Domain: all real numbers
Range: y>-3
Answer:
Option A
Step-by-step explanation:
Given:
- a. 3x-5= 3x + 5
- b. 3x-5= 3x - 5
- c. 3x - 5 = 2x+5
- d. 3x-5 = 2x + 10
To find:
- Which one of the linear equations have no solution.
Solution:
a) 3x-5= 3x + 5
Add 5 to both sides
3x-5= 3x + 5
3x - 5 + 5 = 3x + 5 + 5
Simplify
(Add the numbers)
3x - 5 + 5 = 3x + 5 + 5
3x = 3x + 5 + 5
(Add the numbers)
3x = 3x + 5 + 5
3x = 3x + 10
Subtract 3x from both sides
3x = 3x + 10
3x - 3x = 3x + 10 - 3
Simplify
(Combine like terms)
3x -3x = 3x + 10 - 3
0 = 3x + 10 - 3
(Combine like terms)
0 = 3x + 10 - 3
0 = 10
The input is a contradiction: it has no solutions
b) 3x-5= 3x - 5
Since both sides equal, there are infinitely many solutions.
c) 3x - 5 = 2x+5
Add 5 to both sides
3x = 2x + 5 + 5
Simplify 2x + 5 + 5 to 2x + 10
3x = 2x + 10
Subtract 2x from both sides
3x - 2x = 10
Simplify 3x - 2x to x.
x = 10
d) 3x-5 = 2x + 10
Add 5 to both sides
3x = 2x + 10 + 5
Simplify 2x + 10 + 5 to 2x + 15
3x = 2x + 15
Subtract 2x from both sides
3x - 2x = 15
Simplify 3x -2x to x.
x = 15
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Answer:
As you can see all c and d both have solutions, eliminating them as options. Option B has infinite solutions leaving Option A which has no solutions.
Therefore, <u><em>Option A</em></u> is the linear equation that has no solution.
Answer:
B shows a translation of three units to the right
