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Solve a system of equations with a parameter a:

2x + y = 3
ax − 3y = 4

Determine for which values of the parameter this system has (i) a unique solution, (ii) no solutions, (iii) infinitely many solutions.

User Paulrezmer
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1 Answer

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Final answer:

To solve the system of equations with the parameter a: 2x + y = 3 and ax − 3y = 4, we can use the elimination method. The system has a unique solution if 6 - 2a ≠ 0, no solution if 6 - 2a = 0 and 1 ≠ 0, and infinitely many solutions if 6 - 2a = 0 and 1 = 0.

Step-by-step explanation:

To solve the system of equations with the parameter a: 2x + y = 3 and ax − 3y = 4, we can use the method of substitution or elimination. Let's use the elimination method.

Multiply the first equation by 3 and the second equation by 2 to make the coefficients of y in both equations the same:

6x + 3y = 9

2ax − 6y = 8

Now, subtract the second equation from the first equation:

6x + 3y − (2ax − 6y) = 9 - 8

6x + 3y - 2ax + 6y = 1

Combine like terms:

6x - 2ax + 9y = 1

Factor out x from the first two terms and y from the last term:

x(6 - 2a) + 9y = 1

This is a linear equation in the form y = mx + b, where m is the slope.

If the slope is not zero (m ≠ 0), then the system has a unique solution. If the slope is zero (m = 0), then the system has infinitely many solutions.

Therefore, for what values of the parameter a will the system have (i) a unique solution, (ii) no solution, or (iii) infinitely many solutions?

If 6 - 2a ≠ 0, then the system has a unique solution.

If 6 - 2a = 0 and 1 ≠ 0, then the system has no solution.

If 6 - 2a = 0 and 1 = 0, then the system has infinitely many solutions.

User Davidisdk
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