Question

Ms. Rodriguez purchased markers to use on her class whiteboard. She bought a total of 16 markers and three times as many blue markers as red markers. Set up the two equations that can be used to find the number of markers of each color that Ms. Rodriguez purchased. Let constraint 1 refer to the equation that represents the total number of markers. Let constraint 2 refer to the equation that describes the ratio of the number of blue markers to red markers.
Only constraint ____
would be met if Ms. Rodriguez purchased 3 red markers and 9 blue markers.

Only constraint ____
would be met if Ms. Rodriguez purchased 6 red markers and 10 blue markers.

Answer 1

Answer:

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Hold on I think I figured it out it might be wrong so don't take my word for this but, constraint 1... (R+B=12), constraint 2... (R+B=16)

Step-by-step explanation:

Answer 2

Constraint $1$ is $\boldsymbol{x+y=16}$

Constraint $2$ is $\boldsymbol{\frac{x}{y}=3}$

Only constraint $2$ would be met if Ms. Rodriguez purchased 3 red markers and 9 blue markers.

Only constraint $1$ would be met if Ms. Rodriguez purchased 6 red markers and 10 blue markers.

The linear equations in two variables are of the highest exponent order of $\mathbf{1}$ and have one, none, or infinitely many solutions. The standard form of a two-variable linear equation is $\mathbf{ax+by+c=0}$ where $x,y$ are the two variables.

Let $\mathbf{x,y}$ denote number of blue markers, red markers respectively.

Total number of markers $=\mathbf{16}$

$\mathbf{x+y=16}$ (constraint $1$)

Number of blue markers is three times number of red markers.

$\mathbf{x=3y}$

$\mathbf{\frac{x}{y}=3}$ (constraint $2$)

Ms. Rodriguez purchased $3$ red markers and $9$ blue markers.

Put values of $x,y$ as $9,3$ respectively in the following equations:

$\mathbf{x+y=16}$

      $\mathbf{\frac{x}{y}=3}$

$9+3\neq 16$

     $\frac{9}{3}=3$

Only constraint $2$ is met.

Ms. Rodriguez purchased $6$ red markers and $10$ blue markers.

Put values of $x,y$ as $10,6$ respectively in the following equations:

$\mathbf{x+y=16}$

      $\mathbf{\frac{x}{y}=3}$

$10+6=16$

     $\frac{10}{6}\neq 3$

Only constraint $1$ is met.

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