Linear Equations III
Joseph Lee
Metropolitan Community College
Joseph Lee
Linear Equations III
Example 1.a
Solve.
3x − 4 = 7x − 12
Joseph Lee
Linear Equations III
Example 1.a
Solve.
3x − 4 = 7x − 12
Solution.
3x − 4 = 7x − 12
Joseph Lee
Linear Equations III
Example 1.a
Solve.
3x − 4 = 7x − 12
Solution.
3x − 4 = 7x − 12
−4 = 4x − 12
Joseph Lee
Linear Equations III
Example 1.a
Solve.
3x − 4 = 7x − 12
Solution.
3x − 4 = 7x − 12
−4 = 4x − 12
8 = 4x
Joseph Lee
Linear Equations III
Example 1.a
Solve.
3x − 4 = 7x − 12
Solution.
3x − 4 = 7x − 12
−4 = 4x − 12
8 = 4x
2= x
Joseph Lee
Linear Equations III
Example 1.b
Solve.
7x − 4 = −3x + 1
Joseph Lee
Linear Equations III
Example 1.b
Solve.
7x − 4 = −3x + 1
Solution.
7x − 4 = −3x + 1
Joseph Lee
Linear Equations III
Example 1.b
Solve.
7x − 4 = −3x + 1
Solution.
7x − 4 = −3x + 1
10x − 4 = 1
Joseph Lee
Linear Equations III
Example 1.b
Solve.
7x − 4 = −3x + 1
Solution.
7x − 4 = −3x + 1
10x − 4 = 1
10x = 5
Joseph Lee
Linear Equations III
Example 1.b
Solve.
7x − 4 = −3x + 1
Solution.
7x − 4 = −3x + 1
10x − 4 = 1
10x = 5
x=
Joseph Lee
1
2
Linear Equations III
Example 1.c
Solve.
2(4 − 5x) = −4x
Joseph Lee
Linear Equations III
Example 1.c
Solve.
2(4 − 5x) = −4x
Solution.
2(4 − 5x) = −4x
Joseph Lee
Linear Equations III
Example 1.c
Solve.
2(4 − 5x) = −4x
Solution.
2(4 − 5x) = −4x
8 − 10x = −4x
Joseph Lee
Linear Equations III
Example 1.c
Solve.
2(4 − 5x) = −4x
Solution.
2(4 − 5x) = −4x
8 − 10x = −4x
8 = 6x
Joseph Lee
Linear Equations III
Example 1.c
Solve.
2(4 − 5x) = −4x
Solution.
2(4 − 5x) = −4x
8 − 10x = −4x
8 = 6x
4
= x
3
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
6x − 8 = 4x − 2x − 1
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
6x − 8 = 4x − 2x − 1
6x − 8 = 2x − 1
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
6x − 8 = 4x − 2x − 1
6x − 8 = 2x − 1
4x − 8 = −1
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
6x − 8 = 4x − 2x − 1
6x − 8 = 2x − 1
4x − 8 = −1
4x = 7
Joseph Lee
Linear Equations III
Example 1.d
Solve.
2(3x − 4) = 4x − (2x + 1)
Solution.
2(3x − 4) = 4x − (2x + 1)
6x − 8 = 4x − 2x − 1
6x − 8 = 2x − 1
4x − 8 = −1
4x = 7
x=
Joseph Lee
7
4
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
8 − 4x + 12 = 6x − 15 − 5x
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
8 − 4x + 12 = 6x − 15 − 5x
−4x + 20 = x − 15
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
8 − 4x + 12 = 6x − 15 − 5x
−4x + 20 = x − 15
20 = 5x − 15
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
8 − 4x + 12 = 6x − 15 − 5x
−4x + 20 = x − 15
20 = 5x − 15
35 = 5x
Joseph Lee
Linear Equations III
Example 1.e
Solve.
8 − 4(x − 3) = 3(2x − 5) − 5x
Solution.
8 − 4(x − 3) = 3(2x − 5) − 5x
8 − 4x + 12 = 6x − 15 − 5x
−4x + 20 = x − 15
20 = 5x − 15
35 = 5x
7= x
Joseph Lee
Linear Equations III
Example 2.a
Solve.
1
(x − 3) = 3x
4
Joseph Lee
Linear Equations III
Example 2.a
Solve.
Solution.
1
(x − 3) = 3x
4
1
(x − 3) = 3x
4
Joseph Lee
Linear Equations III
Example 2.a
Solve.
Solution.
1
(x − 3) = 3x
4
1
(x − 3) = 3x
4
1
4 (x − 3) = 4(3x)
4
Joseph Lee
Linear Equations III
Example 2.a
Solve.
Solution.
1
(x − 3) = 3x
4
1
(x − 3) = 3x
4
1
4 (x − 3) = 4(3x)
4
x − 3 = 12x
Joseph Lee
Linear Equations III
Example 2.a
Solve.
Solution.
1
(x − 3) = 3x
4
1
(x − 3) = 3x
4
1
4 (x − 3) = 4(3x)
4
x − 3 = 12x
−3 = 11x
Joseph Lee
Linear Equations III
Example 2.a
Solve.
Solution.
1
(x − 3) = 3x
4
1
(x − 3) = 3x
4
1
4 (x − 3) = 4(3x)
4
x − 3 = 12x
−3 = 11x
−
Joseph Lee
3
= x
11
Linear Equations III
Example 2.b
Solve.
1
1
3
x − = x −2
4
2
3
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
1
1
3
x−
= 12
x −2
12
4
2
3
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
1
1
3
x−
= 12
x −2
12
4
2
3
9x − 6 = 4x − 24
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
1
1
3
x−
= 12
x −2
12
4
2
3
9x − 6 = 4x − 24
5x − 6 = −24
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
1
1
3
x−
= 12
x −2
12
4
2
3
9x − 6 = 4x − 24
5x − 6 = −24
5x = −18
Joseph Lee
Linear Equations III
Example 2.b
Solve.
Solution.
1
1
3
x − = x −2
4
2
3
3
1
1
x − = x −2
4
2
3
1
1
3
x−
= 12
x −2
12
4
2
3
9x − 6 = 4x − 24
5x − 6 = −24
5x = −18
x= −
Joseph Lee
18
5
Linear Equations III
Example 2.c
Solve.
x −7
2x + 1
=
3
5
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
2x + 1
x −7
= 15
15
3
5
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
2x + 1
x −7
= 15
15
3
5
5(x − 7) = 3(2x + 1)
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
2x + 1
x −7
= 15
15
3
5
5(x − 7) = 3(2x + 1)
5x − 35 = 6x + 3
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
2x + 1
x −7
= 15
15
3
5
5(x − 7) = 3(2x + 1)
5x − 35 = 6x + 3
−35 = x + 3
Joseph Lee
Linear Equations III
Example 2.c
Solve.
Solution.
x −7
2x + 1
=
3
5
x −7
2x + 1
=
3
5
2x + 1
x −7
= 15
15
3
5
5(x − 7) = 3(2x + 1)
5x − 35 = 6x + 3
−35 = x + 3
−38 = x
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
9x − 3 + 18 = 4x + 8
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
9x − 3 + 18 = 4x + 8
9x + 15 = 4x + 8
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
9x − 3 + 18 = 4x + 8
9x + 15 = 4x + 8
5x + 15 = 8
Joseph Lee
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
9x − 3 + 18 =
9x + 15 =
5x + 15 =
5x =
Joseph Lee
4x + 8
4x + 8
8
−7
Linear Equations III
Example 2.d
Solve.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
Solution.
1
1
(3x − 1) + 3 = (2x + 4)
2
3
1
1
6 (3x − 1) + 3 = 6 (2x + 4)
2
3
3(3x − 1) + 18 = 2(2x + 4)
9x − 3 + 18 =
9x + 15 =
5x + 15 =
5x =
x=
Joseph Lee
4x + 8
4x + 8
8
−7
− 57
Linear Equations III
Example 2.e
Solve.
2
1
5
(x − 3) = (x + 1) −
6
3
2
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
5(x − 3) = 4(x + 1) − 3
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
5(x − 3) = 4(x + 1) − 3
5x − 15 = 4x + 4 − 3
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
5(x − 3) = 4(x + 1) − 3
5x − 15 = 4x + 4 − 3
5x − 15 = 4x + 1
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
5(x − 3) = 4(x + 1) − 3
5x − 15 = 4x + 4 − 3
5x − 15 = 4x + 1
x − 15 = 1
Joseph Lee
Linear Equations III
Example 2.e
Solve.
Solution.
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
(x − 3) = (x + 1) −
6
3
2
2
1
5
6 (x − 3) = 6 (x + 1) −
6
3
2
5(x − 3) = 4(x + 1) − 3
5x − 15 = 4x + 4 − 3
5x − 15 = 4x + 1
x − 15 = 1
x = 16
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Solution.
8x − (3 + 4x) = 2(2x − 2)
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Solution.
8x − (3 + 4x) = 2(2x − 2)
8x − 3 − 4x = 4x − 4
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Solution.
8x − (3 + 4x) = 2(2x − 2)
8x − 3 − 4x = 4x − 4
4x − 3 = 4x − 4
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Solution.
8x − (3 + 4x) = 2(2x − 2)
8x − 3 − 4x = 4x − 4
4x − 3 = 4x − 4
−3 = −4
Joseph Lee
Linear Equations III
Example 3.a
Solve.
8x − (3 + 4x) = 2(2x − 2)
Solution.
8x − (3 + 4x) = 2(2x − 2)
8x − 3 − 4x = 4x − 4
4x − 3 = 4x − 4
−3 = −4
The equation represents a contradiction, as −3 6= −4. Therefore,
there is no solution to this equation. That is to say, the left side
of the equation will never equal the right side of the equation for
any value of x.
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Solution.
−5 + 3x + 9 = 4x − (x − 4)
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Solution.
−5 + 3x + 9 = 4x − (x − 4)
3x + 4 = 4x − x + 4
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Solution.
−5 + 3x + 9 = 4x − (x − 4)
3x + 4 = 4x − x + 4
3x + 4 = 3x + 4
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Solution.
−5 + 3x + 9 = 4x − (x − 4)
3x + 4 = 4x − x + 4
3x + 4 = 3x + 4
4= 4
Joseph Lee
Linear Equations III
Example 3.b
Solve.
−5 + 3x + 9 = 4x − (x − 4)
Solution.
−5 + 3x + 9 = 4x − (x − 4)
3x + 4 = 4x − x + 4
3x + 4 = 3x + 4
4= 4
The equation represents a tautology, as 4 = 4. Therefore, the
solution to this equation is all real numbers. That is to say, the
left side of the equation will equal the right side of the equation for
any value of x.
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Solution.
9x − 6 − 2x = 2(7x − 3)
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Solution.
9x − 6 − 2x = 2(7x − 3)
7x − 6 = 14x − 6
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Solution.
9x − 6 − 2x = 2(7x − 3)
7x − 6 = 14x − 6
−6 = 7x − 6
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Solution.
9x − 6 − 2x = 2(7x − 3)
7x − 6 = 14x − 6
−6 = 7x − 6
0 = 7x
Joseph Lee
Linear Equations III
Example 3.c
Solve.
9x − 6 − 2x = 2(7x − 3)
Solution.
9x − 6 − 2x = 2(7x − 3)
7x − 6 = 14x − 6
−6 = 7x − 6
0 = 7x
0= x
Joseph Lee
Linear Equations III