Consider the following simple examples, (a) 3x+10=x+4 (b) z
z−5 +1 3 = −5
5−z (c) Y =C+I
Equation (a) contains thevariablex, whereas (b) has the variablez, and equation (c) has the three variablesY,C, andI.
Tosolvean equation means to find all values of the variables for which the equation is satisfied. For equation (a) this is easy. In order to isolate the unknownxon one side of the equation, we add−xto both sides. This gives 2x+10=4. Adding−10 to both sides of this equation yields 2x =4−10= −6. Dividing by 2 we get the solutionx= −3.
If any value of a variable makes an expression in an equation undefined, that value is not allowed.Thus, the choicez=5 is not allowed in (b) because it makes the expressions z/(z−5)and−5/(5−z)undefined, because they are 5/0 and−5/0, respectively. As we shall show in Example 3 below, equation (b) has no solutions.
Equation (c) has many solutions, one of which isY =1000,C =700, andI =300.
For problem (a) the solution procedure was probably well known. The method we used is summed up in the following frame, noting that two equations which have exactly the same solutions are calledequivalent.
To get equivalent equations, do the following on both sides of the equality sign:
(A) add (or subtract) the same number,
(B) multiply (or divide) by the same number =0.
When faced with more complicated equations involving parentheses and fractions, we usu- ally begin by multiplying out the parentheses, and then we multiply both sides of the equation by the lowest common denominator for all the fractions. Here is an example.
E X A M P L E 1 Solve the equation 6p−12(2p−3)=3(1−p)−76(p+2).
Solution: First multiply out all the parentheses: 6p−p+32 =3−3p−76p−73. Second, multiply both sides by 6 to clear all the fractions: 36p−6p+9=18−18p−7p−14.
Third, gather terms: 55p= −5. Thusp= −5/55= −1/11.
The next two examples show that sometimes a surprising degree of care is needed to find the right solutions.
E X A M P L E 2 Solve the equation x+2
x−2 − 8
x2−2x = 2 x.
Solution: Since x2 −2x = x(x −2), the common denominator isx(x −2). We see that x = 2 and x = 0 both make the equation absurd, because then at least one of the denominators becomes 0. Providedx = 0 andx = 2, we can multiply both sides of the equation by the common denominatorx(x−2)to obtain
x+2
x−2 ãx(x−2)− 8
x(x−2)ãx(x−2)= 2
x ãx(x−2)
Cancelling common factors reduces this to(x+2)x−8=2(x−2)orx2+2x−8=2x−4 and sox2 =4. Equations of the formx2 =a, wherea >0, have two solutionsx =√
a andx = −√
a. In our case,x2=4 has solutionsx =2 andx = −2. Butx=2 makes the original equation absurd, soonlyx = −2is a solution.
E X A M P L E 3 Solve the equation z z−5+1
3 = −5 5−z.
Solution: We see thatzcannot be 5. Remembering this restriction, multiply both sides by 3(z−5). This gives
3z+z−5=15
which has the unique solutionz=5. Because we had to assumez=5, we must conclude that no solution exists for the original equation.
S E C T I O N 2 . 1 / H O W T O S O L V E S I M P L E E Q U A T I O N S 37 Often, solving a problem in economic analysis requires formulating an appropriate algebraic equation.
E X A M P L E 4 A firm manufactures a commodity that costs $20 per unit to produce. In addition, the firm has fixed costs of $2000. Each unit is sold for $75. How many units must be sold if the firm is to meet a profit target of $14 500?
Solution: If the number of units produced and sold is denoted byQ, then the revenue of the firm is 75Qand the total cost of production is 20Q+2000. Because profit is the difference between total revenue and total cost, it can be written as 75Q−(20Q+2000). Because the profit target is 14 500, the equation
75Q−(20Q+2000)=14 500
must be satisfied. It is easy to find the solutionQ=16 500/55=300 units.
P R O B L E M S F O R S E C T I O N 2 . 1
In Problems 1–3, solve each of the equations.
1. (a) 5x−10=15 (b) 2x−(5+x)=16−(3x+9) (c) −5(3x−2)=16(1−x) (d) 4x+2(x−4)−3=2(3x−5)−1 (e) 23x= −8 (f) (8x−7)5−3(6x−4)+5x2=x(5x−1) (g) x2+10x+25=0 (h) (3x−1)2+(4x+1)2=(5x−1)(5x+1)+1
2. (a) 3x+2=11 (b) −3x=21 (c) 3x=14x−7
(d) x−3
4 +2=3x (e) 1
2x+1 = 1
x+2 (f) √
2x+14=16
⊂SM⊃3. (a) x−3
x+3=x−4
x+4 (b) 3
x−3− 2
x+3= 9
x2−9 (c) 6x 5 −5
x =2x−3
3 +8x
15 4. Solve the following problems by first formulating an equation in each case:
(a) The sum of twice a number and 5 is equal to the difference between the number and 3. Find the number.
(b) The sum of three successive natural numbers is 10 more than twice the smallest of them.
Find the numbers.
(c) Jane receives double pay for every hour she works over and above 38 hours per week. Last week, she worked 48 hours and earned a total of $812. What is Jane’s regular hourly wage?
(d) James has invested $15 000 at an annual interest rate of 10%. How much additional money should he invest at the interest rate of 12% if he wants the total interest earned by the end of the year to equal $2100?
(e) When Mr. Barnes passed away, 2/3 of his estate was left to his wife, 1/4 was shared by his children, and the remainder, $100 000, was donated to a charity. How big was Mr. Barnes’s estate?
⊂SM⊃5. Solve the following equations:
(a) 3y−1 4 −1−y
3 +2=3y (b) 4 x+ 3
x+2= 2x+2 x2+2x+ 7
2x+4 (c)
2− z 1−z
1+z = 6
2z+1 (d) 1
2 p
2 −3 4
−1 4
1−p 3
−1
3(1−p)= −1 3 6. A person hasyeuros to spend on three kinds of fruit, namely apples, bananas, and cherries.
She decides to spend13yeuros on each kind. The prices in euros per kilo are 3 for apples, 2 for bananas, and 6 for cherries. What is the total weight of fruit she buys, and how much does she pay per kilo of fruit? (This is an example of “dollar cost” averaging. See Problem 11.5.3.)