Question

The Brady family received 15 pieces of mail on June 4. The mail consisted of letters, bills, magazines, andads. How many letters did they receive if they received three more bills than magazines, the samenumber of letters as magazines, and five more ads than bills?

Answer 1

Let L represent letters

Let B represent bills

Let M represent Magazines

Let A represent Ads

From the question;

If they receive three more bills than magazines implies that

$B=M+3$

If they receive the same number of letters as magazines implies

$L=M$

If they receive five more ads than bills, it implies that

$\begin{gathered} A=B+5 \\ \text{Thus, since }B=M+3 \\ A=M+3+5=M+8 \end{gathered}$

Recall that, the sum of all the mails is 15, i.e

$L+B+M+A=15$

Substitute the values of L, B, and A in terms of M to find the value of M

$M+M+3+M+M+8=15$

Collect like terms

$\begin{gathered} M+M+M+M+3+8=15 \\ 4M+11=15 \\ 4M=15-11 \\ 4M=4 \end{gathered}$

Divide both sides by 4

$\begin{gathered} \frac{4M}{4}=\frac{4}{4} \\ M=1 \end{gathered}$

If, M =4, Find the number of L

$\begin{gathered} \text{ Recall,} \\ L=M \\ \text{Thus,} \\ L=1 \end{gathered}$

Hence, they receive 1 letter