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.
(1 point)
The intersection point is __.
To find the intersection point of the two equations, we can solve the system of equations simultaneously.
Given equations:
1) x - 5y = 7
2) 3x - 2y = -4
To solve for x, we can rewrite equation 1 as x = 5y + 7 and then substitute into equation 2:
3(5y + 7) - 2y = -4
15y + 21 - 2y = -4
13y + 21 = -4
13y = -25
y = -25/13
Now we can substitute this value back into equation 1 to solve for x:
x = 5(-25/13) + 7
x = -125/13 + 91/13
x = -34/13
Therefore, the intersection point of the two equations is (-34/13, -25/13).
thats wrong
I apologize for the mistake in my previous response.
Let's revisit the given equations:
1) x - 5y = 7
2) 3x - 2y = -4
We can solve this system of equations by elimination or substitution method.
First, let's use the elimination method to solve for x:
Multiply equation 1 by 3 and equation 2 by 1:
3(x - 5y) = 3(7)
3x - 15y = 21
3x - 2y = -4
Subtract the two equations:
-13y = 25
y = -25/13
Now, substitute y = -25/13 back into equation 1:
x - 5(-25/13) = 7
x + 125/13 = 7
x = 7 - 125/13
x = (91 - 125) / 13
x = -34/13
Therefore, the correct intersection point of the two equations is (-34/13, -25/13). Thank you for bringing this to my attention.
-34/13, -25/13 is wrong
I apologize for the error in my previous responses. Let's correct it:
Given equations:
1) x - 5y = 7
2) 3x - 2y = -4
Solving by elimination:
-3(x - 5y) = -3(7) (multiply Eq. 1 by -3)
-3x + 15y = -21
3x - 2y = -4 (Eq. 2)
Adding the two equations gives:
13y = -25
y = -25/13
Substitute y = -25/13 back into equation 1:
x - 5(-25/13) = 7
x + 125/13 = 7
x = 7 - 125/13
x = 91/13 - 125/13
x = -34/13
Therefore, the correct intersection point of the two equations is (-34/13, -25/13). Thank you for your patience.