Chapter 1.4, Problem 39E

a.

To determine

To write: A system of equation to represent the situation.

Answer to Problem 39E

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=32(ii)\\ x=2y+2z(iii)\end{array}$

Explanation of Solution

Given information: A florist must make bridesmaid bouquets for wedding

Calculation: let $x$ represent number of rose in each bouquet , $y$ represent number of lilies in each bouquet and $z$ represent number of irises in each bouquet.

Set up a system of three equations:

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=32(ii)\\ x=2y+2z(iii)\end{array}$

b.

To determine

To solve: The system to find how many of each type of flower should be in each bouquet.

Answer to Problem 39E

Number of rose is $8$ number of lilies is $4$ number of irises is $4$ .

Explanation of Solution

Given information: The system of three equations are given:

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=32(ii)\\ x=2y+2z(iii)\end{array}$

Calculation: The system of three equations are given:

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=32(ii)\\ x=2y+2z(iii)\end{array}$

Put $x=2y+2z$ in equation $(i)\mathrm{and}(ii)$ we get:

  $\begin{array}{l}2y+2z+y+z=12\\ 3y+3z=12(iv)\\ 2.5(2y+2z)+4y+2z=32\\ 5y+5z+4y+2z=32\\ 9y+7z=32(v)\end{array}$

Equation $(iv)$ multiplies by $3$ and subtract from $(v)$ we get:

  $\begin{array}{l}9y+7z=32\\ 9y+9z=36\\ ---\\ \overline{\begin{array}{l}-2z=-4\\ z=\frac{-4}{-2}=2\\ \mathrm{put}z=2\mathrm{in}\mathrm{equation}(iv)\\ 3y+3(2)=12\\ 3y+6=12\\ 3y=12-6=6\\ y=\frac{6}{3}=2\\ \mathrm{put}y=2\mathrm{and}z=2\mathrm{in}\mathrm{equation}(i)\\ x+2+2=12\\ x+4=12\\ x=12-4=8\end{array}}\end{array}$

Therefore number of rose is $8$ number of lilies is $4$ and number of irises is $4$ .

c.

To determine

Does the problem still have exactly one solution? If so find the solution. If not, give three possible solutions.

Answer to Problem 39E

If the bouquet must have all the three types of flowers, the only possible solutions are: $\begin{array}{l}8\mathrm{roses},1\mathrm{lily},3\mathrm{iries}\\ 8\mathrm{roses},2\mathrm{lilies},2\mathrm{iries}\\ 8\mathrm{roses},3\mathrm{lilies},1\mathrm{iris}\end{array}$ .

Explanation of Solution

Given information: The system of three equations are given:

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=\frac{u}{5}(ii)\\ x=2y+2z(iii)\end{array}$

Calculation: The system of three equations are given:

  $\begin{array}{l}x+y+z=12(i)\\ 2.5x+4y+2z=\frac{u}{5}(ii)\\ x=2y+2z(iii)\end{array}$

Put $x=2y+2z$ in equation $(i)\mathrm{and}(ii)$ we get:

  $\begin{array}{l}2y+2z+y+z=12\\ 3y+3z=12(iv)\\ 2.5(2y+2z)+4y+2z=\frac{u}{5}\\ 5y+5z+4y+2z=\frac{u}{5}\\ 9y+7z=\frac{u}{5}(v)\end{array}$

Equation $(iv)$ multiplies by $3$ and subtract from $(v)$

  $\begin{array}{l}9y+7z=\frac{u}{5}\\ 9y+9z=36\\ ---\\ \overline{\begin{array}{l}-2z=\frac{u}{5}-36\\ z=\frac{u-180}{-10}=\frac{180-u}{10}\end{array}}\\ \mathrm{put}z=\frac{180-u}{10}\mathrm{in}\mathrm{equation}(v)\\ 9y+7(\frac{180-u}{10})=\frac{u}{5}\\ 9y=\frac{u}{5}-7(\frac{180-u}{10})\\ 9y=\frac{2u-1260+7u}{10}\\ y=\frac{9(u-140)}{90}=\frac{u-140}{10}\\ \mathrm{put}y\mathrm{and}z\mathrm{in}\mathrm{equation}(iii)\\ x=2(\frac{u-140}{10}+\frac{180-u}{10})\\ x=2(\frac{u-140+180-u}{10})\\ x=\frac{2(40)}{10}=8\end{array}$

So when the budget varies the number of roses is constant while the number of lilies and irises depend on the budget, therefore there is more than one solution.

  $\begin{array}{l}z=0\mathrm{then}\frac{180-u}{10}=0\\ 180-u=0\\ u=180\mathrm{then}y=\frac{180-140}{10}=\frac{40}{10}=4\\ z=1\mathrm{then}\frac{180-u}{10}=1\\ 180-u=10\\ 180-10=u\\ u=170\\ u=170\mathrm{then}y=\frac{170-140}{10}=\frac{30}{10}=3\\ z=2\mathrm{then}\frac{180-u}{10}=2\\ 180-u=20\\ u=180-20=160\mathrm{then}\\ y=\frac{160-140}{10}=\frac{20}{10}=2\end{array}$

When determining the solutions for $y,z$ we must remember that they are integer numbers. As $y+z=4$ we have to check if $(0,4),(1,3),(2,2),(3,1),(4,0)$ can be solutions.

If the bouquet must have all the three types of flowers, the only possible solutions are:

  $\begin{array}{l}8\mathrm{roses},1\mathrm{lily},3\mathrm{iries}\\ 8\mathrm{roses},2\mathrm{lilies},2\mathrm{iries}\\ 8\mathrm{roses},3\mathrm{lilies},1\mathrm{iris}\end{array}$