This system of equations has infinitely many solutions.
This can be seen by adding the two equations together, which eliminates the y term:
8x - 6y + 8x + 6y = 48 + 48
16x = 96
x = 6
Now substitute x back into one of the equations to solve for y:
8(6) - 6y = 48
48 - 6y = 48
-6y = 0
y = 0
Therefore, the solution to the system of equations is x = 6 and y = 0.
How many solutions does the system of equations have???
8x−6y=48
8x+6y=48
7 answers
Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30
4x−6y=−48
−2x+4y=30
To solve this system of equations using elimination with multiplication, we want to manipulate the equations so that when we add them together, one variable is eliminated.
1. Multiply the first equation by 2 to match the coefficient of x in the second equation:
8x - 12y = -96
2. The second equation remains the same:
-2x + 4y = 30
3. Add the two equations together:
8x - 12y + (-2x + 4y) = -96 + 30
6x - 8y = -66
4. Divide this new equation by 2 to simplify:
3x - 4y = -33
Now we have a new equation:
4. 3x - 4y = -33
5. Rewrite the second original equation:
-2x + 4y = 30
6. Add equations 4 and 5:
3x - 4y + (-2x + 4y) = -33 + 30
x = -3
7. Substitute x back into the second equation:
-2(-3) + 4y = 30
6 + 4y = 30
4y = 24
y = 6
Therefore, the solution to the system of equations is x = -3 and y = 6.
1. Multiply the first equation by 2 to match the coefficient of x in the second equation:
8x - 12y = -96
2. The second equation remains the same:
-2x + 4y = 30
3. Add the two equations together:
8x - 12y + (-2x + 4y) = -96 + 30
6x - 8y = -66
4. Divide this new equation by 2 to simplify:
3x - 4y = -33
Now we have a new equation:
4. 3x - 4y = -33
5. Rewrite the second original equation:
-2x + 4y = 30
6. Add equations 4 and 5:
3x - 4y + (-2x + 4y) = -33 + 30
x = -3
7. Substitute x back into the second equation:
-2(-3) + 4y = 30
6 + 4y = 30
4y = 24
y = 6
Therefore, the solution to the system of equations is x = -3 and y = 6.
Solve the system of equations.
3x−4y=40
5x−5y=55
3x−4y=40
5x−5y=55
To solve this system of equations, we can use the method of substitution or elimination. Here, we will use the elimination method.
Given equations:
1. 3x - 4y = 40
2. 5x - 5y = 55
We can see that if we multiply Equation 1 by 5 and Equation 2 by 3, the coefficients of y in both equations will become opposites, allowing us to eliminate y.
1. Multiplying Equation 1 by 5:
5(3x - 4y) = 5(40)
15x - 20y = 200
2. Multiplying Equation 2 by 3:
3(5x - 5y) = 3(55)
15x - 15y = 165
Now our equations are:
3. 15x - 20y = 200
4. 15x - 15y = 165
Subtracting Equation 4 from Equation 3:
(15x - 20y) - (15x - 15y) = 200 - 165
15y - 20y = 35
-5y = 35
y = -7
Now, substitute y = -7 back into Equation 1 to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
Therefore, the solution to the system of equations is x = 4 and y = -7.
Given equations:
1. 3x - 4y = 40
2. 5x - 5y = 55
We can see that if we multiply Equation 1 by 5 and Equation 2 by 3, the coefficients of y in both equations will become opposites, allowing us to eliminate y.
1. Multiplying Equation 1 by 5:
5(3x - 4y) = 5(40)
15x - 20y = 200
2. Multiplying Equation 2 by 3:
3(5x - 5y) = 3(55)
15x - 15y = 165
Now our equations are:
3. 15x - 20y = 200
4. 15x - 15y = 165
Subtracting Equation 4 from Equation 3:
(15x - 20y) - (15x - 15y) = 200 - 165
15y - 20y = 35
-5y = 35
y = -7
Now, substitute y = -7 back into Equation 1 to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
Therefore, the solution to the system of equations is x = 4 and y = -7.
How many solutions does the system have?
12x−15y=18
4x−5y=6
12x−15y=18
4x−5y=6
To determine the number of solutions for the system of equations, we need to check if the lines represented by the equations are parallel, intersecting, or coincident.
Given equations:
1. 12x - 15y = 18
2. 4x - 5y = 6
We can simplify these equations by dividing them throughout by a common factor to make the coefficients easier to compare.
1. 4x - 5y = 6
2. 4x - 5y = 6
Since the equations are the same when simplified, this system has infinitely many solutions due to the two equations representing the same line. This means that any point on the line satisfying one equation will also satisfy the other equation.
Therefore, the system of equations has infinitely many solutions.
Given equations:
1. 12x - 15y = 18
2. 4x - 5y = 6
We can simplify these equations by dividing them throughout by a common factor to make the coefficients easier to compare.
1. 4x - 5y = 6
2. 4x - 5y = 6
Since the equations are the same when simplified, this system has infinitely many solutions due to the two equations representing the same line. This means that any point on the line satisfying one equation will also satisfy the other equation.
Therefore, the system of equations has infinitely many solutions.