Answer:
The second, fourth and B option are correct.
Step-by-step explanation:
In order to solve this problem, we are going to define the following variables :
[tex]X:[/tex] ''Minutes she used her pay-as-you-go phone for a call''
[tex]Y:[/tex] ''Minutes of data she used''
Then, we are going to make a linear system of equations to find the values of [tex]X[/tex] and [tex]Y[/tex].
''This month, she used a total of 85 minutes'' ⇒
[tex]X+Y=85[/tex] (I)
(I) is the first equation of the system.
''The bill was $31'' ⇒
[tex](\frac{7}{20})X+(\frac{2}{5})Y=31[/tex] (II)
(II) is the second equation of the system.
The system of equations will be :
[tex]\left \{ {{X+Y=85} \atop {(\frac{7}{20})X+(\frac{2}{5})Y=31}} \right.[/tex]
The second option ''The system of equations is [tex]X+Y=85[/tex] and [tex](\frac{7}{20})X+(\frac{2}{5})Y=31[/tex] .'' is correct
Now, to solve the system, we can eliminate the x-variable from the equations by multiplying the equation with the fractions by 20 and multiplying the other equation by -7. Then, we can sum them to obtain the value of [tex]Y[/tex] :
[tex]X+Y=85[/tex] (I)
[tex](\frac{7}{20})X+(\frac{2}{5})Y=31[/tex] (II) ⇒
[tex](-7)X+(-7)Y=-595[/tex] (I)'
[tex]7X+8Y=620[/tex] (II)'
If we sum (I)' and (II)' ⇒
[tex](-7)X+(-7)Y+7X+8Y=-595+620[/tex] ⇒ [tex]Y=25[/tex]
If we replace this value of [tex]Y[/tex] in (I) ⇒
[tex]X+Y=85\\X+25=85\\X=60[/tex]
The fourth option ''To eliminate the x-varible from the equations, you can multiply the equation with the fractions by 20 and multiply the other equation by -7'' is correct.
With the solution of the system :
[tex]\left \{ {{X=60} \atop {Y=25}} \right.[/tex]
We answer that the option ''B-She used 60 minutes for calling and 25 minutes for data'' is correct.
