Question

beau sold hardcover books for $15 each and paperback books for $10 each. he earned a total of $350 from selling these books. let $x$ be the number of hardcover books beau sold and $y$ be the number of paperback books. the system of equations that models this scenario is: $x + y=30$, $15x + 10y = 350$. how many hardcover and paperback books did beau sell? 20 hardcovers and 10 paperbacks; 10 hardcovers and 20 paperbacks; 15 hardcovers and 15 paperbacks; 30 hardcovers and 0 paperbacks

Explanation:

Step1: Solve the system of equations

We have the system

$$\begin{cases}x + y=30\\15x + 10y=350\end{cases}$$

. From the first - equation $y = 30 - x$.

Step2: Substitute $y$ into the second equation

Substitute $y = 30 - x$ into $15x+10y = 350$. We get $15x+10(30 - x)=350$. Expand the left - hand side: $15x+300 - 10x=350$.

Step3: Simplify the equation

Combine like terms: $(15x - 10x)+300 = 350$, which gives $5x+300 = 350$. Subtract 300 from both sides: $5x=350 - 300=50$.

Step4: Solve for $x$

Divide both sides of $5x = 50$ by 5: $x=\frac{50}{5}=10$.

Step5: Solve for $y$

Substitute $x = 10$ into $y = 30 - x$, then $y=30 - 10 = 20$.

Answer:

10 hardcover books and 20 paperback books.