Jamal factored it completely.
in Jennifer's answer, 3x+3 has a common factor of 3 which can be factored to become 3(x+1) which is what Jamal did.
Alright.
For 7, you'll want to put congruent sides equal to each other, assuming they are parallelograms. So, you'll get the two equations:
3x+2=23
2y-7=9
Solve using GEMDAS/PEMDAS, and you'll get these answers.
3x+2=23
3x=21
x=7
2y-7=9
2y=2
y=1
For 8, you'll want to do the exact same thing, formatting the numbers to equal each other. You'll get these two equations:
3y+5=14
2x-5=17
Solving them would make:
3y+5=14
3y=9
y=3
2x-5=17
2x=22
x=11
For 9, you have to remember that the angle opposite of one angle in a defined parallelogram are congruent. Thus:
130=2h
5k=50
solve them and you get
h=65
k=10
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Hope that helped. Good luck.
|x - 5| = 16
x - 5 = +/- 16
x - 5 = 16 or x - 5 = -16
<u> +5</u> <u>+5 </u> <u> +5 </u> <u>+5 </u>
x = 21 or x = -11
Answer: x = {21, -11}
