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For what value of k, the system of equations x+2y=5 3x+ky−15=0 has no solution?

User Bones
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1 Answer

3 votes

Answer:

Explanation:

AI-generated answer

To determine the value of k for which the system of equations has no solution, we need to examine the coefficients of x and y in the two equations. Let's solve the system step by step:

1. Given equations:

x + 2y = 5 (equation 1)

3x + ky - 15 = 0 (equation 2)

2. We can rewrite equation 2 in terms of x and y:

3x + ky - 15 = 0

ky = -3x + 15

y = (-3x + 15)/k

3. Now, let's substitute the value of y from equation 1 into equation 2:

x + 2((-3x + 15)/k) = 5

4. Simplify the equation:

x - 6x/k + 30/k = 5

5. Combine like terms:

(kx - 6x + 30)/k = 5

6. Multiply both sides of the equation by k to eliminate the fraction:

kx - 6x + 30 = 5k

7. Rearrange the equation:

kx - 6x = 5k - 30

(k - 6)x = 5k - 30

8. For the system to have no solution, the coefficients of x on both sides must be equal, but the constant terms must be different.

9. Setting the coefficients equal:

k - 6 = 0

10. Solve for k:

k = 6

Therefore, the system of equations has no solution when k = 6.

User Rukamakama
by
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