addison earns a fixed hourly rate working as a sales clerk. if she works on a holiday, she earns a different…

addison earns a fixed hourly rate working as a sales clerk. if she works on a holiday, she earns a different hourly rate than she earns on a regular day. in one week, she earns $188.50 by working 5 hours on a holiday and 16 hours during regular days. a different week, she earns $254.00 by working 8 hours on a holiday and 20 hours during regular days. how much more is addisons holiday hourly rate than her regular hourly rate? $2.00 $8.50 $10.50 $19.00

addison earns a fixed hourly rate working as a sales clerk. if she works on a holiday, she earns a different hourly rate than she earns on a regular day. in one week, she earns $188.50 by working 5 hours on a holiday and 16 hours during regular days. a different week, she earns $254.00 by working 8 hours on a holiday and 20 hours during regular days. how much more is addisons holiday hourly rate than her regular hourly rate? $2.00 $8.50 $10.50 $19.00

Answer

Explanation:

Step1: Set up equations

Let $x$ be the holiday - hourly rate and $y$ be the regular - hourly rate. We have the system of equations: $5x + 16y=188.5$ and $8x + 20y = 254$.

Step2: Multiply the first equation

Multiply the first equation $5x + 16y=188.5$ by 4 and the second equation $8x + 20y = 254$ by 5 to make the coefficients of $x$ the same for elimination. The first equation becomes $20x+64y = 754$ and the second becomes $40x + 100y=1270$.

Step3: Eliminate $x$

Multiply the new - first equation $20x+64y = 754$ by 2, we get $40x + 128y=1508$. Subtract the equation $40x + 100y=1270$ from $40x + 128y=1508$. $(40x + 128y)-(40x + 100y)=1508 - 1270$. $28y=238$, so $y=\frac{238}{28}=8.5$.

Step4: Find $x$

Substitute $y = 8.5$ into the first original equation $5x+16\times8.5 = 188.5$. $5x+136 = 188.5$. $5x=188.5 - 136=52.5$. $x=\frac{52.5}{5}=10.5$.

Step5: Calculate the difference

Find the difference between $x$ and $y$: $x - y=10.5-8.5 = 2$.

Answer:

$2.00$