Asked by Makayla
Given the following system of linear equations, do Part A and Part B below.
{3x = 3y - 3
{y = x + 1
Part A: Using the method of your choice (graphing, substitution or addition), find the solution to the system of linear equations. Include all of your work in your final answer.
Part B: In two or more complete sentences, explain why you chose the specific method you used to find the solution to the system.
WRITER
All Answers 1
Answered by
GPT-5 mini
AI
Part A — work and solution:
Given
1) 3x = 3y − 3
2) y = x + 1
Substitute equation (2) into (1):
3x = 3(x + 1) − 3
3x = 3x + 3 − 3
3x = 3x
This is an identity (true for every x), so the two equations are equivalent and not independent. Therefore there are infinitely many solutions: every point on the line y = x + 1 satisfies both equations.
Solution set: {(x, y) | y = x + 1}, or parametrically {(t, t + 1) | t ∈ ℝ}.
Part B — explanation of method:
I used substitution because the second equation is already solved for y, so substituting y = x + 1 directly into the first equation is quickest and reveals whether the system has a unique solution, no solution, or infinitely many. In this case the substitution produced an identity, showing the two equations represent the same line and therefore infinitely many solutions.
Given
1) 3x = 3y − 3
2) y = x + 1
Substitute equation (2) into (1):
3x = 3(x + 1) − 3
3x = 3x + 3 − 3
3x = 3x
This is an identity (true for every x), so the two equations are equivalent and not independent. Therefore there are infinitely many solutions: every point on the line y = x + 1 satisfies both equations.
Solution set: {(x, y) | y = x + 1}, or parametrically {(t, t + 1) | t ∈ ℝ}.
Part B — explanation of method:
I used substitution because the second equation is already solved for y, so substituting y = x + 1 directly into the first equation is quickest and reveals whether the system has a unique solution, no solution, or infinitely many. In this case the substitution produced an identity, showing the two equations represent the same line and therefore infinitely many solutions.
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