Q26. - Algebra 2

Question

Jonathan and members of his Spanish club are going to a Costa Rica over spring break. Before his trip, he purchases 10 travelers checks in denominations of \(20,\)50 and \(100, totaling \)370, He has twice as many \(20 checks as \)50 checks how many of each type of denominations of travelers checks does he have?

Step-by-Step Solution

Answer

The $20 denominations of checks are 6, and $50 denominations of checks are 3, and $100 denominations of checks is 1.

Step 1 – First construct a system of equations corresponding to the given situation.

Let $x$ be the number of $20$ denominations of checks, and $y$ be the number of $50$ denominations of checks, and $z$ be the number of $100$ denominations of checks.

 Total 10 travelers checks are purchased in denominations of $20,50$, and $100$.

So,$x+y+z=10$

 The total amount spent is $370$.

This can be mathematically written as $20x+50y+100z=370$.

 Jonathan has twice as many $20$ checks as $50$ checks.

This can be mathematically written as $x=2y\Rightarrow x-2y=0$.

 So, the system of equations will be:

$\begin{array}{c}x+y+z=10\\ 20x+50y+100z=370\\ x-2y=0\end{array}$

Step 2 – Use the elimination method to get the system of equations in two variables.

Multiply the $equation x+y+z=10$ by 2 and add the new resultant equation to the equation $x-2y=0$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}x+y+z=10\\ x-2y=0\end{array}}\begin{array}{c}\text{multiply by 2}\\ \end{array}\underset{\_}{\begin{array}{l}2x+2y+2z=20\\ x-2y=0\end{array}}\\ 3x+0+2z=20\end{array}$

So, the resultant equation is $3x+2z=20$

Multiply the $equation x+y+z=10$ by 50 and subtract the new resultant equation from the equation $20x+50y+100z=370$

$\begin{array}{l}\underset{\_}{\begin{array}{l}x+y+z=10\\ 20x+50y+100z=370\end{array}}\begin{array}{c}\text{multiply by 50}\\ \end{array}\underset{\_}{\begin{array}{l}50x+50y+50z=500\\ (-)20x+50y+100z=370\end{array}}\\ 30x+0-50z=130\end{array}$

.So, the resultant equation is $30x-50z=130$.

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Step 3 – Solve the system of equations in two variables

Multiply $3x+2z=20$by 25 and add the new resultant equation to $30x-50z=130$

$\begin{array}{l}\underset{\_}{\begin{array}{l}3x+2z=20\\ 30x-50z=130\end{array}}\begin{array}{c}\text{multiply by 25}\\ \end{array}\underset{\_}{\begin{array}{l}75x+50z=500\\ 30x-50z=130\end{array}}\\ 105x+0=630\end{array}$

Solve $105x=630$for $x$

$\begin{array}{c}105x=630\\ \frac{105x}{105}=\frac{630}{105}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} divide both sides by }105\\ x=6\end{array}$

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Step 4 – Find the values of   y and z .

Substitute $x=6 in x-2y=0$ and find the value of $y$.

$\begin{array}{c}x-2y=0\\ 6-2y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }6\text{ for }x\\ -2y=-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }6\text{ from both sides}\\ y=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by 2}\end{array}$

Substitute $x=6 in 3x+2z=20$ and find the value of $z$

$\begin{array}{c}3x+2z=20\\ 3\left(6\right)+2z=20\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }6\text{ for }x\\ 18+2z=20\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ 2z=2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }18\text{\hspace{0.17em}from both sides}\\ z=1\text{ Divide both sides by }2\end{array}$

Hence, the solution of the given system of equations is$\left(x,y,z\right)=\left(6,3,1\right)$.

Hence, the $20$ denominations of checks are 6, and $50$ denominations of checks are 3, and $100$denominations of checks is 1.

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