Wolfrich lived in Portugal and Brazil for a total period of $14$ months in order to learn Portuguese. $\newline$ He learned an average of $130$ new words per month when he lived in Portugal and an average of $150$ new words per month when he lived in Brazil. In total, he learned $1920$ new words. $\newline$ How long did Wolfrich live in Portugal, and how long did he live in Brazil? $\newline$ Wolfrich lived in Portugal for months and lived in Brazil for months. $\newline$ Show Calculator

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Q. Wolfrich lived in Portugal and Brazil for a total period of

14

months in order to learn Portuguese.

\newline

He learned an average of

130

new words per month when he lived in Portugal and an average of

150

new words per month when he lived in Brazil. In total, he learned

1920

new words.

\newline

How long did Wolfrich live in Portugal, and how long did he live in Brazil?

\newline

Wolfrich lived in Portugal for

months and lived in Brazil for

months.

\newline

Show Calculator

Equations Setup: Let's denote the number of months Wolfrich lived in Portugal as $P$ and the number of months he lived in Brazil as $B$ . We know that the total time spent in both countries is $14$ months, so we can write the first equation as: $\newline$ $P + B = 14$
Substitution Method: We also know that Wolfrich learned an average of $130$ words per month in Portugal and $150$ words per month in Brazil, totaling $1920$ words. We can write the second equation based on the number of words learned as: $\newline$ $130P + 150B = 1920$
Equation Simplification: We now have a system of two linear equations: $\newline$ $1$ ) $P + B = 14$ $\newline$ $2$ ) $130P + 150B = 1920$ $\newline$ We can solve this system using the substitution or elimination method. Let's use the substitution method by expressing $P$ in terms of $B$ from the first equation: $\newline$ $P = 14 - B$
Solve for B: Substitute $P = 14 - B$ into the second equation: $\newline$ $130(14 - B) + 150B = 1920$ $\newline$ Now, distribute $130$ into the parentheses: $\newline$ $1820 - 130B + 150B = 1920$
Find P: Combine like terms by subtracting $130B$ from $150B$ : $\newline$ $1820 + 20B = 1920$ $\newline$ Now, subtract $1820$ from both sides to isolate the term with $B$ : $\newline$ $20B = 1920 - 1820$ $\newline$ $20B = 100$
Find P: Combine like terms by subtracting $130B$ from $150B$ : $1820 + 20B = 1920$ Now, subtract $1820$ from both sides to isolate the term with $B$ : $20B = 1920 - 1820$ $20B = 100$ Divide both sides by $20$ to solve for $B$ : $B = \frac{100}{20}$ $150B$ $0$ So, Wolfrich lived in Brazil for $150B$ $1$ months.
Find P: Combine like terms by subtracting $130B$ from $150B$ : $\newline$ $1820 + 20B = 1920$ $\newline$ Now, subtract $1820$ from both sides to isolate the term with $B$ : $\newline$ $20B = 1920 - 1820$ $\newline$ $20B = 100$ Divide both sides by $20$ to solve for $B$ : $\newline$ $B = \frac{100}{20}$ $\newline$ $150B$ $0$ $\newline$ So, Wolfrich lived in Brazil for $150B$ $1$ months.Now that we know $B$ , we can find $150B$ $3$ using the first equation: $\newline$ $150B$ $4$ $\newline$ Subtract $150B$ $1$ from both sides to solve for $150B$ $3$ : $\newline$ $150B$ $7$ $\newline$ $150B$ $8$ $\newline$ So, Wolfrich lived in Portugal for $150B$ $9$ months.