = 12/3
x = 4
Therefore, the solution to the system of equations is (4, -7).
The correct response is (4, −7) left parenthesis 4 comma negative 7 right parenthesis.
3x - 4y = 40 (Equation 1)
5x - 5y = 55 (Equation 2)
We can solve this system of equations by using the method of substitution or elimination. Let's use the method of substitution.
From Equation 1, we can isolate x:
3x = 4y + 40
x = (4y + 40)/3
Now, substitute this expression for x into Equation 2:
5((4y + 40)/3) - 5y = 55
(20y + 200)/3 - 5y = 55
(20y + 200) - 15y = 165
20y - 15y = 165 - 200
5y = -35
y = -35/5
y = -7
Now substitute the value of y back into Equation 1 to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 40 - 28
3x = 12
x = 12/3
x = 4
Therefore, the solution to the system of equations is (4, -7).
The correct response is (4, −7) left parenthesis 4 comma negative 7 right parenthesis.
how many solutions does the system have? 12x−15y=18 4x−5y=6 (1 point) responses two solutions two solutions no solution no solution one solution one solution infinitely many solutions infinitely many solutions
To determine the number of solutions for the system of equations:
12x - 15y = 18 (Equation 1)
4x - 5y = 6 (Equation 2)
We can notice that the second equation is a multiple of the first equation, dividing both sides by 3 results in the same equation:
4(12x - 15y) = 4(18)
48x - 60y = 72
So, the two equations are equivalent and represent the same line. This means that the system has infinitely many solutions.
Therefore, the correct response is infinitely many solutions.
12x - 15y = 18 (Equation 1)
4x - 5y = 6 (Equation 2)
We can notice that the second equation is a multiple of the first equation, dividing both sides by 3 results in the same equation:
4(12x - 15y) = 4(18)
48x - 60y = 72
So, the two equations are equivalent and represent the same line. This means that the system has infinitely many solutions.
Therefore, the correct response is infinitely many solutions.
Solve the system of equations. 6x–5y=27 3x+5y=36 (1 point)
To solve the system of equations:
6x - 5y = 27 (Equation 1)
3x + 5y = 36 (Equation 2)
We can solve this system of equations using the method of elimination. By adding the two equations, the y terms will cancel out:
(6x - 5y) + (3x + 5y) = 27 + 36
Simplifying further:
6x - 5y + 3x + 5y = 63
(6x + 3x) + (-5y + 5y) = 63
9x = 63
x = 63 / 9
x = 7
Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use Equation 1:
6(7) - 5y = 27
42 - 5y = 27
-5y = 27 - 42
-5y = -15
y = -15 / -5
y = 3
Therefore, the solution to the system of equations is (7, 3).
The correct response is (7, 3)
6x - 5y = 27 (Equation 1)
3x + 5y = 36 (Equation 2)
We can solve this system of equations using the method of elimination. By adding the two equations, the y terms will cancel out:
(6x - 5y) + (3x + 5y) = 27 + 36
Simplifying further:
6x - 5y + 3x + 5y = 63
(6x + 3x) + (-5y + 5y) = 63
9x = 63
x = 63 / 9
x = 7
Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use Equation 1:
6(7) - 5y = 27
42 - 5y = 27
-5y = 27 - 42
-5y = -15
y = -15 / -5
y = 3
Therefore, the solution to the system of equations is (7, 3).
The correct response is (7, 3)