Question 1204237
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The two responses you have received show basically the same solution, using two equations with two unknowns solved using substitution.<br>
The other standard algebraic way of solving a system of two equations in two unknowns is elimination.<br>
Let x be the number of hours worked at $85 per hour and y be the number of hours worked at $105 per hour.  Then<br>
the total time was 20 hours: {{{x+y=20}}}
the total charge was $1800: {{{85x+105y=1800}}}<br>
Simplify the second equation by dividing by the greatest common factor, 5.<br>
{{{17x+21y=360}}}<br>
Multiply the first equation by 17:<br>
{{{17x+17y=340}}}<br>
Eliminate x by subtracting one equation from the other.<br>
{{{4y=20}}}
{{{y=5}}}<br>
Use y=5 in the first equation {{{x+y=20}}} to find x=15.<br>
ANSWER: The mechanic who charges $85 per hour worked 15 hours on the car; the mechanic who charged $105 per hour worked 5 hours.<br>
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Here is a quick and easy informal solution that uses exactly the same calculations as the formal elimination method above.<br>
If all 20 hours were by the mechanic who charged $85 per hour, the total charge would be $1700.  The actual charge was $100 more than that.  Since the second mechanic charged $20 more per hour than the first, the number of hours he worked was $100/$20 = 5; so the first mechanic worked 15 hours.<br>
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And here is another, very different, informal way of solving any 2-part "mixture" problem like this.<br>
The average charge per hour was $1800/20 = $90.  Use a number line if it helps to observe/calculate that $90 is one-fourth of the way from $85 to $105.  That means the mechanic who charged $105 per hour worked one-fourth of the total of 20 hours -- i.e., 5 hours; meaning the other mechanic worked 15 hours.<br>