liner equation

Relevant wiki: System of Linear Equations (Simultaneous Equations)

$4b+5p=32$

$5b+4p=31$

Subtracting the second equation from the first, we get:

$-b+p=1$

$p=b+1$

Substituting into the first equation, we get:

$4b+5(b+1)=32$

$9b+5=32$

$9b=27\Rightarrow\boxed{b=3}$

By adding one book and removing a pencil, the total price is $1 less. Repeat until no pencils are left and you have 9 books which total to $27. Divide to get the individual price of $3.

Relevant wiki: System of Linear Equations (Simultaneous Equations)

$B=book$ and $P=pencil$

$4B+5P=32$ $\implies$ $\color{#D61F06}\boxed{1}$

$5B+4P=31$ $\implies$ $\color{#D61F06}\boxed{2}$

Method 1: Substitution Method

In $\color{#D61F06}\boxed{1}$ , solve $P$ in terms of $B$ then substitute in $\color{#D61F06}\boxed{2}$ .

$4B+5P=32$ $\implies$ $P=\dfrac{32-4B}{5}$

$5B+4P=31$ $\implies$ $5B+4\left(\dfrac{32-4B}{5}\right)=31$ $\implies$ $5B+\dfrac{128}{5}-\dfrac{16B}{5}=31$ $\implies$ $\dfrac{9B}{5}=\dfrac{27}{5}$ $\implies$ $\color{plum}\large{\boxed{B=3}}$

Method 2: Elimination Method

$(-4 \times \color{#D61F06}\boxed{1}$ $)+(5 \times \color{#D61F06}\boxed{2}$ $)$ , we get

$9B=27$

$\color{plum}\large{{\boxed{B=3}}}$

Method 3: By Determinants

$B=\dfrac{ \begin{vmatrix} 32 & 5 \\ 31 & 4 \end{vmatrix}}{ \begin{vmatrix} 4 & 5 \\ 5 & 4 \end{vmatrix}}=\dfrac{(32)(4)-(31)(5)}{(4)(4)-(5)(5)}=\dfrac{-27}{-9}=$ $\color{plum}\large{\boxed{3}}$

4 books and 5 pencils cost $32 while 5 books and 4 pencils cost $31 so adding them, we see that 9 books and 9 pencils cost $63, meaning a book and a pencil together cost $7. Then 4 books and 4 pencils cost $28 and we subtract this from 5 books and 4 pencils which cost $31, we see that 1 book costs $3.

Solution by elimination method:

Suppose $b =$ price of a book and $p =$ price of a pencil.

$4b + 5p = 32 .........(1)$

$5b + 4p = 31 .........(2)$

Multiply the first equation by 4 and multiply second equation by 5.

$16b + 20p = 128$

$-25b - 20p = -155$

After subtracting, we get

$9b = 27$

$b = \boxed{3}$

One book = $3 and One pencil = $4 $4(\$3) + 5(\$4) = \$32$

$5(\$3) + 4(\$4) = \$31$

A very math-looking approach then!

$\text{books}=b$ and $\text{pencils}=p.$

$\begin{aligned} \left(\begin{array}{cc} 4 & 5 \\ 5 & 4 \end{array}\right) \left(\begin{array}{c} b \\ p \end{array}\right) & =\left(\begin{array}{c} 32 \\ 31 \end{array}\right) \\\\ \left(\begin{array}{c} b \\ p \end{array}\right) & =\dfrac{1}{4\cdot4-5\cdot5}\left(\begin{array}{cc} 4 & -5 \\ -5 & 4 \end{array}\right)\left(\begin{array}{c} 32 \\ 31 \end{array} \right) \\\\ & =-\dfrac{1}{9}\left(\begin{array}{c} -27 \\ -36 \end{array} \right) \\\\ & =\left(\begin{array}{c} 3 \\ 4 \end{array} \right) \end{aligned}$

Therefore, one book costs $\$3.$

...and one pencil costs $\$4,$ apparently. Damn, that's a one expensive pencil.

$\text{suppose,books=b and pencils=p}$

$\text{now,}$ $4b+5p=32$ .....................[1]

$\text{and}$ $5b+4p=31$ .......................[2]

$\text{now,multiplying [1] by 4 and multiplying [2] by 5 and then doing [2]-[1], we get that,}$

$25b+20p-16b-20p=155-128$

or, $9b=27$

or, $b=3$

$\text{so, the cost of a book is 3}$ $

4x + 5y = 32

5x + 4y = 31

Add both equations

9x + 9y = 63

9(x+y) = 63

x + y = 7

The answers are 2, 3, 4 or 9, we can rule out 9, since it's too big

1) if x =2 then y=5

4 * 2 + 5 * 5 = 33, so X ISN'T 2

2) if x=3 then y=4

4 * 3 + 5 * 4 = 32

also 5 * 3 + 4 * 4 = 31

The second equation contains an even number of pencils and an odd number of books which sum to an odd number. So the price of a book is odd. 9 is too large, so that leaves us with 3.

Given that the answers are all whole numbers it is faster to just solve for 5, then 4 then 3 to find a solution.

All I want to add to previous solutions is: How does a pencil cost more than a book?!?

Add the two data - 9 of each cost $63, so one of each costs $7. Take 4 of each from the second datum: $(31-28) = $3 Ockham's razor?

By adding one book and removing one pencil you reduce the amount by $1.So if you remove all four pecils and add four more books that will reduce the amount in the second group by $4, to $27.So if 9 books cost $27,one book costs $3.

4b+5p=32$

5b+4p=31$

Difference between 1 book's value and 1 pen's value is 1$ in favour of pen

Sum up two equations

9b+9p=63$

1b+1p=7$

1b=3$

1p=4$

4b + 5p = 32

5b + 4p = 31

Note that in these kinds of problems with alrternating coefficients, a good first attempt is to directly add them which usually results in some sort of factoring allowing a "unit sum" to occur.

Result: 9b + 9p = 63

Factored: b + p = 7

Here can be noted that each coefficient is 1, ie a unit and multiplying will be an effective way to reduce an above alternating equation. As we're looking for b find the equation with more b than p.

Equation 2:

5b + 4p = 31

-(4b + 4p = 28)

b = 3.

The same could be done to find p.

First time posting, would love any criticism or comments.

It is not always necessary to solve simultaneous equations by following a set algorithm. Sometimes you can just see the answer by inspection, and sometimes you can exploit peculiar features of the equations. In this problem the interesting pattern of '5' and '4' leads to a quick solution -

Let b and p be the cost of a book and pencil respectively. Adding the two top rows together and dividing by 9 shows that

$b+p=7$

Subtracting the top row from the second shows that

$b-p=-1$

Adding these two equations gives

$2b=6$

And so

$b=\boxed{3}$