Question

1. which equation, together with ( y = -1.5x + 3 ), makes a system with one solution?

a. ( y = -1.5x + 6 )
b. ( y = -1.5x )
c. ( 2y = -3x + 6 )
d. ( 2y + 3x = 6 )
e. ( y = -2x + 3 )

1. the system ( x - 6y = 4 ), ( 3x - 18y = 4 ) has no solution.

a. change one constant or coefficient to make a new system with one solution.
b. change one constant or coefficient to make a new system with an infinite number of solutions.

Explanation:

Question 3

Step1: Identify slope of given line

The given line is $y = -1.5x + 3$, slope $m_1 = -1.5$.

Step2: Analyze each option's slope

A. $y=-1.5x+6$, slope $m=-1.5$ (parallel, no solution)
B. $y=-1.5x$, slope $m=-1.5$ (parallel, no solution)
C. Rewrite $2y=-3x+6$ as $y=-1.5x+3$, same line (infinite solutions)
D. Rewrite $2y+3x=6$ as $y=-1.5x+3$, same line (infinite solutions)
E. $y=-2x+3$, slope $m=-2
eq -1.5$ (intersects, one solution)

Step1: Rewrite original equations

First equation: $x - 6y = 4$
Second equation: $3x - 18y = 4$, rewrite as $x - 6y = \frac{4}{3}$

Step2: Adjust for one solution

For one solution, slopes must differ. Change the coefficient of $x$ or $y$ in the second equation so its slope is not equal to $\frac{1}{6}$. For example, change $3x$ to $2x$:
New second equation: $2x - 18y = 4$ (slope $\frac{2}{18}=\frac{1}{9}
eq \frac{1}{6}$)

Step1: Rewrite original equations

First equation: $x - 6y = 4$
Second equation: $3x - 18y = 4$, rewrite as $x - 6y = \frac{4}{3}$

Step2: Adjust for infinite solutions

For infinite solutions, equations must be scalar multiples. Multiply the first equation by 3: $3x - 18y = 12$. Change the constant term of the second equation from 4 to 12.

Answer:

E. $y = -2x + 3$

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Question 4a