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During summer vacation, Maya spent some time in a village with her grandparents. One day she went to the garden to pick some apples and pears.

External credits: @pikisuperstar

Answer

a Graph:

Solution: $a=10,$ $p=8$

b $a=10,$ $p=8$

c The solutions are the same. In this case, the Substitution Method is more useful for a number of reasons.

d The system is consistent and independent.

Solution

a In order to solve the system of equations by graphing, both equations should be written in slope-intercept form. The second equation is already in the required form, so only the first equation needs to be rewritten.

$5a+4p=82$

SubEqn

$\text{LHS}-5a=\text{RHS}-5a$

$4p=82-5a$

DivEqn

$\text{LHS}/4=\text{RHS}/4$

$p=20.5-1.25a$

Now, graph both equations on the same coordinate plane. To graph the first equation, first plot the $y\text{-}$intercept of $20.5.$ Next, by using the slope of $\text{-}1.25,$ move $1$ unit to the right and $1.25$ units down, or $4$ units to the right and $4\cdot 1.25=5$ units down to plot the second point.

Draw a line through the two plotted points to obtain the graph of the first equation.

The second equation can be graphed by following the same process.

The solution of the system of equations is represented by the point of intersection of the lines.

The lines intersect at $(10,8).$ Therefore, $a=10$ and $p=8,$ which indicates that Maya collected $10$ apples and $8$ pears.

b Now, the system of equations will be solved by using the Substitution Method. The variable $p$ is already isolated in the second equation, so $18-a$ can be substituted for $p$ in the first equation.

$\{\begin{array}{ll}5a+4p=82& \text{(I)}\\ p=18-a& \text{(II)}\end{array}$

Substitute

$\text{(I): }$ $p=18-a$

$\{\begin{array}{l}5a+4(18-a)=82\\ p=18-a\end{array}$

Distr

$\text{(I): }$ Distribute $4$

$\{\begin{array}{l}5a+72-4a=82\\ p=18-a\end{array}$

SubTerm

$\text{(I): }$ Subtract term

$\{\begin{array}{l}a+72=82\\ p=18-a\end{array}$

SubEqn

$\text{(I): }$ $\text{LHS}-72=\text{RHS}-72$

$\{\begin{array}{l}a=10\\ p=18-a\end{array}$

Substitute

$\text{(II): }$ $a=10$

$\{\begin{array}{l}a=10\\ p=18-10\end{array}$

SubTerms

$\text{(II): }$ Subtract terms

$\{\begin{array}{l}a=10\\ p=8\end{array}$

The solution of the system of equations is $a=10$ and $p=8.$

c From the two previous parts, both methods of solving the system of equations gave the same solution. Therefore, both methods of solving are correct.

$$\begin{array}{ccc}\text{Graphing}& & \text{Substitution}\\ \text{Method}& & \text{Method}\\ \searrow & & \swarrow \\ & (10,8)& \end{array}$$

However, in this case, the Substitution Method can be more convenient because it is shorter and gives the exact solution. By comparison, the graphing method requires the equations to be in slope-intercept form and does not always result in finding the exact solution.

d In order to interpret the system of equations in terms of consistency and independence, start by recalling the definitions of these concepts.

The given system of equations has exactly one solution corresponding to each variable, so it is a consistent and independent system.

Maya's grandparents own a small farmyard where they raise sheep and chickens. Maya was curious how many of each her grandparents have, so she asked them about it.

External credits: @brgfx

Her grandfather really likes riddles, so he told her that their sheep and chickens have a total of $102$ heads and $252$ legs and asked Maya to calculate the number of each animal herself.

Maya's grandparents also grow some carrots and potatoes. Last year they harvested $6$ pounds of potatoes and $4$ pounds of carrots per square yard of garden. In total they grew $185$ pounds of these vegetables.

External credits: @freepik

To their big surprise, this year they managed to harvest $9$ pounds of potatoes and $6$ pounds of carrots per square yard of garden, for a total of $277.5$ pounds of vegetables.

Solution

a First, the equation describing the harvest of the last year will be written. The variables $p$ and $c$ denote the numbers of square yards of garden where potatoes and carrots grow, respectively. Maya's grandparents grew $6$ pounds of potatoes per acre, so the product $6p$ represents the total number of pounds of potatoes.

$$\begin{array}{c}6p=\text{Total pounds of potatoes}\end{array}$$

Similarly, $4c$ represents the total number of pounds of carrots they grew.

$$\begin{array}{c}4c=\text{Total pounds of carrots}\end{array}$$

Maya's grandparents grew a total of $185$ pounds of vegetables last year. Therefore, the sum of $6p$ and $4c$ is equal to $185.$

$$\begin{array}{c}\text{Last Year}\\ 6p+4c=185\end{array}$$

The equation describing the harvest of this year can now be formed by using a similar process. This year the grandparents managed to grow $9$ pounds of potatoes per square yard and $6$ pounds of carrots per square yard. Therefore, $9p$ and $6c$ are the total number of pounds of potatoes and carrots they harvested, respectively.

$$\begin{array}{rl}9p& =\text{Total pounds of potatoes}\\ 6c& =\text{Total pounds of carrots}\end{array}$$

The sum of these expressions is $277.5,$ which is the total amount of vegetables that Maya's grandparents grew this year.

$$\begin{array}{c}\text{This Year}\\ 9p+6c=277.5\end{array}$$

A system of equations can be written using these two linear equations.

$$\begin{array}{c}\{\begin{array}{l}6p+4c=185\\ 9p+6c=277.5\end{array}\end{array}$$

b To solve the system of equations using the Substitution Method, isolate one variable in one equation and substitute the corresponding expression into the other equation. For example, $c$ can be isolated on the left-hand side of the first equation.

$\{\begin{array}{ll}6p+4c=185& \text{(I)}\\ 9p+6c=277.5& \text{(II)}\end{array}$

SubEqn

$\text{(I): }$ $\text{LHS}-6p=\text{RHS}-6p$

$\{\begin{array}{l}4c=185-6p\\ 9p+6c=277.5\end{array}$

DivEqn

$\text{(I): }$ $\text{LHS}/4=\text{RHS}/4$

$\{\begin{array}{l}c=46.25-1.5p\\ 9p+6c=277.5\end{array}$

Substitute

$\text{(II): }$ $c=46.25-1.5p$

$\{\begin{array}{l}c=46.25-1.5p\\ 9p+6(46.25-1.5p)=277.5\end{array}$

Distr

$\text{(II): }$ Distribute $6$

$\{\begin{array}{l}c=46.25-1.5p\\ 9p+277.5-9p=277.5\end{array}$

SubTerm

$\text{(II): }$ Subtract term

$\{\begin{array}{l}c=46.25-1.5p\\ 277.5=277.5\end{array}$

As shown, solving Equation (II) for $p$ resulted in a true statement. This indicates that these two equations do not provide enough information to calculate the exact solution of the system. Therefore, it has infinitely many solutions.

c To interpret the system in terms of consistency and independence, first recall the definitions of these concepts.

In this case, the system of equations has solutions, so it is a consistent system. Additionally, since it has infinitely many solutions, it is a dependent system.

To keep herself busy and earn some extra cash, Maya found a part time job at a local restaurant. One week she is trained to work in the kitchen and another she works as a waitress. One day, each person working in the kitchen cooked $24$ dishes, while each waiter served $120$ dishes.

External credits: @pikisuperstar, @macrovector

At the end of the day, when the kitchen was closing, Maya noticed that $2$ cooked dishes did not get served. The number of people in the kitchen was $4$ more than $5$ times the number of waiters.