Solutions to Problem Set 3

Math 101-06

9-8-2016

[Formulas]

1. Solve the equation $V = \pi r^2
   h$ for h. (This is the formula for the volume V of a cylinder in terms of its radius r and its height h.)

$$\eqalign{ V & = \pi r^2 h \cr \noalign{\vskip2pt} \dfrac{1}{\pi r^2} \cdot V & = \dfrac{1}{\pi r^2} \cdot \pi r^2 h \cr \noalign{\vskip2pt} \dfrac{V}{\pi r^2} & = h \quad\halmos \cr}$$


2. Solve the equation $A = x^2 + 4
   x y$ for y. (This is the surface area of a rectangular box with height y and a square bottom and top with sides of length x.)

$$\eqalign{ A & = x^2 + 4 x y \cr A - x^2 & = x^2 + 4 x y - x^2 \cr A - x^2 & = 4 x y \cr \noalign{\vskip2pt} \dfrac{1}{4 x} \cdot (A - x^2) & = \dfrac{1}{4 x} \cdot 4 x y \cr \noalign{\vskip2pt} \dfrac{A - x^2}{4 x} & = y \quad\halmos \cr}$$


3. Solve $3 a + k a + 5 b = 0$ for a.

$$\eqalign{ 3 a + k a + 5 b & = 0 \cr 3 a + k a + 5 b - 5 b & = 0 - 5 b \cr 3 a + k a & = -5 b \cr a (3 + k) & = -5 b \cr \noalign{\vskip2pt} \dfrac{1}{3 + k} \cdot a (3 + k) & = \dfrac{1}{3 + k} \cdot (-5 b) \cr \noalign{\vskip2pt} a & = \dfrac{-5 b}{3 + k} \quad\halmos \cr}$$

You may also write the answer as $a = \dfrac{5 b}{3 + k}$ .


4. Solve $17 x = a x + 11$ for x.

$$\eqalign{ 17 x & = a x + 11 \cr 17 x - a x & = a x + 11 - a x \cr 17 x - a x & = 11 \cr x (17 - a) & = 11 \cr \noalign{\vskip2pt} \dfrac{1}{17 - a} \cdot x (17 - a) & = \dfrac{1}{17 - a} \cdot 11 \cr \noalign{\vskip2pt} x & = \dfrac{11}{17 - a} \quad\halmos \cr}$$


5. Solve the equation $4 a b + 7
   a = 13 - 3 b$ for b.

$$\eqalign{ 4 a b + 7 a & = 13 - 3 b \cr 4 a b + 3 b & = 13 - 7 a \cr b(4 a + 3) & = 13 - 7 a \cr \noalign{\vskip2pt} \dfrac{b(4 a + 3)}{4 a + 3} & = \dfrac{13 - 7 a}{4 a + 3} \cr \noalign{\vskip2pt} b & = \dfrac{13 - 7 a}{4 a + 3} \quad\halmos \cr}$$


6. Solve the equation $10 x - 3 b
   y = 6 y^2 - b x$ for x.

$$\eqalign{ 10 x - 3 b y & = 6 y^2 - b x \cr 10 x - 3 b y + 3 b y + b x & = 6 y^2 - b x + 3 b y + b x \cr 10 x + b x & = 6 y^2 + 3 b y \cr (10 + b) x & = 6 y^2 + 3 b y \cr \noalign{\vskip2pt} \dfrac{(10 + b) x}{10 + b} & = \dfrac{6 y^2 + 3 b y}{10 + b} \cr \noalign{\vskip2pt} x & = \dfrac{6 y^2 + 3 b y}{10 + b} \quad\halmos \cr}$$


7. Solve the equation $3 (x + y)
   = a (x - y)$ for x.

$$\eqalign{ 3 (x + y) & = a (x - y) \cr 3 x + 3 y & = a x - a y \cr 3 x + 3 y - 3 y - a x & = a x - a y - 3 y - a x \cr 3 x - a x & = -a y - 3 y \cr x (3 - a) & = -a y - 3 y \cr \noalign{\vskip2pt} \dfrac{1}{3 - a} \cdot x (3 - a) & = \dfrac{1}{3 - a} \cdot (-a y - 3 y) \cr \noalign{\vskip2pt} x & = \dfrac{-a y - 3 y}{3 - a} \quad\halmos \cr}$$


8. Solve the equation $A =
   \dfrac{7}{B - 5}$ for B.

$$\eqalign{ A & = \dfrac{7}{B - 5} \cr \noalign{\vskip2pt} (B - 5) \cdot A & = (B - 5) \cdot \dfrac{7}{B - 5} \cr \noalign{\vskip2pt} A (B - 5) & = 7 \cr \noalign{\vskip2pt} \dfrac{1}{A} \cdot A (B - 5) & = \dfrac{1}{A} \cdot 7 \cr \noalign{\vskip2pt} B - 5 & = \dfrac{7}{A} \cr \noalign{\vskip2pt} B - 5 + 5 & = \dfrac{7}{A} + 5 \cr \noalign{\vskip2pt} B & = \dfrac{7}{A} + 5 \quad\halmos \cr}$$


Perhaps the most valuable result of all education is the ability to make yourself do the thing you have to do, when it ought to be done, whether you like it or not. - Thomas Huxley


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