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# Linear Equations

6.2  Solving Linear Equations

Equations of the form a x+b=0 are called linear equations in the variable x. In this section we will be concerned with the problem of solving linear equations, and equations that reduce to linear equations.

We deﬁne two equations as equivalent if they have the same solution set. The following two operations on an equation always result in a new equation which is equivalent to the original one. These operations, sometimes called elementary transformations, are:

T.1 The same expression representing a real number may be added to both sides of an equation.

T.2 The same expression representing a nonzero real number may be multiplied into both sides of an equation.

Using these operations we may transform an equation whose solution set is not obvious through a series of equivalent equations to an equation that has an obvious solution set.

Example 1.  Solve the equation

(a) 2 x-3=4+x

Add x to both sides to obtain

x+2 x-3= x+4+x  (T.1)

or  x-3=4

Add 3 to both sides to obtain

x-3+3=4+3  (T.1)

or  x=7

Since 2 x-3=4+x is equivalent to x-3=4, which, in turn, is equivalent to x=7, whose solution set is obviously {7}, we know that the solution set of (a) is {7}.

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Example 2.  Solve the equation

(b) 12 x+23=52 x-1

Add  (12) x to both sides to obtain

23=52 x-12 x-1  (T.1)

or  23=2 x-1

Add 1 to both sides to obtain

1+23=2 x  (T.1)

or  53=2 x

Multiply both sides by 12 to obtain

56=x  (T.2)

Thus, the solution set of (b) is {56}.

Every linear equation can be solved in the same way as in the above examples. In fact, let us consider the general linear equation

a x+b=0

Add b to both sides to obtain

a x= b

Multiply both sides by 1a to obtain

x= (ba)

if a a0. The general linear equation, therefore, has as its solution set {ba}, if a0. Thus each linear equation has at most one solution.

The next two examples are of equations that reduce to linear equations.

Example 3.  Solve the equation

23+4 y (5 y+4)=9+10 y (2 y+3)

We expand both sides to obtain

23+20 y2+16 y=9+20 y2+30 y

Add 20 y2 to both sides to obtain

23+16 y=9+30 y

We now solve as in the previous examples.

23+16 y=9+30 y

23-9=30 y-16 y

14=14 y

y=1

Thus the solution set is {1}.

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Example 4.  Solve the equation

(c) 2 xx-1=2x-1+1

The replacement set of (c) is all real numbers except 1. Assuming that x1, we multiply both sides of (c) by x-1 to obtain

(d) 2 x=2+x-1,x1

Solving the equation 2 x=2+x-1, we obtain 1 as the only solution Since 1 is not in the replacement of (d), (d) has no solution. Furthermore, (c) is equivalent to (d), therefore (c) has no solution.

6.3  Solving Literal Equations

An equation containing more than one variable, or containing symbols representing constants such as a,b, and c, can be solved for one of the symbols in terms of the remaining symbols by applications of the operations T.1 and T.2 in the preceding section. The student will encounter such problems in other courses.

Example 1.  Solve c x-3 a=b for x.

Add 3 a to both sides.

c x=b+3 a

Multiply both sides by 1c.

x=b+3 ac

This last equation expresses x in terms of the other symbols.

Example 2.  Solve 3 a y-2 b=2 c y for y.

Add 2 b to both sides.

3 a y=2 c y+2 b

Add  2 c y to both sides.

3 a y-2 c y=2 b

Factor out y.

(3 a-2 c) y=2 b

Multiply both sides by 13 a-2 c

y=2 b3 a-2 c

Example 3.  Solve ax+b2 x=c for x.

Multiply both sides by 2 x.

2 a+b=2 c x

2 c x=2 a+b

Multiply by 12 c.

x=2 a+b2 c

We conclude this section by including two more examples similar to those that the student may encounter in other areas.

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Example 4.  Solve A=P (1+r t) for r.

Apply the distributive law.

A=P+P r t

A-P=P r t

Multiply both sides by 1P t.

A-PP t=r

Example 5.  Solve 1R=1r1+1r2 for r1.

Add the two terms on the right—hand side.

1R=r2+r1r1 r2

Multiply by R r1 r2.

r1 r2=R (r2+r1)

r1 r2=R r2+R r1

Add  R r1 to both sides.

r1 r2-R r1=R r2

Factor out r1.

r1 (r2-R)=R r2

Multiply by 1r2-R.

r1=R r2r2-R

6.4  Solving Statement Problems

One of the fundamental applications of algebra is solving problems that have been stated in words. A statement problem is a word description of a situation that involves both known and unknown quantities. In this section each problem will be solvable by means of one equation involving one unknown.

Our problem is to choose the unknown and to determine the equation that it must satisfy. Although there is no single approach to all of the problems, the following suggestions are sometimes helpful:

1. Read the problem carefully until the situation is thoroughly understood.

2. Determine what quantities are asked for, then choose the one that seems to be the best to use as the unknown.

3. Establish the relationship between the unknown and the other quantities in the problem.

4. Find the information that tells which two quantities are equal.

5. Use the information in (4) to write the equation.

6. Solve the equation and check the solution to see that it satisﬁes the original problem.

At this stage, the emphasis will be on translating statement problems into equations. Although some of the problems can be solved almost by inspection, the practice that we obtain in setting up equations will prove helpful in working more difficult problems.

Example 1.  If 2 times a certain integer is added to the next consecutive integer the result is 34. Find the integers.
Step 2. Let x be the ﬁrst integer.
Step 3. Then x+1 is the next consecutive integer.
Step 4.  2 times a certain integer plus the next consecutive integer is 34.
Step 5.  2 x+(x+1)=34

Step 6. Solve.

2 x+(x+1)=34

3 x+1=34

3 x=33

x=11

Check.  2·11+(11+1)=34

Example 2.  Bob and Joe together earned \$60. Both were paid at the same rate, but Bob worked three times as long as Joe. How much did each receive?

Step 2. Let x be the number of dollars that Joe received.

Step 3. Then 3 x is the number of dollars that Bob received

Step 4. Bob and Joe together earned \$60.
Step 5. 3 x+x=60

Step 6. Solve.

3 x+x=60

4 x=60

x=15

3 x=45

Check  3·15+15=60

Example 3.  The sum of the digits of a two digit number is 12. If the digits are reversed the number is decreased by 36. What is the number?

Step 2. Let x be the tens digit.

Step 3. Then 12-x is the units digit.

Step 4. If the digits are reversed then the number is decreased by 36

Step 5. 10 (12-x)+x=10 x+(12-x)-36

Step 6. Solve.

10 (12-x)+x=10 x+(12-x)-36

=120-10 x+x=10 x+12-x-36

=120-9 x=9 x-24

=144=18 x

=x=8

=12-x=4

Therefore the number is84.

Check.  84-36=48

Example 4.  How many pounds of candy valued at 48¢ per pound should be added to 50 pounds of candy valued at 80¢ per pound in order for the store owner to be able to sell the candy at 60¢ per pound?
Step 2. Let x be the number of pounds of 48¢ per pound candy.
Step 3. Then 50+x would be the pounds of candy he would have at 60¢ per pound.
Step 4. The amount of candy at 48¢ per pound times 48¢, plus the amount of candy at 80¢ per pound times 80¢, must be equal to the amount of candy at 60¢ per pound times 60¢.
Step 5. (48¢/lb)(x lbs) + (80¢/lb) (50 lbs) = (60¢/lb) [(50+x)lbs]

Step 6. Solve.

48 x+80·50=60 (50+x)

48 x+4000=3000+60 x

1000=12 x

x=(83 13)lbs

Check.  (83+13) 48+80·50=60 (50+83+13)

Problems involving velocities (or speeds) will use the formula

d=r t

where d is the distance traveled, r is the rate, and t is the time. When the formula is used, d and r must be expressed in the same unit of distance, whiler and t must be expressed in the same unit of time.

Example 5.  A group of students drove to a lake in the north woods to ﬁsh. They traveled 380 miles in 7 hours, of which 4 hours were on a paved highway and the remaining time was on a dirt road. If the average speed on the dirt road was 25 miles per hour less than the average speed on the highway, then ﬁnd for each part of the trip the average speed and the distance traveled.

Step 2. Let x be the speed on the dirt road.

Step 3. Then x+25 is the speed on the highway.

Step 4. The distance traveled on the highway plus the distance traveled on the dirt road is equal to 380 miles.

Step 5. Since d=r t, we have

[(x+25) m ih r] (4 h r s)+[x m ih r] (3 h r s)=380 m i

Step 6. Solve.

(x+25) 4+3 x=380

4 x+100+3 x=380

7 x=280

x=40 miles per hour

x+25=65miles per hour

Check.  (40+25) 4+40·3=380

Work problems which involve the rate of performance can often be solved by ﬁrst ﬁnding the fractional part of the task done by each person or machine in one unit of time, and then ﬁnding an equation that relates these various fractional parts.

Example 6.  A boy can cut a lawn in 4 hours while the father can cut it in 3 hours. How long would it take them to cut the same lawn working together?

Step 2. Let x be the number of hours that it would take them to cut the lawn Working together.

Step 3. Choose one hour as our unit of time. Now the boy can cut 14 of the lawn in one hour, the father can cut 13 of the lawn in one hour, and togeﬂner they can cut 1x of the lawn in one hour.

Step 4. The amount cut by the boy in one hour plus the amount cut by the father in one hour is equal to the amount they can cut together in one hour.

Step 5. 13+14=1x

Step 6. Solve.

13+14=1x

712=1x

x=127 hours

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