Question

look at the systems of equations.
( y = \frac{1}{2}x - 2 )
( y = \frac{1}{2}x + 3 )
( y = 2x - 1 )
( y = 3x - 4 )
( 2x - y = 7 )
( 6x - 3y = 14 )
( x - y = 6 )
( 2x - 2y = 12 )
( 2x - y = 7 )
( 6x - 3y = 14 )
( y = 2x - 1 )
( y = 3x - 4 )
( y = \frac{1}{2}x - 2 )
( y = \frac{1}{2}x + 3 )
select all that apply.
click or tap each correct answer.

Explanation:

Step1: Analyze first system (parallel lines)

The system is $y = \frac{1}{2}x - 2$ and $y = \frac{1}{2}x + 3$. They have the same slope $\frac{1}{2}$ and different y-intercepts, so no solution.

Step2: Analyze second system (intersecting lines)

The system is $y = 2x - 1$ and $y = 3x - 4$. Set equal:
$$2x - 1 = 3x - 4$$
Solve for $x$:
$$-1 + 4 = 3x - 2x \implies x = 3$$
Substitute $x=3$ into $y=2x-1$:
$$y = 2(3)-1 = 5$$
This system has one unique solution $(3,5)$.

Step3: Analyze third system (dependent equations)

The system is $2x - y = 7$ and $6x - 3y = 14$. Multiply first equation by 3:
$$6x - 3y = 21$$
This conflicts with $6x - 3y = 14$, so no solution.

Step4: Analyze fourth system (intersecting lines)

The system is $x - y = 6$ and $2x - 2y = 12$. Multiply first equation by 2:
$$2x - 2y = 12$$
This matches the second equation, so infinitely many solutions (all points on $x-y=6$).

Answer:

• System 1 ($y = \frac{1}{2}x - 2$; $y = \frac{1}{2}x + 3$): No solution
• System 2 ($y = 2x - 1$; $y = 3x - 4$): One solution $(3, 5)$
• System 3 ($2x - y = 7$; $6x - 3y = 14$): No solution
• System 4 ($x - y = 6$; $2x - 2y = 12$): Infinitely many solutions