Solve the system of two linear equations. −3x−4y=−23 3x−5y=5 (
We have the following system of linear equations:
−3x − 4y = −23
3x − 5y = 5
To solve it, let's first use the addition method (also known as the elimination method) to eliminate one variable. We can do this by adding the two equations together, as they're set up to eliminate x:
(-3x − 4y) + (3x − 5y) = -23 + 5
The x terms cancel each other out:
-3x + 3x = 0
-4y - 5y = -9y
And on the right side, we get:
-23 + 5 = -18
Now we have:
-9y = -18
Now, we can solve for y by dividing both sides of the equation by -9:
−9y / -9 = -18 / -9
y = 2
We found that y = 2. Now let's use this value to solve for x. We can plug y = 2 into either one of the original equations, but let's use the first one:
−3x − 4y = −23
−3x − 4(2) = −23
−3x − 8 = −23
Now add 8 to both sides to isolate -3x:
−3x = −23 + 8
−3x = −15
Finally, divide both sides by -3 to solve for x:
−3x / -3 = −15 / -3
x = 5
So the solution to the system of equations is:
x = 5
y = 2
Our ordered pair solution is (5, 2).
How many solutions does the system of equations have?
8x−6y=48
8x+6y=48 (1 point)
To determine how many solutions the system of equations has, we can either solve the system or analyze the coefficients of x and y to understand their relationship. Let's do the latter first.
Here are the given equations:
8x − 6y = 48
8x + 6y = 48
Both equations seem to have the same coefficients for x and the same constant term on the right side, but the coefficients for y have opposite signs. If we subtract the second equation from the first, we'll get:
(8x − 6y) − (8x + 6y) = 48 − 48
The x terms will cancel out:
8x − 8x = 0
−6y − 6y = −12y
The right side becomes zero:
48 − 48 = 0
So the resulting equation is:
−12y = 0
This means:
y = 0
Now, substituting y = 0 into either of the original equations to solve for x, we'll use the first equation:
8x − 6y = 48
8x − 6(0) = 48
8x = 48
Now we can solve for x by dividing by 8:
8x / 8 = 48 / 8
x = 6
So, we have the solution:
x = 6
y = 0
Now that we've found a single solution (6, 0) to the system, we can say that the system has exactly one unique solution. This is also supported by the fact that the two lines represented by the equations are not parallel (they have the same slope) and not the same line (they are not identical), so they intersect at exactly one point, which is (6, 0).
how many solutions does it have
The system of equations has one unique solution, which is the point of intersection of the two lines represented by the equations. We found that solution to be (x, y) = (6, 0).
To clarify, a system of linear equations can have:
1. Exactly one solution when the lines intersect at a single point (they are not parallel or identical).
2. No solution if the lines are parallel and do not intersect.
3. Infinitely many solutions if the lines are coincident (identical in every point).
In the case of the given system:
8x − 6y = 48
8x + 6y = 48
We found these two lines intersect at one point, which confirms there is exactly one solution.
mination with multiplication.
4x−6y=−48
−2x+4y=30(1 point)
Responses
(−33,−9)
left parenthesis negative 33 comma negative 9 right parenthesis
(−3, 6)
left parenthesis negative 3 comma 6 right parenthesis
(−21,−6)
left parenthesis negative 21 comma negative 6 right parenthesis
(3, 9)
left parenthesis 3 comma 9 right parenthesis
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To solve the system of equations using elimination with multiplication:
4x − 6y = −48
−2x + 4y = 30
We can multiply the second equation by 2 to make the coefficient of x in both equations the same (but with opposite signs) to eliminate x when we add the equations together.
Multiplying the second equation by 2 gives us:
−4x + 8y = 60
Now we have:
4x − 6y = −48
−4x + 8y = 60
Adding these two equations together will eliminate the x variable:
(4x − 6y) + (−4x + 8y) = −48 + 60
Which simplifies to:
0x + 2y = 12
Now we can solve for y:
2y = 12
y = 12 / 2
y = 6
With the value of y, we can now substitute it back into one of the original equations to find x. Let's use the first equation:
4x − 6y = −48
Substitute y = 6 into the equation:
4x − 6(6) = −48
4x − 36 = −48
Add 36 to both sides to isolate 4x:
4x = −48 + 36
4x = −12
Now solve for x:
x = −12 / 4
x = −3
So the solution to the system of equations is:
x = −3
y = 6
This corresponds to the second option:
(−3, 6)
Solve the system of equations.
3x−4y=40
5x−5y=55 (1 point)
Responses
(14, 3)
left parenthesis 14 comma 3 right parenthesis
(8,−3)
left parenthesis 8 comma negative 3 right parenthesis
(18, 7)
left parenthesis 18 comma 7 right parenthesis
(4,−7)
To solve the system of equations:
3x − 4y = 40
5x − 5y = 55
we can use either the substitution method or the elimination method. Let's use the elimination method for this example.
First, let's attempt to eliminate one variable. However, in this case, the coefficients of x and y are not lining up well for immediate elimination. So, we need to multiply each equation by a certain number to get either the x or y coefficients to match up.
Let's try to match the y coefficients. To eliminate y, we can multiply the first equation by 5 and the second equation by 4, so both equations have a term of 20y (but with opposite signs).
Multiplying the first equation by 5 gives us:
15x − 20y = 200
Multiplying the second equation by 4 gives us:
20x − 20y = 220
Now we have:
15x − 20y = 200
20x − 20y = 220
Now let's subtract the first new equation from the second new equation to eliminate y:
20x − 20y - (15x − 20y) = 220 - 200
This simplifies to:
20x − 20y - 15x + 20y = 220 - 200
5x = 20
Now we can solve for x:
x = 20 / 5
x = 4
With the value of x, we can now substitute it back into one of the original equations to find y. Let's use the first equation:
3x − 4y = 40
Substitute x = 4 into the equation:
3(4) − 4y = 40
12 − 4y = 40
Subtract 12 from both sides to isolate −4y:
−4y = 40 − 12
−4y = 28
Now solve for y:
y = 28 / (−4)
y = −7
The solution to the system of equations is:
x = 4
y = −7
This corresponds to the option:
(4, −7)
How many solutions does the system have?
12x−15y=18
4x−5y=6 (1 point)
Responses
two solutions
two solutions
one solution
one solution
infinitely many solutions
infinitely many solutions
no solution