Digital SAT Math Practice Questions - Medium : Systems of two linear equations in two variables - New Syllabus
DSAT MAth Practice questions – all topics
- Algebra Weightage: 35% Questions: 13-15
- Linear equations in one variable
- Linear equations in two variables
- Linear functions
- Systems of two linear equations in two variables
- Linear inequalities in one or two variables
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SAT MAth and English – full syllabus practice tests
▶️ Answer/Explanation
Answer: A
A system has no solution if lines are parallel and distinct. In slope-intercept form \( y = mx + b \), parallel lines have the same slope \( m \) but different intercepts \( b \).
For A: \( y = 6x + 3 \) and \( y = 6x + 9 \) have slope \( m = 6 \) and intercepts \( b = 3 \) and \( b = 9 \), so no solution.
For B: \( y = 10 \) and \( y = 10x + 10 \) intersect at \( (0, 10) \), so has solution.
For C: \( y = 14x + 14 \) and \( y = 10x + 14 \) have different slopes, intersect at \( (0, 14) \), so has solution.
For D: \( x = 3 \) and \( y = 10 \) intersect at \( (3, 10) \), so has solution.
▶️ Answer/Explanation
Answer: A
Solution is the intersection point, approximately \((\frac{3}{2}, 0)\) from the graph.
For A: \( y = 0 \) is a horizontal line (y = 0), \( x = \frac{3}{2} \) is a vertical line (x = 1.5), intersecting at \((\frac{3}{2}, 0)\).
For B: \( y = \frac{3}{2} \) and \( x = 0 \) intersect at \( (0, \frac{3}{2}) \).
For C: \( y = 0 \) and \( x = 1 \) intersect at \( (1, 0) \).
For D: \( y = 1 \) and \( x = 0 \) intersect at \( (0, 1) \).
▶️ Answer/Explanation
Answer: A
System: \( y = 4x + 1 \) and \( y = 4x + 3 \)
Both have slope 4, but different intercepts (1 and 3).
Parallel lines with different intercepts do not intersect.
Thus, zero solutions.