You work no more than 12 hours a week at your two jobs. The first job pays you $8 an hour, and the second job pays you $10 an hour. You must earn at least $100 each week. Which graph represents this situation (let x represent hours at job 1 and y represent hours at job 2)?

Respuesta :

Answer:

10 hours at job (1)

2 hours at job (2)

Step-by-step explanation:

As per the given information, one earns ($8) dollars at one of their jobs, and ($10) hours at the other. One must earn a total of ($100) dollars, and can work no more than (12) hours. Let (x) be the hours worked at job 1 and (y) be the hours worked at job two.

Since one can work no more than (12) hours, the sum of (x) and (y) must be (12), therefore the following equation can be formed;

[tex]x+y=12[/tex]

One earns ($8) dollars at one of their jobs and ($10) at the other, but one earns a total of (100) one can form an equation to represent this situation.  Multiply the hours worked by the money earn per hour for each job, add up the result and set it equal to (100).

[tex]8x+10y=100[/tex]

Now set up these equations in a system;

[tex]\left \{ {{x+y=12} \atop {8x+10y=100}} \right.[/tex]

Use the process of elimination to solve this system. The process of elimination is a method of solving a system of equations. One must first manipulate one of the equations in the system such that one of the variable coefficients is the additive inverse of the other. That way, when one adds the equation, the variable cancels, one can solve for the other variable then back solve to find the value of the first variable,

[tex]\left \{ {{x+y=12} \atop {8x+10y=100}} \right.[/tex]

Manipulate,

[tex]= \left \{ {{(*-8)(x+y=12)} \atop {8x+10y=100}} \right.\\\\[/tex]

Simplify,

[tex]= \left \{ {{-8x-8y=-96} \atop {8x+10y=100}} \right.\\[/tex]

Add,

[tex]=2y=4[/tex]

Inverse operations,

[tex]y=2[/tex]

Backsolve for (x), use equation one to achieve this,

[tex]x+y=12\\[/tex]

Substitute,

[tex]x+2=12[/tex]

Inverse operations,

[tex]x=10[/tex]

ACCESS MORE