SOLUTION: find two numbers such that five times the larger plus three times the smaller is 47 and four times the larger minus twice the smaller is 20

Question 164824: find two numbers such that five times the larger plus three times the smaller is 47 and four times the larger minus twice the smaller is 20
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
lets call the numbers a and b with a being the larger and b the smaller.
5a+3b=47 and 4a-2b=20 you can solve by substitution or similtaneously. I will solve by substitution take the 1st formula and solve for a or b. I will solve for a so subtract 3b from each side and then divide by 5 and you get
a=(47-3b)/5 now substituting this for a in the 2nd formula we get (4(47-3b)/5)
-2b=20 if we multiply each term by 5 we get 4(47-3b)-10b=100 multiplying we get 188-12b-10b=100 simplify 22b=88 b=4 substituting in any original equation(I pick the 1st one we get 5a+3(4)=47 so 5a= 35 therefore a = 6
answer larger number a=6
smaller number b=4
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