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KatRina [158]
3 years ago
14

Help me with this please

Mathematics
1 answer:
alina1380 [7]3 years ago
6 0
If you assign variables to the problem, it can make things a lot simpler. Lets say chairs are x and tables are y. Therefore you have:
2x+6y=40 
5x+3y=25
Now you can isolate the variable of one equation and put it into another (it doesn't matter which. I'm going to manipulate the top equation to plug into the bottom one).
2x=40-6y 
x=20-3y
Now I plug into bottom equatioin:
5(20-3y) + 3y=25
100-15y+3y=25
100-12y=25
-12y=-75
y=$6.25
Now you can plug in y in either equation to get x. 
2x+6(6.25)=40
37.5+2x=40
2x=2.5
x=1.25
So it costs $6.25 for each table and $1.25 for each chair. If you think about it, it would make sense for the table to cost more for the chair. 
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3 years ago
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2 years ago
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