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1.7 Solving Linear Equations

Learning Objectives

  1. Use the properties of equality to solve basic linear equations.
  2. Identify and solve conditional linear equations, identities, and contradictions.
  3. Clear fractions from equations.
  4. Set up and solve linear applications.

Solving Basic Linear Equations

An equationStatement indicating that two algebraic expressions are equal. is a statement indicating that two algebraic expressions are equal. A linear equation with one variableAn equation that can be written in the standard form ax+b=0, where a and b are real numbers and a0., x, is an equation that can be written in the standard form ax+b=0 where a and b are real numbers and a0. For example,

3x12=0

A solutionAny value that can replace the variable in an equation to produce a true statement. to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation 3x12=0 is x and the solution is x=4. To verify this, substitute the value 4 in for x and check that you obtain a true statement.

3x12=03(4)12=01212=00=0

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example 1

Is a=12 a solution to 10a+5=25?

Solution:

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

10a+5=10(12)+5=5+5=1025

Answer: No, a=12 does not satisfy the equation.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equationsEquations with the same solution set. as equations with the same solution set.

3x5=163x=21x=7}Equivalentequations

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, {7}. To obtain equivalent equations, use the following properties of equalityProperties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers.. Given algebraic expressions A and B, where c is a nonzero number:

Addition property of equality:

IfA=B,thenA+c=B+c

Subtraction property of equality:

IfA=B,thenAc=Bc

Multiplication property of equality:

IfA=B,thencA=cB

Division property of equality:

IfA=B,thenAc=Bc

Note: Multiplying or dividing both sides of an equation by 0 is carefully avoided. Dividing by 0 is undefined and multiplying both sides by 0 results in the equation 0 = 0.

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form ax+b=c, then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Example 2

Solve: 7x2=19.

Solution:

7x2=197x2+2=19+2Add2tobothsides.7x=217x7=217Dividebothsidesby7.x=3

Answer: The solution is 3.

Example 3

Solve: 56=8+12y.

Solution:

When no sign precedes the term, it is understood to be positive. In other words, think of this as 56=+8+12y. Therefore, we begin by subtracting 8 on both sides of the equal sign.

568=8+12y848=12y4812=12y124=y

It does not matter on which side we choose to isolate the variable because the symmetric propertyAllows you to solve for the variable on either side of the equal sign, because x=5 is equivalent to 5=x. states that 4=y is equivalent to y=4.

Answer: The solution is 4.

Example 4

Solve: 53x+2=8.

Solution:

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient 53.

53x+2=853x+22=82Subtract2onbothsides.53x=103553x=35(10)2Multiplybothsidesby35.1x=3(2)x=6

Answer: The solution is −6.

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Try this! Solve: 23x+12=56.

Answer: x=2

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

  • Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
  • Step 2: Use the appropriate properties of equality to combine like terms on opposite sides of the equal sign. The goal is to obtain the variable term on one side of the equation and the constant term on the other.
  • Step 3: Divide or multiply as needed to isolate the variable.
  • Step 4: Check to see if the answer solves the original equation.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

Note: At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Example 5

Solve: 4a+2a=1.

Solution:

First combine the like terms on the left side of the equal sign.

4a+2a=1       Combinesame-sideliketerms.5a+2=1Subtract2onbothsides.5a=1Dividebothsidesby5.a=15=15

Always use the original equation to check to see if the solution is correct.

4a+2a=4(15)+215=45+215515=4+10+15=55=1

Answer: The solution is 15.

Given a linear equation in the form ax+b=cx+d, we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Example 6

Solve: 2y3=5y+11.

Solution:

Subtract 5y on both sides so that we can combine the terms involving y on the left side.

2y35y=5y+115y7y3=11

From here, solve using the techniques developed previously.

7y3=11Add3tobothsides.7y=14y=147Dividebothsidesby7.y=2

Answer: The solution is −2.

Solving will often require the application of the distributive property.

Example 7

Solve: 12(10x2)+3=7(12x).

Solution:

Simplify the linear expressions on either side of the equal sign first.

12(10x2)+3=7(12x)          Distribute.5x+1+3=714x         Combinesame-sideliketerms.5x+4=714x         Combineopposite-sideliketerms.9x=3         Solve.x=39=13

Answer: The solution is 13.

Example 8

Solve: 5(3a)2(52a)=3.

Solution:

Begin by applying the distributive property.

5(3a)2(52a)=3155a10+4a=35a=3a=2

Here we point out that a is equivalent to 1a; therefore, we choose to divide both sides of the equation by −1.

a=21a1=21a=2

Alternatively, we can multiply both sides of a=2 by negative one and achieve the same result.

a=2(1)(a)=(1)(2)a=2

Answer: The solution is 2.

Try this! Solve: 63(4x1)=4x7.

Answer: x=1

There are three different types of equations. Up to this point, we have been solving conditional equationsEquations that are true for particular values.. These are equations that are true for particular values. An identityAn equation that is true for all possible values. is an equation that is true for all possible values of the variable. For example, x=xIdentity has a solution set consisting of all real numbers, . A contradictionAn equation that is never true and has no solution. is an equation that is never true and thus has no solutions. For example, x+1=xContradiction has no solution. We use the empty set, Ø, to indicate that there are no solutions.

If the end result of solving an equation is a true statement, like 0 = 0, then the equation is an identity and any real number is a solution. If solving results in a false statement, like 0 = 1, then the equation is a contradiction and there is no solution.

Example 9

Solve: 4(x+5)+6=2(2x+3).

Solution:

4(x+5)+6=2(2x+3)4x+20+6=4x+64x+26=4x+626=6

Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.

Answer: Ø

Example 10

Solve: 3(3y+5)+5=10(y+2)y.

Solution:

3(3y+5)+5=10(y+2)y9y+15+5=10y+20y9y+20=9y+209y=9y0=0

Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.

Answer:

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example 11

Solve: 13x+15=15x1.

Solution:

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the LCD(3,5)=15.

15(13x+15)=15(15x1)Multiplybothsidesby15.1513x+1515=1515x151Simplify.5x+3=3x15Solve.2x=18x=182=9

Answer: The solution is −9.

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

Expression

Equation

12x+53

12x+53=0

We simplify expressions and solve equations. If you multiply an expression by 6, you will change the problem. However, if you multiply both sides of an equation by 6, you obtain an equivalent equation.

Incorrect

Correct

12x+536(12x+53)=3x+10

12x+53=06(12x+53)=603x+10=0

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Key Words

Translation

Sum, increased by, more than, plus, added to, total

 +

Difference, decreased by, subtracted from, less, minus

 −

Product, multiplied by, of, times, twice

Quotient, divided by, ratio, per

÷

Is, total, result

 =

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “let x represent…” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

Example 12

When 6 is subtracted from twice the sum of a number and 8 the result is 5. Find the number.

Solution:

Let n represent the unknown number.

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

twice the sum of a number and 8

2(n+8)

the sum of twice a number and 8

2n+8

The key was to focus on the phrase “twice the sum,” this prompted us to group the sum within parentheses and then multiply by 2. After translating the sentence into a mathematical statement we then solve.

2(n+8)6=52n+166=52n+10=52n=5n=52

Check.

2(n+8)6=2(52+8)6=2(112)6=116=5

Answer: The number is 52.

General guidelines for setting up and solving word problems follow.

  • Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
  • Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
  • Step 3: Translate and set up an algebraic equation that models the problem.
  • Step 4: Solve the resulting algebraic equation.
  • Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Example 13

A rectangle has a perimeter measuring 92 meters. The length is 2 meters less than 3 times the width. Find the dimensions of the rectangle.

Solution:

The sentence “The length is 2 meters less than 3 times the width,” gives us the relationship between the two variables.

Let w represent the width of the rectangle.

Let 3w2 represent the length.

The sentence “A rectangle has a perimeter measuring 92 meters” suggests an algebraic set up. Substitute 92 for the perimeter and the expression 3w2 for the length into the appropriate formula as follows:

P=2l+2w92=2(3w2)+2w

Once you have set up an algebraic equation with one variable, solve for the width, w.

92=2(3w2)+2wDistribute.92=6w4+2wCombineliketerms.92=8w4Solveforw.96=8w12=w

Use 3w2 to find the length.

l=3w2=3(12)2=362=34

To check, make sure the perimeter is 92 meters.

P=2l+2w=2(34)+2(12)=68+24=92

Answer: The rectangle measures 12 meters by 34 meters.

Example 14

Given a 438% annual interest rate, how long will it take $2,500 to yield $437.50 in simple interest?

Solution:

Let t represent the time needed to earn $437.50 at 438%. Organize the information needed to use the formula for simple interest, I=prt.

Given interest for the time period:

I=$437.50

Given principal:

p=$2,500

Given rate:

r=438%=4.375%=0.04375

Next, substitute all of the known quantities into the formula and then solve for the only unknown, t.

I=prt437.50=2500(0.04375)t437.50=109.375t437.50109.375=109.375t109.3754=t

Answer: It takes 4 years for $2,500 invested at 438% to earn $437.50 in simple interest.

Example 15

Susan invested her total savings of $12,500 in two accounts earning simple interest. Her mutual fund account earned 7% last year and her CD earned 4.5%. If her total interest for the year was $670, how much was in each account?

Solution:

The relationship between the two unknowns is that they total $12,500. When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

     Let x represent the amount invested in the mutual fund.

     Let 12,500 − x represent the remaining amount invested in the CD.

     Organize the data.

Interest earned in the mutual fund:

I=prt=x0.071=0.07x

Interest earned in the CD:

I=prt=(12,500x)0.0451=0.045(12,500x)

Total interest:

$670

The total interest is the sum of the interest earned from each account.

mutualfundinterest+CDinterest=totalinterest0.07x+0.045(12,500x)=670

This equation models the problem with one variable. Solve for x.

0.07x+0.045(12,500x)=6700.07x+562.50.045x=6700.025x+562.5=6700.025x=107.5x=107.50.025x=4,300

Use 12,500x to find the amount in the CD.

12,500x=12,5004,300=8,200

Answer: Susan invested $4,300 at 7% in a mutual fund and $8,200 at 4.5% in a CD.

Key Takeaways

  • Solving general linear equations involves isolating the variable, with coefficient 1, on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
  • If solving a linear equation leads to a true statement like 0 = 0, then the equation is an identity and the solution set consists of all real numbers, .
  • If solving a linear equation leads to a false statement like 0 = 5, then the equation is a contradiction and there is no solution, Ø.
  • Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
  • Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.

Topic Exercises

    Part A: Solving Basic Linear Equations

      Determine whether or not the given value is a solution.

    1. 5x+4=1;   x=1

    2. 4x3=7;   x=1

    3. 3y4=5;   y=93

    4. 2y+7=12;   y=52

    5. 3a6=18a;   a=3

    6. 5(2t1)=2t;   t=2

    7. axb=0;   x=ba

    8. ax+b=2b;   x=ba

      Solve.

    1. 5x3=27

    2. 6x7=47

    3. 4x+13=35

    4. 6x9=18

    5. 9a+10=10

    6. 53a=5

    7. 8t+5=15

    8. 9t+12=33

    9. 23x+12=1
    10. 38x+54=32
    11. 13y5=2
    12. 25y6=8
    13. 7y=22

    14. 6y=12

    15. Solve for x: axb=c

    16. Solve for x: ax+b=0

    Part B: Solving Linear Equations

      Solve.

    1. 6x5+2x=19

    2. 72x+9=24

    3. 12x29x=5x+8

    4. 163x22=84x

    5. 5y69y=32y+8

    6. 79y+12=3y+1111y

    7. 3+3a11=5a82a

    8. 23a=5a+78a

    9. 13x32+52x=56x+14

    10. 58+15x34=310x14

    11. 1.2x0.52.6x=22.4x

    12. 1.593.87x=3.484.1x0.51

    13. 510x=2x+812x

    14. 8x33x=5x3

    15. 5(y+2)=3(2y1)+10

    16. 7(y3)=4(2y+1)21

    17. 75(3t9)=22

    18. 105(3t+7)=20

    19. 52x=42(x4)

    20. 2(4x5)+7x=5(3x2)

    21. 4(4a1)=5(a3)+2(a2)

    22. 6(2b1)+24b=8(3b1)

    23. 23(x+18)+2=13x13

    24. 25x12(6x3)=43

    25. 1.2(2x+1)+0.6x=4x

    26. 6+0.5(7x5)=2.5x+0.3

    27. 5(y+3)=15(y+1)10y

    28. 3(4y)2(y+7)=5y

    29. 15(2a+3)12=13a+110

    30. 32a=34(1+2a)15(a+5)

    31. 63(7x+1)=7(43x)

    32. 6(x6)3(2x9)=9

    33. 34(y2)+23(2y+3)=3

    34. 5412(4y3)=25(y1)

    35. 2(3x+1)(x3)=7x+1

    36. 6(2x+1)(10x+9)=0

    37. Solve for w: P=2l+2w

    38. Solve for a: P=a+b+c

    39. Solve for t: D=rt

    40. Solve for w: V=lwh

    41. Solve for b: A=12bh

    42. Solve for a: s=12at2

    43. Solve for a: A=12h(a+b)

    44. Solve for h: V=13πr2h

    45. Solve for F: C=59(F32)

    46. Solve for x: ax+b=c

    Part C: Applications

      Set up an algebraic equation then solve.

      Number Problems

    1. When 3 is subtracted from the sum of a number and 10 the result is 2. Find the number.

    2. The sum of 3 times a number and 12 is equal to 3. Find the number.

    3. Three times the sum of a number and 6 is equal to 5 times the number. Find the number.

    4. Twice the sum of a number and 4 is equal to 3 times the sum of the number and 1. Find the number.

    5. A larger integer is 1 more than 3 times another integer. If the sum of the integers is 57, find the integers.

    6. A larger integer is 5 more than twice another integer. If the sum of the integers is 83, find the integers.

    7. One integer is 3 less than twice another integer. Find the integers if their sum is 135.

    8. One integer is 10 less than 4 times another integer. Find the integers if their sum is 100.

    9. The sum of three consecutive integers is 339. Find the integers.

    10. The sum of four consecutive integers is 130. Find the integers.

    11. The sum of three consecutive even integers is 174. Find the integers.

    12. The sum of four consecutive even integers is 116. Find the integers.

    13. The sum of three consecutive odd integers is 81. Find the integers.

    14. The sum of four consecutive odd integers is 176. Find the integers.

      Geometry Problems

    1. The length of a rectangle is 5 centimeters less than twice its width. If the perimeter is 134 centimeters, find the length and width.

    2. The length of a rectangle is 4 centimeters more than 3 times its width. If the perimeter is 64 centimeters, find the length and width.

    3. The width of a rectangle is one-half that of its length. If the perimeter measures 36 inches, find the dimensions of the rectangle.

    4. The width of a rectangle is 4 inches less than its length. If the perimeter measures 72 inches, find the dimensions of the rectangle.

    5. The perimeter of a square is 48 inches. Find the length of each side.

    6. The perimeter of an equilateral triangle is 96 inches. Find the length of each side.

    7. The circumference of a circle measures 80π units. Find the radius.

    8. The circumference of a circle measures 25 centimeters. Find the radius rounded off to the nearest hundredth.

      Simple Interest Problems

    1. For how many years must $1,000 be invested at 512% to earn $165 in simple interest?

    2. For how many years must $20,000 be invested at 614% to earn $3,125 in simple interest?

    3. At what annual interest rate must $6500 be invested for 2 years to yield $1,040 in simple interest?

    4. At what annual interest rate must $5,750 be invested for 1 year to yield $333.50 in simple interest?

    5. If the simple interest earned for 5 years was $1,860 and the annual interest rate was 6%, what was the principal?

    6. If the simple interest earned for 2 years was $543.75 and the annual interest rate was 334%, what was the principal?

    7. How many years will it take $600 to double earning simple interest at a 5% annual rate? (Hint: To double, the investment must earn $600 in simple interest.)

    8. How many years will it take $10,000 to double earning simple interest at a 5% annual rate? (Hint: To double, the investment must earn $10,000 in simple interest.)

    9. Jim invested $4,200 in two accounts. One account earns 3% simple interest and the other earns 6%. If the interest after 1 year was $159, how much did he invest in each account?

    10. Jane has her $6,500 savings invested in two accounts. She has part of it in a CD at 5% annual interest and the rest in a savings account that earns 4% annual interest. If the simple interest earned from both accounts is $303 for the year, then how much does she have in each account?

    11. Jose put last year’s bonus of $8,400 into two accounts. He invested part in a CD with 2.5% annual interest and the rest in a money market fund with 1.5% annual interest. His total interest for the year was $198. How much did he invest in each account?

    12. Mary invested her total savings of $3,300 in two accounts. Her mutual fund account earned 6.2% last year and her CD earned 2.4%. If her total interest for the year was $124.80, how much was in each account?

    13. Alice invests money into two accounts, one with 3% annual interest and another with 5% annual interest. She invests 3 times as much in the higher yielding account as she does in the lower yielding account. If her total interest for the year is $126, how much did she invest in each account?

    14. James invested an inheritance in two separate banks. One bank offered 512% annual interest rate and the other 614%. He invested twice as much in the higher yielding bank account than he did in the other. If his total simple interest for 1 year was $5,760, then what was the amount of his inheritance?

      Uniform Motion Problems

    1. If it takes Jim 114 hours to drive the 40 miles to work, then what is Jim’s average speed?

    2. It took Jill 312 hours to drive the 189 miles home from college. What was her average speed?

    3. At what speed should Jim drive if he wishes to travel 176 miles in 234 hours?

    4. James and Martin were able to drive the 1,140 miles from Los Angeles to Seattle. If the total trip took 19 hours, then what was their average speed?

    Part D: Discussion Board

    1. What is regarded as the main business of algebra? Explain.

    2. What is the origin of the word algebra?

    3. Create an identity or contradiction of your own and share it on the discussion board. Provide a solution and explain how you found it.

    4. Post something you found particularly useful or interesting in this section. Explain why.

    5. Conduct a web search for “solving linear equations.” Share a link to website or video tutorial that you think is helpful.

Answers

  1. No

  2. Yes

  3. No

  4. Yes

  5. 6

  6. 112

  7. 0

  8. 54

  9. 34

  10. −3

  11. −15

  12. x=b+ca

  1. 3

  2. −5

  3. 172

  4. 78

  5. 2.5

  6. Ø

  7. 3

  8. 2

  9. Ø

  10. 53

  11. −81

  12. 1.2

  13. 0

  14. Ø

  15. 65

  16. w=P2l2
  17. t=Dr
  18. b=2Ah
  19. a=2Ahb
  20. F=95C+32
  1. −5

  2. 9

  3. 14, 43

  4. 46, 89

  5. 112, 113, 114

  6. 56, 58, 60

  7. 25, 27, 29

  8. Width: 24 centimeters; length: 43 centimeters

  9. Width: 6 inches; length: 12 inches

  10. 12 inches

  11. 40 units

  12. 3 years

  13. 8%

  14. $6,200

  15. 20 years

  16. He invested $3,100 at 3% and $1,100 at 6%.

  17. Jose invested $7,200 in the CD and $1,200 in the money market fund.

  18. Alice invested $700 at 3% and $2,100 at 5%.

  19. 32 miles per hour

  20. 64 miles per hour

  1. Answer may vary

  2. Answer may vary

  3. Answer may vary