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Question 161503: The sum of two numbers is 15. Three times one of the numbers is 11 less than 5 imes the other. The answer is 7 and 8. I just don't know how to set it up to solve.
Found 3 solutions by Alan3354, gonzo, Electrified_Levi:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The sum of two numbers is 15. Three times one of the numbers is 11 less than 5 imes the other. The answer is 7 and 8. I just don't know how to set it up to solve.
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f = first number
s = second number
f+s = 15
3f + 11 = 5s
Use substitution
f = 15-s
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3*(15-s) + 11 = 5s
45-3s + 11 = 5s
56 = 8s
s = 7
f = 8
Answer by gonzo(654)  (Show Source):
You can put this solution on YOUR website! x = 1 of the numbers
y = the other number
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sum of the numbers equals 15 in equation form becomes
x + y = 15
three times one of the numbers equals 11 less than 5 times the other number in equation form becomes
3*x = 5*y - 11
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two equations to be solved simultaneously are:
x + y = 15
3*x = 5*y - 11
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in the second equation, subtract 5*y from both sides to get
3*x - 5*y = -11
equations are now both in the same form of a*x + b*y = c reproduced below.
x + y = 15
3*x - 5*y = -11
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multiply both sides of the first eqution by 3 to get
3*x + 3*y = 45
3*x - 5*y = -11
subtract the second equation from the first equation to get
0*x + 8*y = 56
effectively removing x from the equation to allow you to solve for y.
this becomes
8*y = 56
which becomes
y = 7
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original first equation is
x + y = 15
substituting 7 for y, we get
x + 7 = 15
subtracting 7 from both sides of the equation, we get
x = 15 - 7
which becomes
x = 8
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second original equation is
3*x = 5*y - 11
substituting 7 for y and 8 for x, we get
3*8 = 5*7 - 11
24 = 35 - 11
24 = 24
proving the values for x and y are correct.
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answers are:
x = 8
y = 7
Answer by Electrified_Levi(103) (Show Source):
You can put this solution on YOUR website! Hi, Hope I can help,
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The sum of two numbers is 15. Three times one of the numbers is 11 less than 5 imes the other. The answer is 7 and 8. I just don't know how to set it up to solve.
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We have to name the two numbers, we will name the first number "x", the second "y"
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There are two equations
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(1) The sum of two numbers is 15. Since we know the numbers ("x" and "y"), we can put the words as an equation
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The sum of two numbers( "x" and "y") is 15. =  =  , this is the first equation
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(2) Three times one of the numbers is 11 less than 5 times the other. We know the numbers ( "x" and "y" ), we can put the words as an equation
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Three times one of the numbers("x") is 11 less than 5 times the other("y") =  =  , this is the second equation
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Our two equations are
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(1)
.
(2)
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These are systems of equations, we can solve these problems several ways.
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This is the way I usually solve these problems
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First solve for a letter in both equations (doesn't matter which one, usually the easiest), we will solve for "x" in both equations
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(1) , to solve "x" we will move "y" to the right side
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=  =  , rearranging, = 
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, our first answer is 
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Lets solve "x" in our second equation
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(2) , to solve "x" we will divide each side by "3"
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=  = 
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, our second answer is 
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Our two answers are
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.
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Since both our answers equal "x", the two answers equal each other
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We can put them in an equation
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 , now we just solve for "y"
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= 
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We will cross multiply
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 =  =  , rearranging
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=  , we will use distributive property
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=  =  =  (since it was a negative "11")
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, we will move (-3y) to the right side
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=  =  , we will now move (-11) to the left side
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=  ,  , divide each side by "8",
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=  =  , rearranging, 
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, "y" =  , we can find "x" by replacing "y" with "7" in one of our two equations
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(1)
.
(2)
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We will use the first equation (replace "y" with "7")
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=  =  =  = 
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, , you can check by replacing "x" with "8", "y" with "7", in both of our equations
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(1) =  =  (True)
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(2) =  =  =  ( True )
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,
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Hope I helped, Levi
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