Chapter 6.3, Problem 27PPE

To determine

Tocalculate: cost of an $8\times 10$ sized-portrait and cost of a wallet-sized portrait

Answer to Problem 27PPE

Cost of $8\times 10$ -sized portrait is $\$12$ and cost of wallet-sized portrait is $\$7$

Explanation of Solution

Calculation:

Let $x$ denotes cost of $8\times 10$ sized-portrait and $y$ denotes cost of wallet-sized portrait.

As two $8\times 10$ sized-portraits and 4 wallet-sized portraits are purchased for $\$52$ ,

  $2x+4y=52(i)$

As three $8\times 10$ sized-portraits and two wallet-sized portraits are purchased for $\$50$ ,

  $3x+2y=50(ii)$

Now, we cannot eliminate a variable in equations $(i),(ii)$ simply by subtracting or adding them.

Both the equations are need to be multiplied by a constant.

Multiply $(i)$ by $-3$

  $\begin{array}{l}-3\left(2x+4y=52\right)\\ \Rightarrow -6x-12y=-156(iii)\end{array}$

Multiply $(ii)$ by $2$

  $\begin{array}{l}2\left(3x+2y=50\right)\\ \Rightarrow 6x+4y=100(iv)\end{array}$

Add $(iii)$ and $(iv)$

  $\begin{array}{l}-8y=-56\\ \Rightarrow y=\frac{-56}{-8}=7\end{array}$

Put $y=7$ in $(ii)$

  $\begin{array}{l}3x+2\left(7\right)=50\\ \Rightarrow 3x+14=50\\ \Rightarrow 3x=50-14\\ \Rightarrow 3x=36\\ \Rightarrow x=\frac{36}{3}=12\end{array}$

So,

Cost of $8\times 10$ -sized portrait is $\$12$ and cost of wallet-sized portrait is $\$7$