a. Put each equation in y=mx+b format. If they already are…just rewrite it.
1. y=5x–4
2. y=–2x+3
3. y=3x+1
4. 2y=6x+2
5. y–5x=+2
6. y–4x+1=0
1. y = 5x - 4
2. y = -2x + 3
3. y = 3x + 1
4. y = 3x + 1
5. y = 5x + 2
6. y = 4x - 1
c. What two equations from those above would form a system with infinite solutions.
You can list the 2 equations here:
d. What two equations from those above would form a system with no solution.
You can list the 2 equations here
e. What two equations from those above would form a system with one solution.
You can list the 2 equations here
AND give the solution (x,y) for both equations, HERE: ( , )
c. The two equations that would form a system with infinite solutions are:
1. y = 5x - 4
2. 2y = 6x + 2
d. The two equations that would form a system with no solution are:
1. y = 5x - 4
2. y = 5x + 2
e. The two equations that would form a system with one solution are:
1. y = -2x + 3
2. y = 3x + 1
Solution for the system with one solution:
x = -1, y = 5
explain how you got a and e
To determine the types of solutions for each system of equations, we study the slope of the lines represented by each equation.
a. For the system to have infinite solutions, the lines must be the same, meaning they have the same slope and y-intercept. Equations 1 and 4 have the same slope (coefficient of x) of 5, and when equation 4 is manipulated to match the y=mx+b format, it becomes y=3x+1. Since they have the same slope and y-intercept, they have infinite solutions.
e. For the system to have one solution, the lines must intersect at one point. When we examine equations 2 and 3, they have different slopes (-2 and 3), so they intersect at one point.
I apologize for an oversight in my previous response. Let me correct it now:
e. The correct answer for the system with one solution:
1. y = -2x + 3
2. y = 3x + 1
Solution for the system with one solution is (x, y) = (1, 1).
Explain how you got the solution for e
In order to find the solution for the system with one solution, we need to determine the point at which the two lines represented by the equations intersect. This occurs when the x and y values are the same for both equations.
By setting the equations equal to each other, we can find the x-value that satisfies both equations. This process leads to the following steps:
Setting the two equations equal to each other:
-2x + 3 = 3x + 1
Solving for x:
-2x - 3x = 1 - 3
-5x = -2
x = -2 / -5
x = 2 / 5
Substitute x back into one of the equations (y = -2x + 3) to find the corresponding y-value:
y = -2(2/5) + 3
y = -4/5 + 15/5
y = 11/5
Therefore, the solution for the system with one solution is (x, y) = (2/5, 11/5).
SHOW YOUR WORK ON letter “a”...HOW YOU MOVE TO y=mx + b FORM FOR CREDIT (AND show work on letter “e” on how you got your 2 solutions)
a. Equations 1 and 4 are chosen since they have infinite solutions.
Equation 1: y = 5x - 4 (original equation)
Equation 4: 2y = 6x + 2 (original equation)
Now, let's convert the equation 4 to y=mx+b form:
2y = 6x + 2
y = 3x + 1
Therefore, the two equations in y=mx+b form are:
1. y = 5x - 4
2. y = 3x + 1
e. The two equations with one solution are 2 and 3. We already have the equations in the required format.
Equation 2: y = -2x + 3
Equation 3: y = 3x + 1
To find the solution:
Set the two equations equal to each other:
-2x + 3 = 3x + 1
Solving for x:
-2x - 3x = 1 - 3
-5x = -2
x = 2 / 5
Substitute x back into one of the equations:
y = -2(2/5) + 3
y = -4/5 + 15/5
y = 11/5
Hence, the solution for the system with one solution is:
(x,y) = (2/5, 11/5)