1. The intersection point is (2, -1).
2. The intersection point is (2, 7).
1. Review the graphs of a system of two linear equations in two variables: x − 5y = 7 and 3x − 2y = −4. Find the solution to both equations.
On a graph there's 3x - 2y = -4 and x - 5y = 3.
The intersection point is (____).
2. Review the graphs of a system of two linear equations in two variables: 4x + 3y = 15 and y = x + 5. Find the solution to both equations.
On a graph there's 4x + 3y = 15 and y = x + 5.
The intersection point is (____).
11 answers
wrong
I apologize for the mistake. Let's correct the solutions:
1. The intersection point for the system of equations x − 5y = 7 and 3x − 2y = −4 is (2, -1).
2. The intersection point for the system of equations 4x + 3y = 15 and y = x + 5 is (2, 7).
1. The intersection point for the system of equations x − 5y = 7 and 3x − 2y = −4 is (2, -1).
2. The intersection point for the system of equations 4x + 3y = 15 and y = x + 5 is (2, 7).
how about for the first one if there's two slopes the first one that starts at (-5, -5) and ends at (5, 2) and has the equation 3x - 2y = -4 and the second one starts at (-5, -2) and ends at (5, 1) and has the equation x - 5y = 3?
Given the two lines with the equations 3x - 2y = -4 and x - 5y = 3, and with the points (-5, -5) and (5, 2) for the first line, the slope of the first line can be calculated as follows:
First line:
- Slope (m) = (y2 - y1) / (x2 - x1)
- Slope = (2 - (-5)) / (5 - (-5)) = 7 / 10 = 0.7
Therefore, the equation 3x - 2y = -4 can be rewritten as y = (3/2)x + 2.
Similarly, calculating the slope of the second line with the points (-5, -2) and (5, 1):
Second line:
- Slope (m) = (y2 - y1) / (x2 - x1)
- Slope = (1 - (-2)) / (5 - (-5)) = 3 / 10 = 0.3
Thus, the equation x - 5y = 3 can be rewritten as y = (1/5)x - 0.6.
To find the intersection point of these two lines, we can solve the system of equations:
(3/2)x + 2 = (1/5)x - 0.6
(3/2)x - (1/5)x = -0.6 - 2
(7/10)x = -2.6
x = -26/7
Substitute x back into one of the equations; let's use the first line equation:
y = (3/2)(-26/7) + 2
y = -39/7 + 2
y = -39/7 + 14/7
y = -25/7
Therefore, the intersection point of the two lines 3x - 2y = -4 and x - 5y = 3 is (-26/7, -25/7).
First line:
- Slope (m) = (y2 - y1) / (x2 - x1)
- Slope = (2 - (-5)) / (5 - (-5)) = 7 / 10 = 0.7
Therefore, the equation 3x - 2y = -4 can be rewritten as y = (3/2)x + 2.
Similarly, calculating the slope of the second line with the points (-5, -2) and (5, 1):
Second line:
- Slope (m) = (y2 - y1) / (x2 - x1)
- Slope = (1 - (-2)) / (5 - (-5)) = 3 / 10 = 0.3
Thus, the equation x - 5y = 3 can be rewritten as y = (1/5)x - 0.6.
To find the intersection point of these two lines, we can solve the system of equations:
(3/2)x + 2 = (1/5)x - 0.6
(3/2)x - (1/5)x = -0.6 - 2
(7/10)x = -2.6
x = -26/7
Substitute x back into one of the equations; let's use the first line equation:
y = (3/2)(-26/7) + 2
y = -39/7 + 2
y = -39/7 + 14/7
y = -25/7
Therefore, the intersection point of the two lines 3x - 2y = -4 and x - 5y = 3 is (-26/7, -25/7).
forget what I said wrong
Understood! Let's reevaluate the system of equations with the given equations and points:
Given equations:
1. 3x - 2y = -4
2. x - 5y = 3
Let's analyze the information provided for the lines:
For the first line:
Start point: (-5, -5)
End point: (5, 2)
The first line can be rewritten as:
1. 3x - 2y = -4 -> y = (3/2)x + 2
For the second line:
Start point: (-5, -2)
End point: (5, 1)
The second line can be rewritten as:
2. x - 5y = 3 -> y = (1/5)x - 0.6
Now, we will find the intersection point by setting the two equations equal to each other:
(3/2)x + 2 = (1/5)x - 0.6
Solving for x:
(3/2)x - (1/5)x = -0.6 - 2
(14/10)x = -2.6
x = -26/7
Substitute x back into the first equation:
y = (3/2)(-26/7) + 2
y = -39/7 + 2
y = -25/7
Therefore, the intersection point of the two lines is at (-26/7, -25/7).
Given equations:
1. 3x - 2y = -4
2. x - 5y = 3
Let's analyze the information provided for the lines:
For the first line:
Start point: (-5, -5)
End point: (5, 2)
The first line can be rewritten as:
1. 3x - 2y = -4 -> y = (3/2)x + 2
For the second line:
Start point: (-5, -2)
End point: (5, 1)
The second line can be rewritten as:
2. x - 5y = 3 -> y = (1/5)x - 0.6
Now, we will find the intersection point by setting the two equations equal to each other:
(3/2)x + 2 = (1/5)x - 0.6
Solving for x:
(3/2)x - (1/5)x = -0.6 - 2
(14/10)x = -2.6
x = -26/7
Substitute x back into the first equation:
y = (3/2)(-26/7) + 2
y = -39/7 + 2
y = -25/7
Therefore, the intersection point of the two lines is at (-26/7, -25/7).
1. Review the graphs of a system of two linear equations in two variables: x − 5y = 7 and 3x − 2y = −4. Find the solution to both equations.
On a graph there's 3x - 2y = -4 and x - 5y = 3.
The intersection point is (____).
2. Review the graphs of a system of two linear equations in two variables: 4x + 3y = 15 and y = x + 5. Find the solution to both equations.
On a graph there's 4x + 3y = 15 and y = x + 5.
The intersection point is (____).
On a graph there's 3x - 2y = -4 and x - 5y = 3.
The intersection point is (____).
2. Review the graphs of a system of two linear equations in two variables: 4x + 3y = 15 and y = x + 5. Find the solution to both equations.
On a graph there's 4x + 3y = 15 and y = x + 5.
The intersection point is (____).
Let's find the solutions to the two systems of equations:
1. x − 5y = 7
3x − 2y = −4
To find the intersection point, we can solve this system of equations. We have the equations:
3x - 2y = -4
x - 5y = 7
Multiplying the second equation by 3 to match the x coefficients, we get:
3x - 15y = 21
Now we have:
3x - 2y = -4
3x - 15y = 21
Subtracting the second equation from the first, we get:
13y = -25
y = -25/13
Substitute y back into the second equation to solve for x:
x - 5(-25/13) = 7
x + 125/13 = 7
x = 7 - 125/13
x = 91/13
Therefore, the intersection point of the first system of equations is (91/13, -25/13).
2. 4x + 3y = 15
y = x + 5
Substitute the value of y from the second equation into the first equation:
4x + 3(x + 5) = 15
4x + 3x + 15 = 15
7x = 0
x = 0
Substitute x back into the second equation to solve for y:
y = 0 + 5
y = 5
Therefore, the intersection point of the second system of equations is (0, 5).
1. x − 5y = 7
3x − 2y = −4
To find the intersection point, we can solve this system of equations. We have the equations:
3x - 2y = -4
x - 5y = 7
Multiplying the second equation by 3 to match the x coefficients, we get:
3x - 15y = 21
Now we have:
3x - 2y = -4
3x - 15y = 21
Subtracting the second equation from the first, we get:
13y = -25
y = -25/13
Substitute y back into the second equation to solve for x:
x - 5(-25/13) = 7
x + 125/13 = 7
x = 7 - 125/13
x = 91/13
Therefore, the intersection point of the first system of equations is (91/13, -25/13).
2. 4x + 3y = 15
y = x + 5
Substitute the value of y from the second equation into the first equation:
4x + 3(x + 5) = 15
4x + 3x + 15 = 15
7x = 0
x = 0
Substitute x back into the second equation to solve for y:
y = 0 + 5
y = 5
Therefore, the intersection point of the second system of equations is (0, 5).
1. Review the graphs of a system of two linear equations in two variables: x − 5y = 7 and 3x − 2y = −4. Find the solution to both equations.
On a graph there's 3x - 2y = -4 and x - 5y = 3.
The intersection point is (____).
On a graph there's 3x - 2y = -4 and x - 5y = 3.
The intersection point is (____).
To find the solution to the system of equations x − 5y = 7 and 3x − 2y = −4, let's first write the equations in slope-intercept form (y = mx + b) by solving for y:
1. x − 5y = 7
x = 5y + 7
2. 3x − 2y = −4
3x = 2y - 4
3x/2 = y - 2
y = 3x/2 + 2
Now, let's graph the equations y = 5x/3 + 7/3 and y = 3x/2 + 2:
The intersection point of the two lines is the solution to the system of equations. Let's find the coordinates of the intersection point by setting the two equations equal to each other:
5y + 7 = 3x/2 + 2
5y = 3x/2 - 5
y = (3/10)x - 1
Now, substitute y back into one of the original equations:
x - 5((3/10)x - 1) = 3
x - (3/2)x + 5 = 3
(1/2)x + 5 = 3
(1/2)x = -2
x = -4
Substitute x back into y = (3/10)x - 1:
y = (3/10)(-4) - 1
y = -12/10 - 1
y = -14/10
y = -7/5
Therefore, the intersection point of the two lines is (-4, -7/5).
1. x − 5y = 7
x = 5y + 7
2. 3x − 2y = −4
3x = 2y - 4
3x/2 = y - 2
y = 3x/2 + 2
Now, let's graph the equations y = 5x/3 + 7/3 and y = 3x/2 + 2:
The intersection point of the two lines is the solution to the system of equations. Let's find the coordinates of the intersection point by setting the two equations equal to each other:
5y + 7 = 3x/2 + 2
5y = 3x/2 - 5
y = (3/10)x - 1
Now, substitute y back into one of the original equations:
x - 5((3/10)x - 1) = 3
x - (3/2)x + 5 = 3
(1/2)x + 5 = 3
(1/2)x = -2
x = -4
Substitute x back into y = (3/10)x - 1:
y = (3/10)(-4) - 1
y = -12/10 - 1
y = -14/10
y = -7/5
Therefore, the intersection point of the two lines is (-4, -7/5).