Answer:
x=3; HJ=21; JK=12
Step-by-step explanation:
One way of the segment addition postulate is that if you have two connected segments on one line, the length of the combined segment is the length of the two smaller segments added together.
[tex]\text{combined segment}=\text{partial segment}+\text{other partial segment}[/tex]
So, the length of HK = the length of HJ + length of JK
In geometry, the length of a line segment is often simply written as the two letters of their endpoints, so the above equation would be more compactly written as: [tex]HK=HJ+JK[/tex]
Now, we're given extra information in the problem stating lengths of each of those segments, but some are in terms of some unknown "x" (instead of just a number). Unfortunate, but it's okay.
One last piece of trickery is that they give [tex]KH=33[/tex], but we have [tex]HK[/tex] in our formula. Looking at the picture, it's the same length to go from K to H, as it is to go from H to K, so [tex]HK=33[/tex] also.
We'll substitute each of the expressions into the equation above, and solve for x, then we'll solve for the lengths of each of the segments.
[tex]HK=HJ+JK[/tex]
[tex]33=(4x+9)+(3x+3)[/tex]
Since each term is separated by addition (no subtraction), we can omit the parentheses from substitution, and since addition is commutative, we can add in any order we want to combine our "like terms" ("like terms" are quantities that can be simplified when added together, for example numbers can be added together and simplified like 1+1 or 3+5... or 9+3. Another example of like terms is terms that have an "x" in them, in our case, the 4x and the 3x).
So, rearranging, we get the following:
[tex]33=4x+3x+9+3[/tex]
... 4x+3x is 7x, and 9+3 is 12, so ...
[tex]33=7x+12[/tex]
To solve for x, observe that x only shows up in the equation once, so all that we need to do is undo things. It is easiest to undo things in the reverse order of the standard order of operations of arithmetic.
Normally, in arithmetic, we multiply before we add. When undoing things in algebra, we undo the addition before we try to undo the multiplication.
To undo addition, we subtract 12 from both sides:
[tex]33=7x+12\\33-12=7x+12-12\\21=7x[/tex]
To undo the 7 multiplied to the x, we need to divide both sides by 7:
[tex]21=7x\\\frac{21}{7}=\frac{7x}{7}\\3=x[/tex]
Knowing the value of x, we can substitute back into the expressions for HJ, and JK, and find the other two requested values:
[tex]HJ=4x+9\\HJ=4(3)+9\\[/tex]
Now that we're just doing arithmetic (simplifying numbers with addition, subtraction, multiplication, etc, on one side of the equation) instead of doing algebra (ex. adding something to both sides of the equation), we proceed in the standard order of operations of arithmetic. Multiplication comes before addition, so:
[tex]HJ=4(3)+9\\HJ=12+9\\HJ=21[/tex]
Similarly, for JK:
[tex]JK=3x+3\\JK=3(3)+3\\JK=9+3\\JK=12[/tex]
Note, as a double-check into the original "segment addition" formula we built:
[tex]HJ=HJ+JK[/tex]
[tex](33)[/tex] ?=? [tex](21)+(12)[/tex] (Are both sides really equal? We don't know yet)
[tex]33=33[/tex] (since 21+12 is 33, the right side IS 33. Yay! They're equal!)
