1. Solve the system of two linear equations.−3x−4y=−23 3x−5y=5 (1 point)
2. How many solutions does the system of equations have?8x−6y=48 8x+6y=48 (1 point)
3. Solve the system of two equations using elimination with multiplication.4x−6y=−48 −2x+4y=30 (1 point)
4. Solve the system of equations.3x−4y=40 5x−5y=55 (1 point)
5. How many solutions does the system have?12x−15y=18 4x−5y=6 (1 point)
1. (5, 2)
2. one
3. ( -3, 6
4. (4, -7)
5. infinitely many solutions
1. To solve the system of equations −3x−4y=−23 and 3x−5y=5, we can use the method of substitution.
First, from the second equation, we can express x in terms of y:
3x = 5 + 5y
x = (5 + 5y) / 3
Substituting this expression for x into the first equation:
-3((5 + 5y) / 3) - 4y = -23
-15/3 - 15y/3 - 4y = -23
-5 - 5y - 4y = -23
-5y - 4y = -23 + 5
-9y = -18
y = -18 / -9
y = 2
Substituting the value of y back into the second equation:
3x - 5(2) = 5
3x - 10 = 5
3x = 15
x = 15 / 3
x = 5
Therefore, the solution to the system of equations is x = 5 and y = 2.
2. The system of equations 8x−6y=48 and 8x+6y=48 has infinitely many solutions. This can be seen by adding the two equations together:
(8x - 6y) + (8x + 6y) = 48 + 48
16x = 96
x = 96 / 16
x = 6
Substituting x = 6 into one of the original equations, such as 8x - 6y = 48, we can solve for y:
8(6) - 6y = 48
48 - 6y = 48
-6y = 0
y = 0
Therefore, the system of equations has infinitely many solutions, with x = 6 and y = 0 being one possible solution.
3. To solve the system of equations 4x−6y=−48 and −2x+4y=30 using elimination with multiplication, we can multiply both equations by appropriate constants to eliminate one of the variables.
First, let's eliminate the x-variable by multiplying the second equation by 2:
4x - 6y = -48
-4x + 8y = 60
Adding the two equations together, the x-terms cancel out:
(4x - 6y) + (-4x + 8y) = -48 + 60
2y = 12
y = 12 / 2
y = 6
Substituting the value of y back into one of the original equations, such as 4x - 6y = -48, we can solve for x:
4x - 6(6) = -48
4x - 36 = -48
4x = -48 + 36
4x = -12
x = -12 / 4
x = -3
Therefore, the solution to the system of equations is x = -3 and y = 6.
4. To solve the system of equations 3x−4y=40 and 5x−5y=55, we can use the method of substitution or elimination. Let's use the method of substitution.
From the first equation, we can express x in terms of y:
3x = 40 + 4y
x = (40 + 4y) / 3
Substituting this expression for x into the second equation:
5((40 + 4y) / 3) - 5y = 55
(200 + 20y) / 3 - 5y = 55
200 + 20y - 15y = 165
5y = 165 - 200
5y = -35
y = -35 / 5
y = -7
Substituting the value of y back into the first equation:
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 x = 4 and y = -7.
5. The system of equations 12x−15y=18 and 4x−5y=6 has infinitely many solutions. This can be seen by multiplying the second equation by 3 and comparing it with the first equation:
12x - 15y = 18
(3)(4x - 5y) = (3)(6)
12x - 15y = 18
As the two equations are identical, they are representing the same line. Therefore, there are infinitely many solutions to the system of equations.
1. To solve the system of equations −3x−4y=−23 and 3x−5y=5, we can use the method of substitution or elimination.
- Starting with the method of substitution:
From the second equation, we can isolate x: 3x = 5y + 5 --> x = (5y + 5)/3
Now we substitute this value of x into the first equation:
-3( (5y + 5)/3) - 4y = -23
-5y - 5 - 4y = -23
-9y - 5 = -23
-9y = -18
y = -18 / -9
y = 2
Next, we substitute the y value we just found back into either equation to get x:
3x - 5(2) = 5
3x - 10 = 5
3x = 15
x = 15 / 3
x = 5
So, the solution to the system is x = 5 and y = 2.
2. To determine the number of solutions in the system of equations 8x−6y=48 and 8x+6y=48, we can compare the coefficients of x and y in both equations.
- The coefficients are the numbers multiplying x and y in each equation.
Here, we can see that the coefficients of x and y are the same in both equations, but the signs are different (positive and negative).
When this happens, it indicates that the lines represented by the equations are parallel and will never intersect. Therefore, the system has no solutions.
3. To solve the system of equations 4x−6y=−48 and −2x+4y=30 using elimination with multiplication:
The goal of this method is to add or subtract the equations in such a way that one variable gets eliminated. We can do this by manipulating the equations to have opposite coefficients for one of the variables.
- Multiply the first equation by 2:
(2)(4x−6y)= (2)(-48) --> 8x - 12y = -96
Now we can add the equations together to eliminate x:
(8x - 12y) + (-2x + 4y) = -96 + 30
6x - 8y = -66
Now, let's solve this equation for one variable. For example, we can isolate x:
6x = 8y - 66
x = (8y - 66)/6
x = (4y - 33)/3
Now we can substitute this expression for x back into either of the original equations to solve for y.
Let's use the first equation, 4x−6y=−48:
4((4y - 33)/3) - 6y = -48
(16y - 132)/3 - 6y = -48
16y - 132 - 18y = -144
-2y - 132 = -144
-2y = -144 + 132
-2y = -12
y = -12 / -2
y = 6
Finally, substitute the value of y back into x = (4y - 33)/3:
x = (4(6) - 33)/3
x = (24 - 33)/3
x = -9/3
x = -3
Therefore, the solution to the system of equations is x = -3 and y = 6.
4. To solve the system of equations 3x−4y=40 and 5x−5y=55:
We can either use substitution or elimination to solve the system.
- Using elimination, let's multiply the first equation by 5 and the second equation by 4 to eliminate the y variable:
(5)(3x - 4y) = (5)(40) --> 15x - 20y = 200
(4)(5x - 5y) = (4)(55) --> 20x - 20y = 220
Now subtract the second equation from the first to eliminate the y variable:
(15x - 20y) - (20x - 20y) = 200 - 220
15x - 20x - 20y + 20y = -20
-5x = -20
x = -20 / -5
x = 4
Substitute the value of x back into either equation, let's use the first equation:
3(4) - 4y = 40
12 - 4y = 40
-4y = 40 - 12
-4y = 28
y = 28 / -4
y = -7
So, the solution to the system is x = 4 and y = -7.
5. To determine the number of solutions in the system of equations 12x−15y=18 and 4x−5y=6:
Again, we can compare the coefficients of x and y in both equations.
Here, we can see that the coefficients of x and y are NOT the same in both equations.
When this happens, it indicates that the lines represented by the equations are not parallel and will intersect at a single point. Therefore, the system has one unique solution.
step by step bot is right
1. The solution to the system of equations is x=4 and y=1.
2. The system of equations has infinitely many solutions.
3. The solution to the system of equations is x=-3 and y=-6.
4. The solution to the system of equations is x=20 and y=-5.
5. The system of equations has one unique solution.