Can you help me on a Algebra 2 problem?

For a weeding, Shereda bought several dozen roses and several dozen carnations. The roses cost $15 per dozen and carnations cost $8 per dozen. Shereda bought total of 17 dozen flowers and paid a total of $192. How many roses did she buy? (please show your work)

Can you help me on a Algebra 2 problem?
R + C = 17

15R + 8C = 192



C = 17 - R



15R + 8(17 - R) = 192

15R - 8R + 136 = 192

7R = 192 - 136 = 56

R = 8

C = 17 - R = 17 - 8 = 9



She bought 9 dozen carnations and 8 dozen roses.
Reply:Simultaneous equations

Roses= R

Carnations= C

R+C=17 (Because u have 17 dozen total)

15R+8C=192 (Dollar cost -15 bucks for roses, 8 for carnations)

So set them against each other and substitute



multiply the top 1 by -8 (to cancel the C's)

-8R-8C=(-136)

15R+8C=192

Combine and cancel

7R=56

R=8, go ahead and plug R back in to the equation



8+C=17

C=9



If you wanna check ur answers, these should work in teh second equation



8(15)+9(8)=192

120+72=192

192=192



So, the answers are right



You have 8 dozen roses and 9 dozen carnations
Reply:r + c = 17

so r = 17 - c



15r + 8c = 192

15(17 - c) + 8c = 192



255 - 15c + 8c = 192



-7c = -63

c = 9 dozen carnations



r = 17 - c = 8 dozen roses
Reply:R+C=17



15R+8C=192



SOLVE FOR ONE OF THE LETTERS

R=17-C PLUG IT IN FOR "R"



15(17-C)+8C=192

255-15C+8C=192 ADD OR SUBTRACT LIKE TERMS

-7C= -63

C= -63/-7 ??C=9 PLUG THIS FOR "C"



R=17-C??R=17-9= R=8 C=9