# Thread: Systems of Equations Applications

1. ## Systems of Equations Applications

Hello All! I have two word problems and I have no idea how to solve. Would be helpful if you could show steps and explain.

1. Three cans of tuna fish and four cans of corned beef costs $12.50. However, six cans of tuna fish and three cans of corned beef cot$15. How much does each type of can cost individually?
3.A school raised funds for its sports teams by selling tickets for a play. A ticket cost $5.75 for adults and$2.25 for students. If there were five times as many adults at the play than students and ticket sales raised $2480, how many adult and student tickers were sold? 2. Originally Posted by simoonlu Hello All! I have two word problems and I have no idea how to solve. Would be helpful if you could show steps and explain. 1. Three cans of tuna fish and four cans of corned beef costs$12.50. However, six cans of tuna fish and three cans of corned beef cot $15. How much does each type of can cost individually? 3.A school raised funds for its sports teams by selling tickets for a play. A ticket cost$5.75 for adults and $2.25 for students. If there were five times as many adults at the play than students and ticket sales raised$2480, how many adult and student tickers were sold?
Label the price of a can of tuna "x"
and the price of a can of corned beef "y".

Then we obtain a pair of simultaneous equations

$3x+4y=12.50$

$6x+3y=15$

Now, eliminate one of x or y to find the other.
You could continue as follows...

$2(3x+4y)=2(12.50)$

$6x+8y=25$

$6x+3y=15$

Subtracting the 2nd line from the first...

$6x+8y-(6x+3y)=25-15$

This allows us to find y, then use the value of y to find x.

That's one way.
How about the 2nd question ? Can you try it ?

3. You can also use the coefficient matrix to solve the equations too.

$\displaystyle
\begin{bmatrix}
3 & 4 & 12.50\\
6 & 3 & 15
\end{bmatrix}$

Using elementary row operations of first multiplying the top row by -2 and adding to the second and then dividing the second row by -5, you will obtain the y answer.

$\displaystyle
\begin{bmatrix}
3 & 4 & 12.50\\
6 & 3 & 15
\end{bmatrix}\mbox{-2row1+row2} \Rightarrow\begin{bmatrix}
3 & 4 & 12.50\\
0 & -5 & -10
\end{bmatrix}$

$\displaystyle
\begin{bmatrix}
3 & 4 & 12.50\\
0 & -5 & -10
\end{bmatrix}\mbox{-1/5 row2}\Rightarrow \begin{bmatrix}
3 & 4 & 12.50\\
0 & 1 & 2
\end{bmatrix}$

Now, plug in y=2 into one of the equations and solve for x.