4. You would need to set up equations. Two per problem. So 5x+2s=98. And 9x+2s=154. Then you would multiply one of them by a negative one to make the 2 student tickets cancel out. I would use the top one. So it turns to -5c-2s=-98. The -2c and 2c cancel out. You put the rest together to get 4X=56. And 56/4 is 14. So the senior tickets (X) are 14$. You would then replace X with fourteen one of the original equations. Again I'm going to use the first one. So 5(14)+2c=98. 5x14 is 70. Subtract 70 from both sides to get 2c=28. Divide both sides by 2 to get C=14. So the student tickets are 14$ and the seniors are also 14$.
5. Same thing on this one. Two equations. 8h+7i=84 and 3h+1i=25. Like going to multiply the second equation by -7 so the i's will cancel out to get -21h-7i=-175. So you will end up with -13h=-91. Divide both sides by -13 to get H=7. Replace H in one of the equations. I'll use second one. So 3(7)+1i=25. 3x7=21 so subtract 21 from both sides to get i=4. So the ivy is 4$ and the hostas are 7$.
Hope that helped a little bit!
2a = x+y move the terms
-x=y-2a change the signs
X =-y+2 write in parametric form
X=-y+2a
Answer:
Step-by-step explanation:
a93 = 93 + 2(52 − 1)
a93 = 93 + 2(52 + 1)
a93 = 52 + 2(93 + 1)
a93 = 52 + 2(93 − 1)
Originally 52 - Get rid of the first 2
a93 = 52 + 2(93 + 1)
a93 = 52 + 2(93 − 1)
2 each week. - Annoying. That doesn't help.
a93 = 52 + 2(93 + 1)
a93 = 52 + 2(93 − 1)
We're not buying any new ones in the first week, so purchases are one down.
a93 = 52 + 2(93 − 1)
I think we made it.
Answer:
i believe that would be -8...
Step-by-step explanation: