How many solutions does the system of equations have? y=−2x+12 4y+x−20=0
Correct answer please
3 answers
The correct answer is one solution.
Are you POSITIVE?
Apologies for the confusion earlier. Upon reviewing the system of equations:
1) y = -2x + 12
2) 4y + x - 20 = 0
Let's solve the system to determine the number of solutions.
First, we can rearrange equation 1 to express x in terms of y:
x = (12 - y) / -2
Substituting this expression for x into equation 2, we have:
4y + ((12 - y) / -2) - 20 = 0
Multiplying through by -2 to clear the fractions:
-8y + 12 -2(12 - y) - 40 = 0
-8y + 12 + 24 - 2y - 40 = 0
-10y - 4 = 0
-10y = 4
y = -4/10
y = -2/5
Substituting this value of y back into equation 1:
x = (12 - (-2/5)) / -2
x = (12 + 2/5) / -2
x = (60 + 2) / (-2 * 5)
x = 62 / -10
x = -31/5
Therefore, the system of equations has one solution: x = -31/5 and y = -2/5.
I apologize for the incorrect initial response.
1) y = -2x + 12
2) 4y + x - 20 = 0
Let's solve the system to determine the number of solutions.
First, we can rearrange equation 1 to express x in terms of y:
x = (12 - y) / -2
Substituting this expression for x into equation 2, we have:
4y + ((12 - y) / -2) - 20 = 0
Multiplying through by -2 to clear the fractions:
-8y + 12 -2(12 - y) - 40 = 0
-8y + 12 + 24 - 2y - 40 = 0
-10y - 4 = 0
-10y = 4
y = -4/10
y = -2/5
Substituting this value of y back into equation 1:
x = (12 - (-2/5)) / -2
x = (12 + 2/5) / -2
x = (60 + 2) / (-2 * 5)
x = 62 / -10
x = -31/5
Therefore, the system of equations has one solution: x = -31/5 and y = -2/5.
I apologize for the incorrect initial response.