A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.
3x-y=0 → 2(3x-y=0) = 6x - 2y = 0
5x+2y=22
The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22
B. So now we combine them:
6x - 2y = 0
+ + +
5x + 2y = 22
= = =
11x + 0 = 22 ← The answer
C. Now that we have the equation 11x = 22, we solve for x
11x = 22 ← Divide both sides by 11
x = 2 ← The answer
D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:
5(2)+2y = 22
10 + 2y = 22
2y = 12
y = 6
<u>Therefore, the solution to this problem is x = 2 and y = 6</u>
63 students. To solve, you first ned to determine the total cost for each student by adding 300 and 5, which equals $305. You then need to see how many times $305 will go into $19215, which shows how many students went. $19215 divided by $305 equals 65.
2)
A: 5
B: 9a & a (from a/6)
C: -5 & ÷4
D: -5 & 9a
I haven't done algebra in a year, so don't think my answers are perfect!
definitions:
term: something separated by a sign/symbol (÷, ×, - +) (a/6 are two separate terms, ÷)
constant terms: variables that can be solved.
unlike terms: terms that don't "go" together, you can't subtract 5 from 9a because there's a variable in the way (eyy that rhymes)
like terms: terms that you can add/subtract/multiply/divide to another term
(another answer to c is 9a & a)
7 to the power of 8 is 5764801 and 7 to the power of -4 is 0.00041649312 so these timed equals
2400.99995<span> </span>
I’m pretty sure that it’s the third answer (q^12)! :)
