Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8
The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.
After doing this, we realize that only -1 works, so the answer is x=-1
Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)
Best wishes:)
Answer:
see below
Step-by-step explanation:
(x) = 7 - 2x
Let t(x) =0
0 = 7-2x
Subtract 7 from each side
-7 = -2x
Divide by -2
-7/-2 =-2x/-2
7/2 =x
h(x) = 4x + 2
Replace x with x+3
h(x+3) = 4(x+3) + 2
Distribute
= 4x+12 +2
=4x+14
The answer is x=3
hope this helps !
Answer:
Another way to name <UST is
<TSU