# Linear Algebra Word Problem

I have been this for hours, yes it is a homework question, but did not turn it in last weeks when I was suppose to, but I have a test coming up, so gotta figure this out not sure how to formulate questions using a system of equations

The density of gold is $19.3g/cm^3$ and the density of silver is $10.5g/cm^3$ a certain crown is made entirely from silver and gold if the total volume of the grown is $220cm^3$, and the weight of the crown is $3823.6g$, what percentage of the total mass is gold?

I am not sure how to get around the units I mean I tried assigning x to be gold and y silver but this $g/cm^3$ how do you equate that to $3823.6g$ and have the unit match up. I mean all I have on my paper is $19.3x + 10.5y = 3823.6$ but I know that is wrong and where do I get the second equation. I am truly stuck and frustrated. my book doesn't have a solutions manual so I wouldn't know if it is right or wrong. Please help me set this up

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Let $g$ be the volume of gold, in $\text{cm}^3$ and $s$ be the volume of silver in the same units. The volume of the crown is $220 \space\text{cm}^3$, so

$$g +s = 220$$

Every $\text{cm}^3$ of gold is $19.3\text{g}$ and the same volume of silver is $10.5\text{g}$. Multiply the individual volume of gold & silver to get the total weight of the crown. $$19.3g +10.5s = 3823.6$$

You now have 2 equations. Use simultaneous equations to solve the individual volume of gold & silver. Finally, find the percentage mass of gold by taking

$$\frac{19.3g}{3823.6g}$$

Update

You look like you're looking for the matrix way of solving. I'll re-write the equations for you in the form:

\begin{align*} g +s &= 220\\ 19.3g +10.5s &= 3823.6 \end{align*}

To put the equation in matrix form,

$$\begin{pmatrix} 1 & 1\\ 19.3 & 10.5 \end{pmatrix} \begin{pmatrix} g \\ s\end{pmatrix} = \begin{pmatrix} 220 \\ 3823.6\end{pmatrix}$$ Find the inverse of the matrix $$\begin{pmatrix} 1 & 1\\ 19.3 & 10.5 \end{pmatrix}$$ which is $$\begin{pmatrix} -1.193 & 0.114\\ 2.193 & 0-.114 \end{pmatrix}$$ And the equation you want to solve will be

$$\begin{pmatrix} -1.193 & 0.114\\ 2.193 & 0-.114 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 19.3 & 10.5 \end{pmatrix} \begin{pmatrix} g \\ s\end{pmatrix} =\begin{pmatrix} -1.193 & 0.114\\ 2.193 & 0-.114 \end{pmatrix} \begin{pmatrix} 220 \\ 3823.6\end{pmatrix}$$

The left hand side will evaluate to be a $2 \times 2$ identity matrix, which you have identified as $I$. The right hand side will give you the values of $g$ and $s$ respectively.

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So I start out with g + s = 220 19.3g + 10.5s = 3823.6 then in matrix form: Row I: 1 1 220, Row II 19.3 10.5 3823.6, Then I solve by doing -10.5*RI + RII which gives so this is Row Echelon Form? I believe I can't do Reduced Row Echelon Form (rref). So I can now just solve for g using the second row and plug into the top to solve for s. Why won't this equation ever reduce to rref? Does it have to be in rref to to have a solution. I appologize I feel like I am asking the lamest questions ever but I want to ace this class. –  rob Feb 24 '13 at 5:54
Hi rob, I have updated my answer. You might want to learn LaTeX to show your working. –  bryansis2010 Feb 24 '13 at 10:35

let the crown be entirely of gold. Then for every $cm^3$you make of silver instead of gold you will lose 8.8 grams off the total crown. if it was all made of gold it would weigh $19.3 \frac{grams}{cm^3}*220 cm^3 = 4246$ grams so therefore the crown weighs $4246-3823= 423$ grams less than the pure gold.

Since or every $cm^3$ you substitute with silver you reduce the weight by 8.8 grams then if x is the number of $cm^3$ of silver then $x*8.8=423$ so $x=\frac{423}{8.8}$which is roughly 48.07

Looking at it in a system of equations (slightly different method) it would look like this, where $x_1$ and $x_2$ are the $cm^3$ of gold and silver respectively: $x_1+x_2=220$ and $19.3x_1+10.5x_2=3823.6$ so now you just have to solve these two equalities.

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how would that look like in a system of equations? –  rob Feb 24 '13 at 5:03