Q 88 - Intermediate Algebra

Question

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $\backslash (116$. Lisa bought two notebooks and three thumb dives for $\backslash )68$. Find the cost of each notebook and

each thumb drive.

Step-by-Step Solution

Answer

The cost of each notebook is $\$4$ and the cost of each thumb drive is $\$20$

Step 1. Given

Troy bought $4$ notebooks and $5$ thumb drives for $\$116$.

Lisa bought $2$ notebooks and $3$ thumb drives for $\$68$

Step 2. Form the linear equations

Let $x$ denote the costa of each notebook and $y$ denote the cost of each thumb drive.

$4$ notebooks and $5$ thumb drives bought for $\$116$

$4x+5y=116$

$2$ notebooks and $3$ thumb drives bought for $\$68$

 $2x+3y=68$

Step 3. Solve the linear equations

Solve the system of linear equations.

$4x+5y=116$

 $2x+3y=68$

Solve the second equation for $x$,

$2x=68-3y$

Divide both sides by $2$,

   $x=34-\frac{3}{2}y$

Step 4. Solve for y

Substitute $x=34-\frac{3}{2}y$ in the first equation, we get,

               $4x+5y=116$

$4(34-\frac{3}{2}y)+5y=116$

    $136-6y+5y=116$

              $136-y=116$

                        $y=136-116$

                         $y=20$

The cost of each thumb drive is $\$20$

Step 5. Solve for x

Substitute $y=20$ in the first equation, we get,

        $4x+5y=116$

   $4x+5\left(20\right)=116$

     $4x+100=116$

               $4x=16$

Divide both sides by $4$,

                 $x=4$

The cost of each notebook is $\$4$

Step 6. Check the answer

Substitute $x=\$4$,

                  $y=\$20$ in the first equaton,

             $4x+5y=\$116$

$4(\$4)+5(\$20)=\$116$

    $\$16+\$100=\$116$

               $\$116=\$116$

Hence the equation holds true.