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Given that the system of equations mx + 2y = 10; 3x – 2y = 0 have the integer solution. Then the possible values of m are
- a)2 and 8
- b)2 and – 8
- c)– 2 and – 8
- d)– 2 and 8
Correct answer is option 'B,C'. Can you explain this answer?
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Most Upvoted Answer
Given that the system of equations mx + 2y = 10; 3x – 2y = 0 hav...
mx + 2y = 10 …(i)
3x – 2y = 0
⇒ 3x = 2y …(ii)
Put in equation (i)
(m + 3)x = 10
x = 10/(m+3)
∴ x and y and integers
∴ m = – 2, 2, – 8
Option (2) or (3) correct.
3x – 2y = 0
⇒ 3x = 2y …(ii)
Put in equation (i)
(m + 3)x = 10
x = 10/(m+3)
∴ x and y and integers
∴ m = – 2, 2, – 8
Option (2) or (3) correct.
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Community Answer
Given that the system of equations mx + 2y = 10; 3x – 2y = 0 hav...
The given system of equations is:
1) mx + 2y = 10
2) 3x - my = 7
To solve this system, we can use the method of substitution.
From equation 1, we can solve for x in terms of y:
mx + 2y = 10
mx = 10 - 2y
x = (10 - 2y)/m
Substituting this expression for x into equation 2, we have:
3(10 - 2y)/m - my = 7
30 - 6y - my = 7
-6y - my = 7 - 30
-6y - my = -23
Factoring out y:
y(-6 - m) = -23
Now we can solve for y:
y = -23/(-6 - m)
y = 23/(m + 6)
Substituting this expression for y back into equation 1, we have:
mx + 2(23/(m + 6)) = 10
mx + 46/(m + 6) = 10
Multiplying through by (m + 6) to clear the fraction:
m(m + 6)x + 46 = 10(m + 6)
m^2x + 6mx + 46 = 10m + 60
m^2x + (6m - 10m) + (46 - 60) = 0
m^2x - 4mx - 14 = 0
This is a quadratic equation in terms of x. To solve for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = m^2, b = -4m, and c = -14. Substituting these values into the quadratic formula, we have:
x = (-(-4m) ± √((-4m)^2 - 4(m^2)(-14))) / (2m^2)
x = (4m ± √(16m^2 + 56m^2)) / (2m^2)
x = (4m ± √(72m^2)) / (2m^2)
x = (4m ± 6m√2) / (2m^2)
x = (2 ± 3√2) / m
Therefore, the solutions to the system of equations are:
x = (2 ± 3√2) / m
y = 23/(m + 6)
1) mx + 2y = 10
2) 3x - my = 7
To solve this system, we can use the method of substitution.
From equation 1, we can solve for x in terms of y:
mx + 2y = 10
mx = 10 - 2y
x = (10 - 2y)/m
Substituting this expression for x into equation 2, we have:
3(10 - 2y)/m - my = 7
30 - 6y - my = 7
-6y - my = 7 - 30
-6y - my = -23
Factoring out y:
y(-6 - m) = -23
Now we can solve for y:
y = -23/(-6 - m)
y = 23/(m + 6)
Substituting this expression for y back into equation 1, we have:
mx + 2(23/(m + 6)) = 10
mx + 46/(m + 6) = 10
Multiplying through by (m + 6) to clear the fraction:
m(m + 6)x + 46 = 10(m + 6)
m^2x + 6mx + 46 = 10m + 60
m^2x + (6m - 10m) + (46 - 60) = 0
m^2x - 4mx - 14 = 0
This is a quadratic equation in terms of x. To solve for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = m^2, b = -4m, and c = -14. Substituting these values into the quadratic formula, we have:
x = (-(-4m) ± √((-4m)^2 - 4(m^2)(-14))) / (2m^2)
x = (4m ± √(16m^2 + 56m^2)) / (2m^2)
x = (4m ± √(72m^2)) / (2m^2)
x = (4m ± 6m√2) / (2m^2)
x = (2 ± 3√2) / m
Therefore, the solutions to the system of equations are:
x = (2 ± 3√2) / m
y = 23/(m + 6)
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Given that the system of equations mx + 2y = 10; 3x – 2y = 0 have the integer solution. Then the possible values of m area)2 and 8b)2 and – 8c)– 2 and – 8d)– 2 and 8Correct answer is option 'B,C'. Can you explain this answer?
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