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Solve Equations Using the Subtraction and Addition Properties of Equality

Solve equations using the Subtraction and Addition Properties of Equality

Verify a Solution of an Equation

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same, so that we end up with a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

Example

Try it.

Determine whether $x=\frac{3}{2}$ is a solution of $4x-2=2x+1$.

Solution

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.



Multiply.
Subtract.

Since $x=\frac{3}{2}$ results in a true equation (4 is in fact equal to 4), $\frac{3}{2}$ is a solution to the equation $4x-2=2x+1$.

Solve Equations Using the Subtraction and Addition Properties of Equality

We are going to use a model to clarify the process of solving an equation. An envelope represents the variable, since its contents are unknown, and each counter represents one. We will set out one envelope and some counters on our workspace, Both sides of the workspace have the same number of counters, but some counters are “hidden” in the envelope. Can you tell how many counters are in the envelope?

What are you thinking? What steps are you taking in your mind to figure out how many counters are in the envelope?

Perhaps you are thinking: “I need to remove the 3 counters at the bottom left to get the envelope by itself. The 3 counters on the left can be matched with 3 on the right and so I can take them away from both sides. That leaves five on the right, so there must be 5 counters in the envelope.” See for an illustration of this process.

What algebraic equation would match this situation? In each side of the workspace represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope $x$.

Let’s write algebraically the steps we took to discover how many counters were in the envelope:


First, we took away three from each side.
Then we were left with five.

Check:

Five in the envelope plus three more does equal eight!

\[5+3=8\]

Example

Try it.

Solve: $y+37=-13.$

Solution

To get y by itself, we will undo the addition of 37 by using the Subtraction Property of Equality.


Subtract 37 from each side to ‘undo’ the addition.
Simplify.
Check:
Substitute $y=-50$

Since $y=-50$ makes $y+37=-13$ a true statement, we have the solution to this equation.

Example

Try it.

Solve: $a-28=-37.$

Solution


Add 28 to each side to ‘undo’ the subtraction.
Simplify.
Check:
Substitute $a=-9$

	The solution to $a-28=-37$ is $a=-9.$

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Solve Equations That Require Simplification

In the previous examples, we were able to isolate the variable with just one operation. Most of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality.

You should always simplify as much as possible before you try to isolate the variable. Remember that to simplify an expression means to do all the operations in the expression. Simplify one side of the equation at a time. Note that simplification is different from the process used to solve an equation in which we apply an operation to both sides.

How to Solve Equations That Require Simplification

Try it.

Solve: $9x-5-8x-6=7.$

Solution

Example

Try it.

Solve: $5(n-4)-4n=-8.$

Solution

We simplify both sides of the equation as much as possible before we try to isolate the variable.


Distribute on the left.
Use the Commutative Property to rearrange terms.
Combine like terms.
Each side is as simplified as possible. Next, isolate $n$.
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Substitute $n=12$.
	The solution to $5(n-4)-4n=-8$ is $n=12.$

Example

Try it.

Solve: $3(2y-1)-5y=2(y+1)-2(y+3).$

Solution

We simplify both sides of the equation before we isolate the variable.


Distribute on both sides.
Use the Commutative Property of Addition.
Combine like terms.
Each side is as simplified as possible. Next, isolate $y$.
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Let $y=-1$.
	The solution to $3(2y-1)-5y=2(y+1)-2(y+3)$ is $y=-1.$

Translate to an Equation and Solve

To solve applications algebraically, we will begin by translating from English sentences into equations. Our first step is to look for the word (or words) that would translate to the equals sign. shows us some of the words that are commonly used.

Equals =

is
is equal to
is the same as
the result is
gives
was
will be

The steps we use to translate a sentence into an equation are listed below.

Example

Try it.

Translate and solve: Eleven more than x is equal to 54.

Solution

Translate.
Subtract 11 from both sides.
Simplify.
Check: Is 54 eleven more than 43? $\begin{array}{lll}43+11 & \overset{?}{=} & 54 \\ 54 & = & 54✓\end{array}$

Example

Try it.

Translate and solve: The difference of $12t$ and $11t$ is $-14$.

Solution

Translate.
Simplify.
Check: $\begin{array}{lll}12(-14)-11(-14) & \overset{?}{=} & -14 \\ -168+154 & \overset{?}{=} & -14 \\ -14 & = & -14✓\end{array}$

Translate and Solve Applications

Most of the time a question that requires an algebraic solution comes out of a real life question. To begin with that question is asked in English (or the language of the person asking) and not in math symbols. Because of this, it is an important skill to be able to translate an everyday situation into algebraic language.

We will start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for. For example, you might use q for the number of quarters if you were solving a problem about coins.

How to Solve Translate and Solve Applications

Try it.

The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?

Solution

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

To Determine Whether a Number is a Solution to an Equation
1. Substitute the number in for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting statement is true.
  - If it is true, the number is a solution.
  - If it is not true, the number is not a solution.
Addition Property of Equality
- For any numbers a, b, and c, if $a=b$, then $a+c=b+c$.
Subtraction Property of Equality
- For any numbers a, b, and c, if $a=b$, then $a-c=b-c$.
To Translate a Sentence to an Equation
1. Locate the “equals” word(s). Translate to an equal sign (=).
2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
To Solve an Application
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Solve Equations Using the Subtraction and Addition Properties of Equality

Verify a Solution of an Equation

In the following exercises, determine whether the given value is a solution to the equation.

Try it.

Is $y=\frac{5}{3}$ a solution of
$6y+10=12y$?

Solution

yes

Try it.

Is $x=\frac{9}{4}$ a solution of
$4x+9=8x$?

Try it.

Is $u=-\frac{1}{2}$ a solution of
$8u-1=6u$?

Solution

Try it.

Is $v=-\frac{1}{3}$ a solution of
$9v-2=3v$?

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, solve each equation using the Subtraction and Addition Properties of Equality.

Try it.

$y+45=-66$

Solution

$y=-111$

Try it.

$y+39=-83$

Try it.

$b+\frac{1}{4}=\frac{3}{4}$

Solution

$b=\frac{1}{2}$

Try it.

$a+\frac{2}{5}=\frac{4}{5}$

Try it.

$p+2.4=-9.3$

Solution

$p=-11.7$

Try it.

$m+7.9=11.6$

Try it.

$m-18=-200$

Solution

$m=-182$

Try it.

$m-12=-12$

Try it.

$x-\frac{1}{3}=2$

Solution

$x=\frac{7}{3}$

Try it.

$x-\frac{1}{5}=4$

Try it.

$y-3.8=10$

Solution

$y=13.8$

Try it.

$x-165=-420$

Solution

$x=-255$

Try it.

$z-101=-314$

Try it.

$z+0.52=-8.5$

Solution

$z=-9.02$

Try it.

$x+0.93=-4.1$

Try it.

$q+\frac{3}{4}=\frac{1}{2}$

Solution

$q=-\frac{1}{4}$

Try it.

$p+\frac{1}{3}=\frac{5}{6}$

Try it.

$p-\frac{2}{5}=\frac{2}{3}$

Solution

$p=\frac{16}{15}$

Try it.

$y-\frac{3}{4}=\frac{3}{5}$

Solve Equations that Require Simplification

In the following exercises, solve each equation.

Try it.

$c+31-10=46$

Solution

$c=25$

Try it.

$m+16-28=5$

Try it.

$9x+5-8x+14=20$

Solution

$x=1$

Try it.

$6x+8-5x+16=32$

Try it.

$-6x-11+7x-5=-16$

Solution

$x=0$

Try it.

$-8n-17+9n-4=-41$

Try it.

$5(y-6)-4y=-6$

Solution

$y=24$

Try it.

$9(y-2)-8y=-16$

Try it.

$8(u+1.5)-7u=4.9$

Solution

$u=-7.1$

Try it.

$5(w+2.2)-4w=9.3$

Try it.

$6a-5(a-2)+9=-11$

Solution

$a=-30$

Try it.

$8c-7(c-3)+4=-16$

Try it.

$6(y-2)-5y=4(y+3)$
$-4(y-1)$

Solution

$y=28$

Try it.

$9(x-1)-8x=-3(x+5)$
$+3(x-5)$

Try it.

$3(5n-1)-14n+9$
$=10(n-4)-6n-4(n+1)$

Solution

$n=-50$

Try it.

$2(8m+3)-15m-4$
$=9(m+6)-2(m-1)-7m$

Try it.

$\text{-}(j+2)+2j-1=5$

Solution

$j=8$

Try it.

$\text{-}(k+7)+2k+8=7$

Try it.

$\text{-}(\frac{1}{4}a-\frac{3}{4})+\frac{5}{4}a=-2$

Solution

$a=-\frac{11}{4}$

Try it.

$\text{-}(\frac{2}{3}d-\frac{1}{3})+\frac{5}{3}d=-4$

Try it.

$8(4x+5)-5(6x)-x$
$=53-6(x+1)+3(2x+2)$

Solution

$x=13$

Try it.

$6(9y-1)-10(5y)-3y$
$=22-4(2y-12)+8(y-6)$

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve it.

Try it.

Nine more than $x$ is equal to 52.

Solution

$x+9=52;x=43$

Try it.

The sum of x and $-15$ is 23.

Try it.

Ten less than m is $-14$.

Solution

$m-10=-14;m=-4$

Try it.

Three less than y is $-19$.

Try it.

The sum of y and $-30$ is 40.

Solution

$y+(-30)=40;y=70$

Try it.

Twelve more than p is equal to 67.

Try it.

The difference of $9x\ \text{and}\ 8x$ is 107.

Solution

$9x-8x=107;107$

Try it.

The difference of $5c\ \text{and}\ 4c$ is 602.

Try it.

The difference of $n$ and $\frac{1}{6}$ is $\frac{1}{2}$.

Solution

$n-\frac{1}{6}=\frac{1}{2};\frac{2}{3}$

Try it.

The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12}$.

Try it.

The sum of $-4n$ and $5n$ is $-82$.

Solution

$-4n+5n=-82;-82$

Try it.

The sum of $-9m$ and $10m$ is $-95$.

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Try it.

Distance Avril rode her bike a total of 18 miles, from home to the library and then to the beach. The distance from Avril’s house to the library is 7 miles. What is the distance from the library to the beach?

Solution

11 miles

Try it.

Reading Jeff read a total of 54 pages in his History and Sociology textbooks. He read 41 pages in his History textbook. How many pages did he read in his Sociology textbook?

Try it.

Age Eva’s daughter is 15 years younger than her son. Eva’s son is 22 years old. How old is her daughter?

Solution

7 years old

Try it.

Age Pablo’s father is 3 years older than his mother. Pablo’s mother is 42 years old. How old is his father?

Try it.

Groceries For a family birthday dinner, Celeste bought a turkey that weighed 5 pounds less than the one she bought for Thanksgiving. The birthday turkey weighed 16 pounds. How much did the Thanksgiving turkey weigh?

Solution

21 pounds

Try it.

Weight Allie weighs 8 pounds less than her twin sister Lorrie. Allie weighs 124 pounds. How much does Lorrie weigh?

Try it.

Health Connor’s temperature was 0.7 degrees higher this morning than it had been last night. His temperature this morning was 101.2 degrees. What was his temperature last night?

Solution

100.5 degrees

Try it.

Health The nurse reported that Tricia’s daughter had gained 4.2 pounds since her last checkup and now weighs 31.6 pounds. How much did Tricia’s daughter weigh at her last checkup?

Try it.

Salary Ron’s paycheck this week was $17.43 less than his paycheck last week. His paycheck this week was $103.76. How much was Ron’s paycheck last week?

Solution

$121.19

Try it.

Textbooks Melissa’s math book cost $22.85 less than her art book cost. Her math book cost $93.75. How much did her art book cost?

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Solve Equations Using the Subtraction and Addition Properties of Equality

We began our work solving equations in previous chapters. It has been a while since we have seen an equation, so we will review some of the key concepts before we go any further.

We said that solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle.

In the earlier sections, we listed the steps to determine if a value is a solution. We restate them here.

Example

Try it.

Determine whether $y=\frac{3}{4}$ is a solution for $4y+3=8y.$

Solution



Multiply.
Add.

Since $y=\frac{3}{4}$ results in a true equation, $\frac{3}{4}$ is a solution to the equation $4y+3=8y.$

We introduced the Subtraction and Addition Properties of Equality in Solving Equations Using the Subtraction and Addition Properties of Equality. In that section, we modeled how these properties work and then applied them to solving equations with whole numbers. We used these properties again each time we introduced a new system of numbers. Let’s review those properties here.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. models the equation $x+3=8.$

The goal is to isolate the variable on one side of the equation. So we ‘took away’ $3$ from both sides of the equation and found the solution $x=5.$

Example

Try it.

Solve: $m+4=-5.$

Solution


Subtract 4 from each side to "undo" the addition.
Simplify.
Check:
Substitute $m=-9$.

		The solution to $m-4=-5$ is $m=-1$.

Condensed: the full section is in OpenStax Prealgebra 2e.

Solve Equations That Need to Be Simplified

In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable.

Example

Try it.

Solve: $3x-7-2x-4=1.$

Solution

The left side of the equation has an expression that we should simplify before trying to isolate the variable.


Rearrange the terms, using the Commutative Property of Addition.
Combine like terms.
Add 11 to both sides to isolate $x$.
Simplify.
Check. Substitute $x=12$ into the original equation.

The solution checks.

Example

Try it.

Solve: $3(n-4)-2n=-3.$

Solution

The left side of the equation has an expression that we should simplify.


Distribute on the left.
Use the Commutative Property to rearrange terms.
Combine like terms.
Isolate n using the Addition Property of Equality.
Simplify.
Check. Substitute $n=9$ into the original equation. The solution checks.

Example

Try it.

Solve: $2(3k-1)-5k=-2-7.$

Solution

Both sides of the equation have expressions that we should simplify before we isolate the variable.


Distribute on the left, subtract on the right.
Use the Commutative Property of Addition.
Combine like terms.
Undo subtraction by using the Addition Property of Equality.
Simplify.
Check. Let $k=-7.$ The solution checks.

Translate an Equation and Solve

In previous chapters, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. reminds us of some of the words that translate to the equal sign.

Equals (=)
is	is equal to	is the same as	the result is	gives	was	will be

Let’s review the steps we used to translate a sentence into an equation.

Now we are ready to try an example.

Example

Try it.

Translate and solve: five more than $x$ is equal to $26.$

Solution

Translate.
Subtract 5 from both sides.
Simplify.
Check: Is $26$ five more than $21$? The solution checks.

Example

Try it.

Translate and solve: The difference of $5p$ and $4p$ is $23.$

Solution

Translate.
Simplify.
Check:
The solution checks.

Translate and Solve Applications

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.

Example

Try it.

The Robles family has two dogs, Buster and Chandler. Together, they weigh $71$ pounds.

Chandler weighs $28$ pounds. How much does Buster weigh?

Solution

Read the problem carefully.
Identify what you are asked to find, and choose a variable to represent it.	How much does Buster weigh? Let $b=$ Buster's weight
Write a sentence that gives the information to find it.	Buster's weight plus Chandler's weight equals 71 pounds.
We will restate the problem, and then include the given information.	Buster's weight plus 28 equals 71.
Translate the sentence into an equation, using the variable $b$.
Solve the equation using good algebraic techniques.
Check the answer in the problem and make sure it makes sense.
Is 43 pounds a reasonable weight for a dog? Yes. Does Buster's weight plus Chandler's weight equal 71 pounds?
$43+28\overset{?}{=}71$
$71=71\ ✓$
Write a complete sentence that answers the question, "How much does Buster weigh?"	Buster weighs 43 pounds

Condensed: the full section is in OpenStax Prealgebra 2e.

Key Concepts

Determine whether a number is a solution to an equation.
1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true.
If it is true, the number is a solution.
If it is not true, the number is not a solution.
Subtraction and Addition Properties of Equality
- Subtraction Property of Equality
  For all real numbers a, b, and c,
  if a = b then $a-c=b-c$.
- Addition Property of Equality
  For all real numbers a, b, and c,
  if a = b then $a+c=b+c$.
Translate a word sentence to an algebraic equation.
1. Locate the “equals” word(s). Translate to an equal sign.
2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
Problem-solving strategy
1. Read the problem. Make sure you understand all the words and ideas.
2. Identify what you are looking for.
3. Name what you are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Solve Equations Using the Subtraction and Addition Properties of Equality

Solve Equations Using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given value is a solution to the equation.

Try it.

Is $y=\frac{1}{3}$ a solution of $4y+2=10y?$

Solution

yes

Try it.

Is $x=\frac{3}{4}$ a solution of $5x+3=9x?$

Try it.

Is $u=-\frac{1}{2}$ a solution of $8u-1=6u?$

Solution

Try it.

Is $v=-\frac{1}{3}$ a solution of $9v-2=3v?$

In the following exercises, solve each equation.

Try it.

$b+\frac{1}{4}=\frac{3}{4}$

Solution

$b=\frac{1}{2}$

Try it.

$a+\frac{2}{5}=\frac{4}{5}$

Try it.

$p+2.4=-9.3$

Solution

p = −11.7

Try it.

$m+7.9=11.6$

Try it.

$x-\frac{1}{3}=2$

Solution

$x=\frac{7}{3}$

Try it.

$x-\frac{1}{5}=4$

Try it.

$y-3.8=10$

Solution

y = 13.8

Try it.

$x-15=-42$

Solution

x = −27

Try it.

$z+5.2=-8.5$

Try it.

$q+\frac{3}{4}=\frac{1}{2}$

Solution

$q=-\frac{1}{4}$

Try it.

$p-\frac{2}{5}=\frac{2}{3}$

Try it.

$y-\frac{3}{4}=\frac{3}{5}$

Solution

$y=\frac{27}{20}$

Solve Equations that Need to be Simplified

In the following exercises, solve each equation.

Try it.

$c+3-10=18$

Try it.

$m+6-8=15$

Solution

m = 17

Try it.

$9x+5-8x+14=20$

Try it.

$6x+8-5x+16=32$

Solution

x = 8

Try it.

$-6x-11+7x-5=-16$

Try it.

$-8n-17+9n-4=-41$

Solution

n = −20

Try it.

$3(y-5)-2y=-7$

Try it.

$4(y-2)-3y=-6$

Solution

y = 2

Try it.

$8(u+1.5)-7u=4.9$

Try it.

$5(w+2.2)-4w=9.3$

Solution

w = −1.7

Try it.

$-5(y-2)+6y=-7+4$

Try it.

$-8(x-1)+9x=-3+9$

Solution

x = −2

Try it.

$3(5n-1)-14n+9=1-2$

Try it.

$2(8m+3)-15m-4=3-5$

Solution

m = −4

Try it.

$-(j+2)+2j-1=5$

Try it.

$-(k+7)+2k+8=7$

Solution

k = 6

Try it.

$6a-5(a-2)+9=-11$

Try it.

$8c-7(c-3)+4=-16$

Solution

c = −41

Try it.

$8(4x+5)-5(6x)-x=53$

Try it.

$6(9y-1)-10(5y)-3y=22$

Solution

y = 28

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

Try it.

Five more than $x$ is equal to $21.$

Try it.

The sum of $x$ and $-5$ is $33.$

Solution

x + (−5) = 33; x = 38

Try it.

Ten less than $m$ is $-14.$

Try it.

Three less than $y$ is $-19.$

Solution

y − 3 = −19; y = −16

Try it.

The sum of $y$ and $-3$ is $40.$

Try it.

Eight more than $p$ is equal to $52.$

Solution

p + 8 = 52; p = 44

Try it.

The difference of $9x$ and $8x$ is $17.$

Try it.

The difference of $5c$ and $4c$ is $60.$

Solution

5c − 4c = 60; c = 60

Try it.

The difference of $n$ and $\frac{1}{6}$ is $\frac{1}{2}.$

Try it.

The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12}.$

Solution

$f-\frac{1}{3}=\frac{1}{12};\ f=\frac{5}{12}$

Try it.

The sum of $-4n$ and $5n$ is $-32.$

Try it.

The sum of $-9m$ and $10m$ is $-25.$

Solution

−9m + 10m = −25; m = −25

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Try it.

Pilar drove from home to school and then to her aunt’s house, a total of $18$ miles. The distance from Pilar’s house to school is $7$ miles. What is the distance from school to her aunt’s house?

Try it.

Jeff read a total of $54$ pages in his English and Psychology textbooks. He read $41$ pages in his English textbook. How many pages did he read in his Psychology textbook?

Solution

Let p equal the number of pages read in the Psychology book. 41 + p = 54. Jeff read 13 pages in his Psychology book.

Try it.

Pablo’s father is $3$ years older than his mother. Pablo’s mother is $42$ years old. How old is his father?

Try it.

Eva’s daughter is $5$ years younger than her son. Eva’s son is $12$ years old. How old is her daughter?

Solution

Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old.

Try it.

Allie weighs $8$ pounds less than her twin sister Lorrie. Allie weighs $124$ pounds. How much does Lorrie weigh?

Try it.

For a family birthday dinner, Celeste bought a turkey that weighed $5$ pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed $16$ pounds. How much did the Thanksgiving turkey weigh?

Solution

21 pounds

Try it.

The nurse reported that Tricia’s daughter had gained $4.2$ pounds since her last checkup and now weighs $31.6$ pounds. How much did Tricia’s daughter weigh at her last checkup?

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Connor’s temperature was $0.7$ degrees higher this morning than it had been last night. His temperature this morning was $101.2$ degrees. What was his temperature last night?

Solution

100.5 degrees

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Melissa’s math book cost $\text{\$22.85}$ less than her art book cost. Her math book cost $\text{\$93.75}.$ How much did her art book cost?

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Ron’s paycheck this week was $\text{\$17.43}$ less than his paycheck last week. His paycheck this week was $\text{\$103.76}.$ How much was Ron’s paycheck last week?

Solution

$121.19

Condensed: the full section is in OpenStax Prealgebra 2e.

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Pratica (40)

Provare ciascuno su carta prima. Rivelare la risposta per controllare; quelli verificati possono essere aperti nel risolutore per ogni passo.

Evaluate $x+4$ when $x=-3$.

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$1$
Evaluate $15-y$ when $y=-5$.

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$20$
Simplify $4(4n+1)-15n$.

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$n+4$
Translate into algebra “5 less than $x$.”

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$x-5$
Determine whether $x=\frac{3}{2}$ is a solution of $4x-2=2x+1$.

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Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
Multiply.
Subtract.
Since $x=\frac{3}{2}$ results in a true equation (4 is in fact equal to 4), $\frac{3}{2}$ is a solution to the equation $4x-2=2x+1$.
Is $y=\frac{4}{3}$ a solution of $9y+2=6y+3$?

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no
Is $y=\frac{7}{5}$ a solution of $5y+3=10y-4$?

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yes
Solve: $y+37=-13.$

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To get y by itself, we will undo the addition of 37 by using the Subtraction Property of Equality.
Subtract 37 from each side to ‘undo’ the addition.
Simplify.
Check:
Substitute $y=-50$
Since $y=-50$ makes $y+37=-13$ a true statement, we have the solution to this equation.
Solve: $x+19=-27$.

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$x=-46$
Solve: $x+16=-34$.

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$x=-50$
Solve: $a-28=-37.$

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Add 28 to each side to ‘undo’ the subtraction.
Simplify.
Check:
Substitute $a=-9$
The solution to $a-28=-37$ is $a=-9.$
Solve: $n-61=-75.$

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$n=-14$
Solve: $p-41=-73.$

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$p=-32$

Solve: $x-\frac{5}{8}=\frac{3}{4}.$

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Use the Addition Property of Equality.
Find the LCD to add the fractions on the right.
Simplify.
Check:
Substitute $x=\frac{11}{8}.$
Subtract.
Simplify.
	The solution to $x-\frac{5}{8}=\frac{3}{4}$ is $x=\frac{11}{8}.$

Solve: $p-\frac{2}{3}=\frac{5}{6}.$

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$p=\frac{9}{6}=\frac{3}{2}$
Solve: $q-\frac{1}{2}=\frac{5}{6}.$

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$q=\frac{4}{3}$
Solve: $n-0.63=-4.2.$

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Use the Addition Property of Equality.
Add.
Check:
Let $n=-3.57$.
Solve: $b-0.47=-2.1.$

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$b=-1.63$
Solve: $c-0.93=-4.6.$

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$c=-3.67$
Solve: $9x-5-8x-6=7.$
Solve: $8y-4-7y-7=4.$

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$y=15$
Solve: $6z+5-5z-4=3.$

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$z=2$

Solve: $5(n-4)-4n=-8.$

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We simplify both sides of the equation as much as possible before we try to isolate the variable.


Distribute on the left.
Use the Commutative Property to rearrange terms.
Combine like terms.
Each side is as simplified as possible. Next, isolate $n$.
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Substitute $n=12$.
	The solution to $5(n-4)-4n=-8$ is $n=12.$

Solve: $5(p-3)-4p=-10.$

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$p=5$
Solve: $4(q+2)-3q=-8.$

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$q=-16$

Solve: $3(2y-1)-5y=2(y+1)-2(y+3).$

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We simplify both sides of the equation before we isolate the variable.


Distribute on both sides.
Use the Commutative Property of Addition.
Combine like terms.
Each side is as simplified as possible. Next, isolate $y$.
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Let $y=-1$.
	The solution to $3(2y-1)-5y=2(y+1)-2(y+3)$ is $y=-1.$

Solve: $4(2h-3)-7h=6(h-2)-6(h-1).$

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$h=6$
Solve: $2(5x+2)-9x=3(x-2)-3(x-4).$

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$x=2$
Translate and solve: Eleven more than x is equal to 54.

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Translate.
Subtract 11 from both sides.
Simplify.
Check: Is 54 eleven more than 43?
$\begin{array}{lll}43+11 & \overset{?}{=} & 54 \\ 54 & = & 54✓\end{array}$
Translate and solve: Ten more than x is equal to 41.

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$x+10=41;x=31$
Translate and solve: Twelve less than x is equal to 51.

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$x-12=51;x=63$
Translate and solve: The difference of $12t$ and $11t$ is $-14$.

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Translate.
Simplify.
Check:
$\begin{array}{lll}12(-14)-11(-14) & \overset{?}{=} & -14 \\ -168+154 & \overset{?}{=} & -14 \\ -14 & = & -14✓\end{array}$
Translate and solve: The difference of $4x$ and $3x$ is 14.

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$4x-3x=14;x=14$
Translate and solve: The difference of $7a$ and $6a$ is $-8$.

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$7a-6a=-8;a=-8$
The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?
Translate into an algebraic equation and solve:
The Pappas family has two cats, Zeus and Athena. Together, they weigh 23 pounds. Zeus weighs 16 pounds. How much does Athena weigh?

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7 pounds
Translate into an algebraic equation and solve:
Sam and Henry are roommates. Together, they have 68 books. Sam has 26 books. How many books does Henry have?

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42 books

Randell paid $28,675 for his new car. This was $875 less than the sticker price. What was the sticker price of the car?

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Step 1. Read the problem.
Step 2. Identify what we are looking for.	"What was the sticker price of the car?"
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $s=$ the sticker price of the car.
Step 4. Translate into an equation. Restate the problem in one sentence.	$28,675 is $875 less than the sticker price
Step 5. Solve the equation.	$\text{\$28,675 is \$875 less than}\ s$ $\begin{array}{lll}28,675 & = & s-875 \\ 28,675+875 & = & s-875+875 \\ 29,550 & = & s\end{array}$
Step 6. Check the answer. Is $875 less than $29,550 equal to $28,675? $\begin{array}{lllll}\begin{array}{lll}29,550-875 & \overset{?}{=} & 28,675 \\ 28,675 & = & 28,675✓\end{array}\end{array}$
Step 7. Answer the question with a complete sentence.	The sticker price of the car was $29,550.

Translate into an algebraic equation and solve:
Eddie paid $19,875 for his new car. This was $1,025 less than the sticker price. What was the sticker price of the car?

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$20,900
Translate into an algebraic equation and solve:
The admission price for the movies during the day is $7.75. This is $3.25 less the price at night. How much does the movie cost at night?

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$11.00

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Simboli utilizzati qui

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Come: Solve Equations Using the Subtraction and Addition Properties of Equality

Solve equations using the Subtraction and Addition Properties of Equality
Solve equations that need to be simplified
Translate an equation and solve
Translate and solve applications
Substitute the number for the variable in the equation.
Simplify the expressions on both sides of the equation.
Determine whether the resulting equation is true.
If it is true, the number is a solution.

Domande che la gente chiede

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Parti di questa pagina sono adattate da OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Condensato e ri-spiegato qui; gli errori sono nostri. Poiché la sua fonte è autorizzata per uso non commerciale, questa lezione è gratuita, non vende nulla e viene condivisa con la stessa licenza.

Altro in Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Translate.
Simplify.
Check: \(\begin{array}{lll}12(-14)-11(-14) & \overset{?}{=} & -14 \\ -168+154 & \overset{?}{=} & -14 \\ -14 & = & -14✓\end{array}\)


Rearrange the terms, using the Commutative Property of Addition.
Combine like terms.
Add 11 to both sides to isolate \(x\).
Simplify.
Check. Substitute \(x=12\) into the original equation.

Read the problem carefully.
Identify what you are asked to find, and choose a variable to represent it.	How much does Buster weigh? Let \(b=\) Buster's weight
Write a sentence that gives the information to find it.	Buster's weight plus Chandler's weight equals 71 pounds.
We will restate the problem, and then include the given information.	Buster's weight plus 28 equals 71.
Translate the sentence into an equation, using the variable \(b\).
Solve the equation using good algebraic techniques.
Check the answer in the problem and make sure it makes sense.
Is 43 pounds a reasonable weight for a dog? Yes. Does Buster's weight plus Chandler's weight equal 71 pounds?
\(43+28\overset{?}{=}71\)
\(71=71\ ✓\)
Write a complete sentence that answers the question, "How much does Buster weigh?"	Buster weighs 43 pounds