SOLUTION: the sum of two numbers is twenty-one. Twice the larger is four times the smaller number. Find the two numbers
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Question 186024: the sum of two numbers is twenty-one. Twice the larger is four times the smaller number. Find the two numbers
Answer by swiemers(10) (Show Source): You can put this solution on YOUR website!
Hi!
To solve the problem, you first need to set up equations for what you know about the problem.
From the first sentence (the sum of two numbers is twenty-one), we could set up the equation:
x + y = 21
From the second sentence (twice the larger is four times the smaller number), we could set up the equation:
2x = 4y
(it doesn't matter which you assign as the "bigger" and "smaller" numbers)
Then we need to combine the two equations to get one of the variables to be eliminated. Before we do that, we want to get both of them in standard form, which is ax + by = c. The first equation is already in standard form, but the second is not. It would become:
4y - 2x = 0 (I just subtracted the "2x" from the left side to bring it over).
Then we need to line them up.
x + y = 21
-2x + 4y = 0
We need to get one of the variables to eliminate. I would suggest elinating the x's. To do so, we need to multiply the top equation by 2.
2[x + y = 21} becomes 2x + 2y = 42
Now we line the equations up again and combine all the like terms (add and subtract going straight down):
2x + 2y = 42
-2x + 4y = 0
0x + 6y = 42
The x cancels out, so you are left with:
6y = 42
Now you solve for y, so you divide both sides by 6 and get:
y = 7
Now that we know the y value, we plug this back into one of the equations to determine the x value. Let's take the first equation, which was:
x + y = 21
We plug in 7 for y and get:
x + 7 = 21
We subtract 7 from both sides and get:
x = 14
So now we have our two answers: x =14, and y = 7. You can plug both of them back into the equations given to check them.
7+14=21 CHECK
2*14=4*7 which is 28=28 CHECK
This was a little difficult to explain via email, so I hope it helps! Let me know if it doesn't! Good luck! Sandy :-)
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