Answer:
5x + 2x.....combine like terms..... = 7x
5x + 2x....subbing in 1 7x - 1....subbing in 1
5(1) + 2(1) = 5 + 2 = 7 7(1) - 1 = 7 - 1 = 6
5x + 2x...subbing in 2 7x - 1...subbing in 2
5(2) + 2(2) = 10 + 4 = 14 7(2) - 1 = 14 - 1 = 13
5x + 2x...subbing in 3 7x - 1...subbing in 3
5(3) + 2(3) = 15 + 6 = 21 7(3) - 1 = 21 - 1 = 20
5x + 2x...subbing in 4 7x - 1....subbing in 4
5(4) + 2(4) = 20 + 8 = 28 7(4) - 1 = 28 - 1 = 27
5x + 2x...subbing in 5 7x - 1...subbing in 5
5(5) + 2(5) = 25 + 10 = 35 7(5) - 1 = 35 - 1 = 34
5x + 2x result values are 1 more then 7x - 1 result values
there are no values that will make the 2 expressions equal....
because 5x + 2x = 7x......and the other one is 7x - 1......so the 7x - 1 values will always be 1 number less...because ur subtracting one
Step-by-step explanation:
Answer:
-3 < x
Step-by-step explanation:
3(x + 2) > X.
Distribute
3x+6 >x
Subtract 3x from each side
3x-3x +6 >x-3x
6> -2x
Divide each side by -2, remembering to flip the inequality
6/-2 < -2x/-2
-3 < x
9514 1404 393
Answer:
- domain: x ∉ {-4, 3}
- range: y ∉ {1}
- horizontal asymptote: y=1
- vertical asymptote: x=3
Step-by-step explanation:
The expression reduces to ...

The domain is limited to values of x where the expression is defined. It is undefined where the denominator is zero, at x=-4 and x=3. The graph of the expression has a "hole" at x=4, where the numerator and denominator factors cancel.
- the domain is all real numbers except -4 and +3
The function approaches the value of 1 as x gets large in magnitude, but it cannot take on the value of 1.
- the range is all real numbers except 1
As discussed in 'range', there is a horizontal asymptote at y=1. That is the value you would get if you were to determine the quotient of the division:*
(x+5)/(x-3) = 1 + (8/(x-3)) . . . . quotient is 1
There is a vertical asymptote at the place where the denominator is zero in the simplified expression: x = 3.
- vertical asymptote at x=3; horizontal asymptote at y=1
_____
* For some rational functions, the numerator has a higher degree than the denominator. In those cases, the quotient may be some function of x. The "end behavior" of the expression will match that function. (Sometimes it is a "slant asymptote", sometimes a higher-degree polynomial.)