Solving and Writing Linear Equations
To solve a linear equation, use the properties of equality and properties of real numbers to find the value of the variable that satisfies the equation.
Example 1
Algebra Solve each equation.
-
5
x
−
3
=
2
5 bold italic x minus 3 equals 2
5
x
−
3
=
2
5
x
=
5
Add
3
to each side
.
x
=
1
Divide each side by
5
.
table with 3 rows and 4 columns , row1 column 1 , 5 x minus 3 , column 2 equals , column 3 2 , column 4 , row2 column 1 , 5 x , column 2 equals , column 3 5 , column 4 cap add , 3 . toeachside . . , row3 column 1 , x , column 2 equals , column 3 1 , column 4 cap divideeachsideby . 5 . , end table
-
1
−
2
(
x
+
1
)
=
x
1 minus 2 open bold italic x plus 1 close equals bold italic x
1
−
2
(
x
+
1
)
=
x
1
−
2
x
−
2
=
x
Use the Distributive Property
.
−
1
−
2
x
=
x
Simplify the left side
.
−
1
=
3
x
Add
2
x
to each side
.
−
1
3
=
x
Divide each side by
3
.
table with 5 rows and 4 columns , row1 column 1 , 1 minus 2 open x plus 1 close , column 2 equals , column 3 x , column 4 , row2 column 1 , 1 minus 2 x minus 2 , column 2 equals , column 3 x , column 4 cap usethecap distributivecap property . . , row3 column 1 , negative 1 minus 2 x , column 2 equals , column 3 x , column 4 cap simplifytheleftside . . , row4 column 1 , negative 1 , column 2 equals , column 3 3 x , column 4 cap add , 2 x . toeachside . . , row5 column 1 , negative , 1 third , column 2 equals , column 3 x , column 4 cap divideeachsideby . 3 . , end table
You will sometimes need to translate word problems into equations. Look for words that suggest a relationship or some type of mathematical operation.
Example 2
Algebra A student has grades of 80, 65, 78, and 92 on four tests. What is the minimum grade she must earn on her next test to ensure an average of 80?
Relate |
average of 80, 65, 78, 92, and next test is 80 |
Pull out the key words and numbers. |
Define |
Let
x
=
the grade on the next test
.
x equals . thegradeonthenexttest . .
|
Let a variable represent what you are looking for. |
Write |
80
+
65
+
78
+
92
+
x
5
=
80
Write an equation.
315
+
x
5
=
80
Combine like terms.
315
+
x
=
400
Multiply each side by
5
.
x
=
85
Subtract
315
from each side.
table with 4 rows and 3 columns , row1 column 1 , fraction 80 plus 65 plus 78 plus 92 plus x , over 5 end fraction , column 2 equals 80 , column 3 cap writeanequation. , row2 column 1 , fraction 315 plus x , over 5 end fraction , column 2 equals 80 , column 3 cap combineliketerms. , row3 column 1 , 315 plus x , column 2 equals 400 , column 3 cap multiplyeachsideby . 5 . , row4 column 1 , x , column 2 equals 85 , column 3 cap subtract . 315 . fromeachside. , end table
|
The student must earn 85 on the next test for an average of 80.
Exercises
Algebra Solve each equation.
-
3
n
+
2
=
17
3 n plus . 2 equals 17
-
5
a
−
2
=
−
12
5 eh minus . 2 equals , negative 12
-
2
x
+
4
=
10
2 x plus . 4 equals 10
-
3
(
n
−
4
)
=
15
3 open n minus 4 . close , equals 15
-
4
+
2
y
=
8
y
4 plus . 2 y equals 8 y
-
−
6
z
+
1
=
13
−
3
z
negative 6 z plus . 1 equals . 13 minus 3 z
-
6
−
(
3
t
+
4
)
=
t
6 minus . open 3 t plus 4 . close , equals t
-
7
=
−
2
(
4
n
−
4.5
)
7 equals , minus . 2 open 4 n minus 4.5 close
-
(
w
+
5
)
−
5
=
(
2
w
+
5
)
open w plus 5 , close minus , 5 equals . open 2 w plus 5 close
-
5
7
p
−
10
=
30
5 sevenths , p minus . 10 equals 30
-
m
−
3
−
3
=
1
m over negative 3 , minus 3 equals 1
-
5
k
+
2
(
k
+
1
)
=
23
5 k plus . 2 open k plus 1 . close , equals 23
- Twice a number subtracted from 35 is 9. What is the number?
- The Johnsons pay
$
9.95
dollars , 9.95 a month plus
$
.035
dollars , .035 per min for local phone service. Last month, they paid
$
12.75
.
dollars , 12.75 , . How many minutes of local calls did they make?