Question

1. which system of equations has exactly one solution?

a) $y=\frac{1}{2}x + 10$
$y=\frac{1}{2}x + 4$
b) $y=\frac{1}{3}x + 9$
$-3x + y = -12$
c) $5x + y = 3$
$5x + y = 2$
d) $4x + 2y = 6$
$y = -2x + 3$

Explanation:

Step1: Define intersecting lines condition

A system has exactly one solution if the slopes of the two lines are different (they are not parallel, so they intersect once). For a line in the form $y=mx+b$, $m$ is the slope. For standard form $Ax+By=C$, slope is $-\frac{A}{B}$.

Step2: Check Option A

Line 1: $y=\frac{1}{2}x+10$, slope $m_1=\frac{1}{2}$
Line 2: $y=\frac{1}{2}x+4$, slope $m_2=\frac{1}{2}$
Slopes are equal (parallel lines), no unique solution.

Step3: Check Option B

Line 1: $y=\frac{1}{3}x+9$, slope $m_1=\frac{1}{3}$
Line 2: $-3x+y=-12$ → $y=3x-12$, slope $m_2=3$
Slopes are different, lines intersect once.

Step4: Verify remaining options (optional)

Option C:
Line1: $5x+y=3$ → $y=-5x+3$, $m_1=-5$
Line2: $5x+y=2$ → $y=-5x+2$, $m_2=-5$ (parallel, no unique solution)
Option D:
Line1: $4x+2y=6$ → $y=-2x+3$, $m_1=-2$
Line2: $y=-2x+3$, $m_2=-2$ (coinciding lines, infinite solutions)

Answer:

B. $y=\frac{1}{3}x+9$; $-3x+y=-12$