Linear and Nonlinear Equations

Equations

An equation is a statement that two expressions are equal. An equation containing variables is in general neither true nor false; rather, its truth depends on the value(s) of the variable(s). For equations in one variable, a value of the variable which makes the equation true is called a solution of the equation. The set of all solutions is called the solution set of the equation. An equation which is true for all those values of the variable for which it is meaningful is called an identity.

Equivalent Equations

Equations are equivalent if they have the same solution sets.

EXAMPLE 5.1 The equations x = -5 and $ x + 5 = 0 $ are equivalent. Each has the solution set $ {-5} $ .

EXAMPLE 5.2 The equations x = 5 and $ x^{2} = 25 $ are not equivalent; the first has the solution set $ {5} $ , while the second has the solution set $ {-5, 5} $ .

The process of solving an equation consists of transforming it into an equivalent equation whose solution is obvious. Operations of transforming an equation into an equivalent equation include the following:

1. ADDING the same number to both sides. Thus, the equations \(a=b\) and \(a+c=b+c\) are equivalent.

2. SUBTRACTING the same number from both sides. Thus, the equations a = b and a - c = b - c are equivalent.

3. MULTIPLYING both sides by the same nonzero number. Thus, the equations $ a = b $ and $ ac = bc(c ) $ are equivalent.

4. DIVIDING both sides by the same nonzero number. Thus, the equations $ a = b $ and $ = (c ) $ are equivalent.

5. SIMPLIFYING expressions on either side of an equation.

Linear Equations

A linear equation is one which is in the form $ ax + b = 0 $ or can be transformed into an equivalent equation in this form. If $ a $ , a linear equation has exactly one solution. If a = 0, the equation has no solutions unless b = 0, in which case the equation is an identity. An equation which is not linear is called nonlinear.

EXAMPLE 5.3 $ 2x + 6 = 0 $ is an example of a linear equation in one variable. It has one solution, -3. The solution set is $ {-3} $ .

EXAMPLE 5.4 $ x^{2} = 16 $ is an example of a nonlinear equation in one variable. It has two solutions, 4 and -4. The solution set is $ {4, -4} $ .

Linear equations are solved by the process of isolating the variable. The equation is transformed into equivalent equations by simplification, combining all variable terms on one side, all constant terms on the other, then dividing both sides by the coefficient of the variable.

EXAMPLE 5.5 Solve the equation $ 3x - 8 = 7x + 9 $ .

\[ \begin{aligned}3x-8&=7x+9\quad&Subtract 7x from both sides.\\-4x-8&=9\quad&Add8to both sides.\\-4x&=17\quad&Divide both sides by-4.\\x&=-\frac{17}{4}\quad&Solution set:\left\{-\frac{17}{4}\right\}\end{aligned} \]

Quadratic Equations

A quadratic equation is one which is in the form $ ax^{2} + bx + c = 0 $ , $ (a ) $ (standard form), or which can be transformed into this form. There are four methods for solving quadratic equations.

1. FACTORING. If the polynomial $ ax^{2} + bx + c $ has linear factors with rational coefficients, write it in factored form, then apply the zero-factor property that AB = 0 only if A = 0 or B = 0.

2. SQUARE ROOT PROPERTY. If the equation is in the form $ A^{2}=b $ , where b is a constant, then its solutions are found as $ A= $ and $ A=- $ , generally written $ A= $ .

3. COMPLETING THE SQUARE.

1. Write the equation in the form $ x^{2} + px = q $ .

2. Add $ p^{2}/4 $ to both sides to form $ x^{2} + px + p^{2}/4 = q + p^{2}/4 $ .

3. The left side is now a perfect square. Write $ (x + p/2)^{2} = q + p^{2}/4 $ and apply the square root property.

4. QUADRATIC FORMULA. The solutions of $ ax^{2} + bx + c = 0 $ , $ (a ) $ can always be written as:

\[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \]

In general, a quadratic equation is solved by first checking whether it is easily factorable. If it is, then the factoring method is used; otherwise the quadratic formula is used.

EXAMPLE 5.6 Solve $ 3x^{2} + 5x + 2 = 0 $

\[ \begin{array}{ll}Ive3x^{2}+5x+2=0&\\\quad3x^{2}+5x+2=0&\quad Polynomial is factorable using integers\ $ 3x+2)(x+1)=0&\quad Apply the zero-factor property\\3x+2=0&\quad or\qquad x+1=0\\x=-\frac{2}{3}&\quad or\qquad x=-1\\\end{array} \]

EXAMPLE 5.7 Solve $ x^{2} + 5x + 2 = 0 $

Polynomial is not factorable, use formula

\[ \begin{aligned}x^{2}+5x+2&=0\quad&Polynomial is not fact\\x&=\frac{-5\pm\sqrt{5^{2}-4\cdot1\cdot2}}{2\cdot1}\quad&a=1,b=5,c=2\\x&=\frac{-5\pm\sqrt{17}}{2}\end{aligned} \]

In the quadratic formula, the quantity $ b^{2}-4ac $ is called the discriminant. The sign of this quantity determines the number of solutions of a quadratic equation:

SIGN OF DISCRIMINANT	NUMBER OF REAL SOLUTIONS
positive	2
zero	1
negative	0

Occasionally complex solutions are of interest. Then the discriminant determines the number and type of solutions:

SIGN OF DISCRIMINANT	NUMBER AND TYPE OF SOLUTIONS
positive	2 real solutions
zero	1 real solution
negative	2 imaginary solutions

EXAMPLE 5.8 For $ x^{2}-8x+25=0 $ , find (a) all real solutions; (b) all complex solutions.

Use the quadratic formula with a = 1, b = -8, c = 25.

\[ \begin{aligned}&x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4\cdot1\cdot25}}{2\cdot1}\\ &x=\frac{8\pm\sqrt{-36}}{2}\\ \end{aligned} \]

No real solution

\[ \begin{aligned}&x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4\cdot1\cdot25}}{2\cdot1}\\ &x=\frac{8\pm\sqrt{-36}}{2}\\ &x=4\pm3i\\ \end{aligned} \]

Many equations which are not at first glance linear or quadratic can be reduced to linear or quadratic equations, or can be solved by a factoring method.

EXAMPLE 5.9 Solve $ x^{3}-5x{2}-4x+20=0 $

\[ \begin{aligned}ve x^{3}-5x^{2}-4x+20&=0\\x^{3}-5x^{2}-4x+20&=0\\x^{2}(x-5)-4(x-5)&=0\ $ x-5)(x^{2}-4)&=0\ $ x-5)(x-2)(x+2)&=0\\x=5or x=2or x&=-2\end{aligned} \]

Factor by grouping

Equations Containing Radicals

Equations containing radicals require an additional operation: In general, the equation a = b is not equivalent to the equation $ a^{n} = b^{n} $ ; however, if n is odd, they have the same real solutions. If n is even, all solutions of a = b are found among the solutions of $ a^{n} = b^{n} $ . Hence it is permissible to raise both sides of an equation to an odd power, and also permissible to raise both sides to an even power if all solutions of the resulting equation are checked to see if they are solutions of the original equation.

EXAMPLE 5.10 Solve $ = x - 4 $

\[ \begin{aligned}\sqrt{x+2}&=x-4&Square both sides.\ $ \sqrt{x+2})^{2}&=(x-4)^{2}\\x+2&=x^{2}-8x+16\\0&=x^{2}-9x+14\\0&=(x-2)(x-7)\\x&=2&or&x=7\\ Check:x&=2:\sqrt{2+2}=2-4?&x&=7:\sqrt{7+2}=7-4?\\2&\neq-2&3&=3\\ Not a solution&7 is the only solution\end{aligned} \]

Applications: Formulas, Literal Equations, and Equations in More Than One Variable

In these situations, letters are used as coefficients rather than particular numbers. However, the procedures for solving for a specified variable are essentially the same; the other variables are simply treated as constants:

EXAMPLE 5.11 Solve $ A = P + Prt $ for P.

This equation is linear in P, the specified variable. Factor out P, then divide by the coefficient of P.

\[ A=P+Prt \]

\[ \frac{A}{1+rt}=P \]

\[ P=\frac{A}{1+rt} \]

EXAMPLE 5.12 Solve $ s = gt^{2} $ for t.

This equation is quadratic in t, the specified variable. Isolate $ t^{2} $ , then apply the square root property.

\[ s=\frac{1}{2}gt^{2} \]

\[ \frac{2s}{g}=t^{2} \]

\[ t=\pm\sqrt{\frac{2s}{g}} \]

Frequently, but not always, in applied situations, only the positive solutions are retained: $ t = $

Applications: Word Problems

Here, a situation is described and questions are posed in ordinary language. It is necessary to form a model of the situation using variables to stand for unknown quantities, construct an equation (later, an inequality or system of equations) that describes the relation among the quantities, solve the equation, then interpret the solution to answer the original questions.

EXAMPLE 5.13 A right triangle has sides whose lengths are three consecutive even integers. Find the lengths of the sides.

Sketch a figure as in Fig. 5-1:

Let x = length of shortest side

$ x + 2 = $

$ x + 4 = $

Figure 5-1

Now apply the Pythagorean theorem: In a right triangle with sides \(a, b, c, a^{2} + b^{2} = c^{2}\). Hence,

\[ \begin{aligned}x^{2}+(x+2)^{2}&=(x+4)^{2}\\x^{2}+x^{2}+4x+4&=x^{2}+8x+16\\2x^{2}+4x+4&=x^{2}+8x+16\\x^{2}-4x-12&=0\ $ x-6)(x+2)&=0\\x=6\quad or\quad x&=-2\end{aligned} \]

The negative answer is discarded. Hence, the lengths of the sides are: $ x = 6 $ , $ x + 2 = 8 $ , and $ x + 4 = 10 $ .

SOLVED PROBLEMS

5.1. Solve: $ -=2- $

\[ \frac{x}{5}-\frac{3x}{4}=2-\frac{x}{8} \]

Multiply both sides by 40, the LCD of all fractions.

\[ 40\cdot\frac{x}{5}-40\cdot\frac{3x}{4}=80-40\cdot\frac{x}{8} \]

\[ \begin{aligned}8x-30x&=80-5x\\-17x&=80\end{aligned} \]

\[ x=-\frac{80}{17} \]

5.2. Solve: $ 2(3x + 4) + 5(6x - 7) = 7(5x - 4) + 1 + x $

Remove parentheses and combine like terms.

\[ \begin{aligned}2(3x+4)+5(6x-7)&=7(5x-4)+1+x\\6x+8+30x-35&=35x-28+1+x\\36x-27&=36x-27\end{aligned} \]

This statement is true for all (real) values of the variable; the equation is an identity.

5.3. Solve: $ 5x = 2x - (1 - 3x) $

Remove parentheses, combine like terms, and isolate the variable.

\[ \begin{aligned}5x&=2x-1+3x\\5x&=5x-1\\0&=-1\end{aligned} \]

The statement is true for no value of the variable; the equation has no solution.

5.4. Solve: $ = 7 $

Multiply both sides by x - 3, the only denominator; then isolate x. Note: $ x $ .

\[ \begin{aligned}(x-3)\frac{x+5}{x-3}&=7(x-3)\\x+5&=7x-21\\-6x&=-26\\x&=\frac{13}{3}\end{aligned} \]

5.5. Solve: $ =5- $

Multiply both sides by $ x + 1 $ , the only denominator. Note: $ x $ .

\[ \begin{aligned}\frac{6}{x+1}&=5-\frac{6x}{x+1}\ $ x+1)\cdot\frac{6}{x+1}&=5(x+1)-(x+1)\frac{6x}{x+1}\\6&=5x+5-6x\\1&=-x\\x&=-1\end{aligned} \]

In this case, since $ x $ , there can be no solution.

5.6. Solve: $ (x + 5)^{2} + (2x - 7)^{2} = 82 $

Remove parentheses and combine like terms; the resulting quadratic equation is factorable.

\[ (x+5)^{2}+(2x-7)^{2}=82 \]

\[ x^{2}+10x+25+4x^{2}-28x+49=82 \]

\[ 5x^{2}-18x-8=0 \]

\[ (5x+2)(x-4)=0 \]

\[ x=-\frac{2}{5}\quad or\quad x=4 \]

5.7. Solve: $ 5x^{2} + 16x + 2 = 0 $

This is not factorable in the integers; use the quadratic formula, with \(a = 5\), \(b = 16\), \(c = 2\).

\[ \begin{aligned}5x^{2}+16x+2&=0\\x&=\frac{-16\pm\sqrt{16^{2}-4\cdot5\cdot2}}{2\cdot5}\\x&=\frac{-16\pm\sqrt{216}}{10}\\x&=\frac{-16\pm6\sqrt{6}}{10}\\x&=\frac{-8\pm3\sqrt{6}}{5}\end{aligned} \]

5.8. Solve $ x^{2}-8x+13=0 $ by completing the square.

\[ \begin{aligned}x^{2}-8x+13&=0\\x^{2}-8x&=-13\\x^{2}-8x+16&=3\ $ x-4)^{2}&=3\\x-4&=\pm\sqrt{3}\\x&=4\pm\sqrt{3}\end{aligned} 訂加 \begin{aligned}\left[\frac{1}{2}(-8)\right]^{2}&=(-4)^{2}=16\\ Add16to both sides.\end{aligned} \]

5.9. Solve: $ + = 4 $

\[ \begin{aligned}\frac{2}{x}+\frac{3}{x+1}&=4\\x(x+1)\frac{2}{x}+x(x+1)\frac{3}{x+1}&=4x(x+1)\\2(x+1)+3x&=4x^{2}+4x\\5x+2&=4x^{2}+4x\\0&=4x^{2}-x-2\end{aligned} \]

This is not factorable in the integers; use the quadratic formula, with \(a = 4\), \(b = -1\), \(c = -2\).

\[ \begin{aligned}&x=\frac{-(-1)\pm\sqrt{(-1)^{2}-4(4)(-2)}}{2\cdot4}\\ &x=\frac{1\pm\sqrt{33}}{8}\\ \end{aligned} \]

5.10. Find all solutions, real and complex, for $ x^{3}-64=0 $ .

First factor the polynomial as the difference of two cubes.

\[ \begin{aligned}x^{3}-4^{3}&=0\ $ x-4)(x^{2}+4x+16)&=0\\x=4or x^{2}+4x+16&=0\end{aligned} \]

Now apply the quadratic formula to the quadratic factor, with a = 1, b = 4, c = 16.

\[ x=\frac{-4\pm\sqrt{4^{2}-4\cdot1\cdot16}}{2\cdot1} \]

\[ x=\frac{-4\pm\sqrt{-48}}{2} \]

\[ x=\frac{-4\pm4i\sqrt{3}}{2} \]

\[ x=-2\pm2i\sqrt{3} \]

Solutions: $ 4, -2 2i. $

5.11. Solve: $ x^{4}-5x{2}-36=0 $

This is an example of an equation in quadratic form. It is convenient, although not necessary, to introduce the substitution $ u = x^{2} $ . Then $ u^{2} = x^{4} $ and the equation becomes:

\[ \begin{aligned}u^{2}-5u-36&=0\ $ u-9)(u+4)&=0\\u=9\quad or\quad u&=-4\end{aligned} \]

This is factorable in the integers.

Now undo the original substitution $ x^{2} = u $ .

\[ \begin{aligned}x^{2}&=9\quad or\quad x^{2}=-4\\x&=\pm3\quad no real solution\end{aligned} \]

5.12. Solve: $ x^{2/3} - x^{1/3} - 6 = 0 $

This equation is in quadratic form. Introduce the substitution $ u = x^{1/3} $ . Then $ u^{2} = x^{2/3} $ and the equation becomes:

\[ \begin{aligned}u^{2}-u-6&=0\ $ u-3)(u+2)&=0\\u&=3\quad or\quad u=-2\end{aligned} \]

Now undo the original substitution $ x^{1/3} = u $ .

\[ \begin{aligned}x^{1/3}&=3\quad or\quad x^{1/3}=-2\\x&=3^{3}\quad x=(-2)^{3}\\x&=27\quad x=-8\end{aligned} \]

5.13. Solve: $ = + 1 $

Square both sides, noting that the right side is a binomial.

\[ \begin{aligned}\sqrt{2x}&=\sqrt{x+1}+1\ $ \sqrt{2x})^{2}&=(\sqrt{x+1}+1)^{2}\\2x&=x+1+2\sqrt{x+1}+1\end{aligned} \]

Now isolate the term containing the square root and square again.

\[ \begin{aligned}x-2&=2\sqrt{x+1}\ $ x-2)^{2}&=(2\sqrt{x+1})^{2}\\x^{2}-4x+4&=4(x+1)\\x^{2}-4x+4&=4x+4\\x^{2}-8x&=0\\x(x-8)&=0\\x&=0\quad or\quad x=8\quad Check:x=0:\sqrt{2\cdot0}=\sqrt{0+1}+1?&x&=8:\sqrt{2\cdot8}=\sqrt{8+1}+1?\\&0\neq1+1&4&=3+1\\&Not a solution&8 is the only solution\end{aligned} \]

5.14. Solve the literal equation $ S = 2xy + 2xz + 2yz $ for y

This equation is linear in y, the specified variable. Since all terms involving y are already on one side, get all terms not involving y on the other side, then divide both sides by the coefficient of y.

\[ S=2xy+2xz+2yz \]

\[ S-2xz=2xy+2yz \]

\[ S-2xz=y(2x+2z) \]

\[ \frac{S-2xz}{2x+2z}=y \]

\[ y=\frac{S-2xz}{2x+2z} \]

5.15. Solve $ + = $ for f.

This equation is linear in f, the specified variable. Multiply both sides by pqf, the LCD of all fractions, then divide both sides by the coefficient of f.

\[ \begin{aligned}\frac{1}{p}+\frac{1}{q}&=\frac{1}{f}\\pqf\cdot\frac{1}{p}+pqf\cdot\frac{1}{q}&=pqf\cdot\frac{1}{f}\\qf+pf&=pq\\f(q+p)&=pq\\f&=\frac{qp}{q+p}\end{aligned} \]

5.16. Solve $ s = gt^{2} - v_{0}t + s_{0} $ for t.

This equation is quadratic in t, the specified variable. Get the equation into standard form for quadratic equations:

\[ s=\frac{1}{2}gt^{2}-\nu_{0}t+s_{0} \]

\[ \frac{1}{2}gt^{2}-v_{0}t+s_{0}-s=0 \]

Now apply the quadratic formula with $ a = g $ , $ b = -v_{0} $ , $ c = s_{0} - s $ .

\[ \begin{aligned}&t=\frac{-(-v_{0})\pm\sqrt{(-v_{0})^{2}-4\big(\frac{1}{2}g\big)(s_{0}-s)}}{2\big(\frac{1}{2}g\big)}\\ &t=\frac{v_{0}\pm\sqrt{v_{0}^{2}-2g(s_{0}-s)}}{g}\\ \end{aligned} \]

5.17. $9000 is to be invested, part at 6% interest, and part at 10% interest. How much should be invested at each rate if a total return of 9% is desired?

Use the formula \(I = Prt\) with \(t\) understood to be one year. Let \(x = \text{amount invested at } 6\%\); a tabular arrangement is helpful:

	P: AMOUNT INVESTED	r: RATE OF INTEREST	I: INTEREST EARNED
First account	x	0.06	0.06x
Second account	9000 - x	0.1	0.1(9000 - x)
Total investment	9000	0.09	0.09(9000)

Since the interest earned is the total of the interest on the two investments, write:

\[ 0.06x+0.1(9000-x)=0.09(9000) \]

Solving yields:

\[ \begin{aligned}0.06x+900-0.1x&=810\\-0.04x&=-90\\x&=2250\end{aligned} \]

Therefore, $2250 should be invested at 6% and 9000 - x = $6750 should be invested at 10%.

5.18. A box with a square base and no top is to be made from a square piece of cardboard by cutting out a 3-inch square from each corner and folding up the sides. If the box is to hold 75 cubic inches, what size piece of cardboard should be used?

Figure 5-2

Sketch a figure (see Fig. 5-2).

Let x = length of side of original piece. Then x - 6 = length of side of box.

Use volume = (length)(width)(height):

\[ \begin{aligned}3(x-6)^{2}&=75\ $ x-6)^{2}&=25\\x-6&=\pm5\\x&=6\pm5\end{aligned} \]

Thus, x = 11 in or x = 1 in. Clearly, the latter does not make sense; hence, the dimensions of the original cardboard must be 11 in square.

5.19. Two people have a walkie-talkie set with a range of $ $ mi. One of them starts walking at noon in an easterly direction, at a rate of 3 mph. Five minutes later the other person starts walking in a westerly direction, at a rate of 4 mph. At what time will they reach the range of the device?

Use distance = (rate)(time). Let t = time since noon. A tabular arrangement is helpful.

	TIME WALKED	RATE OF WALKING	DISTANCE
First person	t	3	3t
Second person	t - 5/60	4	4(t - 5/60)

Since the distances add up to the total distance of $ $ mi, this yields:

\[ \begin{aligned}3t+4\Big(t-\frac{5}{60}\Big)&=\frac{3}{4}\\3t+4t-\frac{1}{3}&=\frac{3}{4}\\7t&=\frac{1}{3}+\frac{3}{4}\\t&=\frac{\frac{1}{3}+\frac{3}{4}}{7}\\t&=\frac{13}{84}\end{aligned} \]

The time will be noon plus $ $ hours, or approximately 12:09 p.m.

5.20. A container is filled with 8 liters of a 20% salt solution. How many liters of pure water must be added to produce a 15% salt solution?

Let x = the number of liters of water added. A tabular arrangement is helpful.

	AMOUNT OF SOLUTION	PERCENTAGE OF SALT	AMOUNT OF SALT
Original solution	8	0.2	(0.2)8
Water	x	0	0
Mixture	8 + x	0.15	0.15(8 + x)

Since the amounts of salt in the original solutions and the added water must add up to the amount of salt in the mixture, this yields:

\[ \begin{aligned}(0.2)8+0&=0.15(8+x)\\1.6&=1.2+0.15x\\0.4&=0.15x\\x&=\frac{0.4}{0.15}\quad or\quad2^{\frac{2}{3}}liters\end{aligned} \]

5.21. Machine A can perform a job in 6 hours, working alone. Machine B can complete the same job in 10 hours, working alone. How long would it take the two machines, working together, to complete the job?

Use quantity of work = (rate)(time). Note that if a machine can do a job in x hours, it performs 1/x of the work in one hour; that is, its rate is 1/x job per hour. Let t = the time worked by each machine. A tabular arrangement is helpful.

	RATE	TIME	QUANTITY OF WORK
Machine A	1/6	t	t/6
Machine B	1/10	t	t/10

Since the quantity of work performed by the two machines totals to one entire job, this yields:

\[ \begin{aligned}\frac{t}{6}+\frac{t}{10}&=1\\30\cdot\frac{t}{6}+30\cdot\frac{t}{10}&=30\\5t+3t&=30\\8t&=30\\t&=\frac{15}{4}\end{aligned} \]

The time would be $ 3 $ hours.

SUPPLEMENTARY PROBLEMS

5.22. Solve: $ 3 - = - (x - 4) + 5 $ Ans. $ - $

5.23. Solve: $ 7(x - 6) - 6(x + 3) = 5(x - 6) - 2(3 + 2x) $ Ans. No solution.

5.24. Solve: $ -= $ Ans. 7

5.25. Find all real solutions:

\[ \textcircled{c}x^{2}-9x=36;(b)3x^{2}=2x+8;(c)4x^{2}+3x+5=0;(d)x^{2}-5=2x+3; \]

\[ (x-8)(x+6)=32;(f)8x^{2}-3x+4=3x^{2}+12;(g)(x-5)^{2}=7;(h)4x^{2}+3x-5=0 \]

Ans. (a) $ {-3,12} $ ; (b) $ {-,2} $ ; (c) no real solutions; (d) $ {-2,4} $ ;

5.26. Solve:

1. $ = -6 $ (b) $ = -6 $ Ans. (a) -45 (b) No solution.

5.27. Find all real solutions:

Ans. (a) $ {-, } $ ; (b) $ {-1, 64} $ ; (c) $ {-2, } $

5.28. Solve: (a) $ x - = 12 $ ; (b) $ + 1 = x $ ; (c) $ - = 2 $

Ans. (a) $ {16} $ ; (b) $ {4} $ ; (c) $ {2, 6} $

5.29. Find all complex solutions for $ x^{3} - 5x^{2} + 4x - 20 = 0 $ Ans. 5, 2i, -2i

5.30. Solve: $ + = $ for q. Ans. $ q = $

5.31. Solve: $ LI^{2} + RI + = 0 $ for I.

\[ Ans.\quad I=\frac{-RC\pm\sqrt{R^{2}C^{2}-4LC}}{2LC} \]

5.32. Solve: $ (x - h)^{2} + (y - k)^{2} = r^{2} $ for y.

\[ Ans.\quad y=k\pm\sqrt{r^{2}-(x-h)^{2}} \]

5.33. Solve for y in terms of x: (a) 3x - 5y = 8; (b) $ x^{2} - 2xy + y^{2} = 4 $ ; (c) $ = 5 $ ; (d) $ x = $

Ans. (a) $ y = $ ; (b) $ y = x + 2 $ or y = x - 2; (c) $ y = x $ ; (d) $ y = 1 $

5.34. A rectangle has perimeter 44 cm. Find its dimensions if its length is 5 cm less than twice its width.

Ans. Width = 9 cm, length = 13 cm

5.35. Solve the walkie-talkie problem (5.19) if the two people start walking at the same time, but the second person walks north.

Ans. Exactly 12:09 p.m.

5.36. A shop wishes to blend coffee priced at $6.50 per pound with coffee priced at $9.00 per pound in order to yield 60 pounds of a blend to sell for $7.50 per pound. How much of each type of coffee should be used?

Ans. 36 pounds of the $6.50-per-pound coffee, 24 pounds of the $9.00-per-pound coffee.

5.37. A container is filled with 8 centiliters of a 30% acid solution. How many centiliters of pure acid must be added to produce a 50% acid solution?

Ans. 3.2 cl

5.38. A chemistry stockroom has two alcohol solutions, a 30% and a 75% solution. How many deciliters of each must be mixed to obtain 90 deciliters of a 65% solution?

Ans. 20 dl of the 30% solution, 70 dl of the 75% solution

5.39. A 6-gallon radiator is filled with a 40% solution of antifreeze in water. How much of the solution must be drained and replaced with pure antifreeze to obtain a 65% solution?

Ans. 2.5 gallons

5.40. Machine A can complete a job in 8 hours, working alone. Working together with machine B, the job can be completed in 5 hours. How long would it take machine B, working alone, to complete the job?

Ans. $ 13 $ hours

5.41. Machine A can do a job, working alone, in 4 hours less than machine B. Working together, they can complete the job in 5 hours. How long would it take each machine, working alone, to complete the job?

Ans. Machine A: 8.4 hours; machine B: 12.4 hours, approximately