Solutions to Problem Set 8

Math 101-06

9-20-2017

[Systems]

1. Solve the following system of equations for x and y.

$$\eqalign{ x - 2 y & = 10 \cr 0.7 x + 0.4 y & = 2 \cr}$$ Multiply the first equation by 2, multiply the second equation by 10, then add the equations:

$$\eqalign{ 2 x - 4 y & = 20 \cr 7 x + 4 y & = 20 \cr \noalign{\vskip2pt \hrule \vskip2pt} 9 x & = 40 \cr \noalign{\vskip2pt} x & = \dfrac{40}{9} \cr}$$

Multiply the first equation by 7, multiply the second equation by 10, then subtract the equations:

$$\eqalign{ 7 x - 14 y & = 70 \cr 7 x + 4 y & = 20 \cr \noalign{\vskip2pt \hrule \vskip2pt} -18 y & = 50 \cr \noalign{\vskip2pt} y & = -\dfrac{25}{9} \cr}$$

The solution is $x =
   \dfrac{40}{9}$ and $y = -\dfrac{25}{9}$ .


2. Solve the following system of equations for x and y.

$$\eqalign{ x + 5 y & = -1 \cr -2 x - 10 y & = 2 \cr}$$ Multiply the first equation by 2 and add it to the second equation:

$$\eqalign{ 2 x + 10 y & = -2 \cr -2 x - 10 y & = 2 \cr \noalign{\vskip2pt \hrule \vskip2pt} 0 & = 0 \cr}$$

The last equation is an identity. Therefore, the system has infinitely many solutions. In fact, the two original equations represent the same line, so the solutions are all the point on the line $x + 5 y = -1$ .


3. Solve the following system of equations for x and y.

$$\eqalign{ 1.1 x + y & = 2 \cr -2 x + 1.3 y & = 5 \cr}$$ I'll begin by multiplying both equations by 10 to clear the decimals first.

$$\eqalign{ 11 x + 10 y & = 20 \cr -20 x + 13 y & = 50 \cr}$$

(You could work with the equations as-is, though the numbers would be a little messier.)

Multiply the first equation by 13, multiply the second equation by 10, then subtract the equations:

$$\eqalign{ 143 x + 130 y & = 260 \cr -200 x + 130 y & = 500 \cr \noalign{\vskip2pt \hrule \vskip2pt} 343 x & = -240 \cr \noalign{\vskip2pt} x & = -\dfrac{240}{343} \cr}$$

Multiply the first equation by 20, multiply the second equation by 11, then add the equations:

$$\eqalign{ 220 x + 200 y & = 400 \cr -220 x + 143 y & = 550 \cr \noalign{\vskip2pt \hrule \vskip2pt} 343 y & = 950 \cr \noalign{\vskip2pt} y & = \dfrac{950}{343} \cr}$$

The solution is $x =
   -\dfrac{240}{343}$ and $y = \dfrac{950}{343}$ .


[Systems word problems]

4. Bonzo has some nickels and some quarters: 52 coins all together. The total value of the coins is $8.20. Find the number of nickels and the number of quarters.

To avoid decimals, I'll do everything in cents.

Let n be the number of nickels. Each one is worth 5 cents.

Let q be the number of quarters. Each one is worth 25 cents.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & Number of coins & & & & Cents per coin & & & & Total value & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & nickels & & n & & $\cdot$ & & 5 & & $=$ & & $5 n$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & quarters & & q & & $\cdot$ & & 25 & & $=$ & & $25 q$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Total & & 52 & & & & & & & & 820 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first column gives

$$n + q = 52, \quad\hbox{so}\quad n = 52 - q.$$

The third column gives

$$5 n + 25 q = 820.$$

Substitute $n = 52 - q$ in the last equation and solve for q:

$$\eqalign{ 5 (52 - q) + 25 q & = 820 \cr 260 - 5 q + 25 q & = 820 \cr 260 + 20 q & = 820 \cr 20 q & = 560 \cr \noalign{\vskip2pt} q & = \dfrac{560}{20} = 28 \cr}$$

Hence, $n = 52 - 28 = 24$ .

Bonzo has 24 nickels and 28 quarters.


5. $1250 is divided among two accounts. The first account pays 5% interest, while the second account pays 6% interest. At the end of one interest period, the two accounts have returned a total of $68.0 in interest. How much was invested in each account?

Let x be the amount invested in the first account, and let y be the amount invested in the second account. \overfullrule=0pt

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & amount invested & & & & interest rate & & & & interest earned & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & first account & & x & & $\cdot$ & & 0.05 & & $=$ & & $0.05 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & second account & & y & & $\cdot$ & & 0.06 & & $=$ & & $0.06 y$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & total & & 1250 & & & & & & & & 68 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first and third columns give two equations:

$$x + y = 1250 \quad\hbox{and}\quad 0.05 x + 0.06 y = 68.$$

Multiply the first equation by 0.05 and subtract it from the second:

$$\matrix{ 0.05 x & + & 0.06 y & = & 68 \cr 0.05 x & + & 0.05 y & = & 62.5 \cr \noalign{\vskip2pt \hrule \vskip2pt} & & 0.01 y & = & 5.5 \cr & & y & = & 550 \cr}$$

Then $x = 1250 - 550 = 700$ . Thus, $700 was invested in the first account and $550 was invested in the second account.


What would life be if we had no courage to attempt anything? - Vincent Van Gogh


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