Two mechanics worked on a car. The first mechanic charged $75 per hour, and the second mechanic charged $50 per hour. The mechan

Question

1 year ago

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Two mechanics worked on a car. The first mechanic charged $75 per hour, and the second mechanic charged $50 per hour. The mechan

ics worked for a
combined total of 20 hours, and together they charged a total of $1125. How long did each mechanic work?

Mathematics

1 answer:

trapecia [35] · Answer 1 · 2023-12-10T23:16:41+03:00

Taking into account the definition of a system of linear equations, the first mechanic worked 5 hours and the second mechanic worked 15 hours.

<h3>System of linear equations</h3>

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.

<h3>Time that each mechanic work</h3>

In this case, a system of linear equations must be proposed taking into account that:

FM: working time of the first mechanic in hours.
SM: working time of the second mechanic in hours.

You know:

The first mechanic charged $75 per hour.
The second mechanic charged $50 per hour.
The mechanics worked for a combined total of 20 hours.
Together they charged a total of $1125.

So, the system of equations to be solved is

$\left \{ {{FM + SM=20} \atop {75FM+50SM=1125}} \right.$

There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.

In this case, isolating the variable FM from the first equation, you obtain:

FM=20 - SM

Now substituting the FM variable in the second equation:

75(20 - SM)+50SM=1125

75×20 - 75SM + 50SM= 1125

1500 -75SM + 50SM= 1125

-75SM + 50SM= 1125 -1500

-25SM= -375

SM= (-375)÷ (-25)

Remembering that FM=20 - SM, the variable FM is calculated as:

FM= 20 -15

Finally, the first mechanic worked 5 hours and the second mechanic worked 15 hours.