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!
Write an equation to find the number of each type of ticket they should sell. Let "x" be # of adult tickets; Let "y" be # of student tickets: Value Equation: 5x+3y=450- b. Graph your equation.y = (-5/3)x+150
c. Use your graph to find two different combinations of tickets sold. I'll leave that to you.
