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Question 466533: Good day!:)
I would like to ask for some help regarding my math homework..It is about Word Problems..
Here's the problem:
Find two rational numbers such that the second number is five more than three times the first number. The sum of numbers is 19.
Can someone please help me?:D
Thank you so much for your help in advance!:)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a rational number is a number that can be expressed as the ratio of 2 integers.
examples:
5 is a rational number because it can be expressed as 5/1 or 10/2 or 15/3, etc.
.25 is a rational number because it can be expressed as 1/4, or 2/8, or 3/12, etc.
your problem states:
Find two rational numbers such that the second number is five more than three times the first number. The sum of numbers is 19.
let x and y be your rational numbers.
the second number is 5 more than 3 times the first number leads to the equation:
y = 3*x + 5
the sum of the numbers being 19 leads to the equation:
x + y = 19
you need to solve these equations simultaneously to get your answer.
since y = 3*x + 5 in the first equation, use this value of y to substitute for y in the second equation to get:
x + y = 19 becomes x + (3*x + 5) = 19
simplify by removing parentheses to get:
x + 3x + 5 = 19
combine like terms to get:
4x + 5 = 19
subtract 5 from both sides of this equation to get:
4x = 14
divide both sides of this equation by 4 to get:
x = 14/4
since you know the value of x, you can now solve for the value of y.
use the equation x + y = 19 to get:
14/4 + y = 19
subtract 14/4 from both sides of this equation to get:
y = 19 - 14/4
put everything on the right side of this equation under a common denominator of 4 to get:
y = (4*19)/4 - 14/4 which becomes y = (76 - 14)/4 which becomes y = 62/4
your answer appear to be:
x = 14/4 which can be reduced to 7/2.
y = 62/4 which can be reduced to 31/2.
they are both rational numbers so all we need to do is confirm that they are the correct solutions for both equations.
the first equation is:
y = 3*x + 5
substituting for x and y, we get:
31/2 = 3 * (7/2) + 5
this becomes 31/2 = 21/2 + 10/2 which becomes 31/2 = 31/2, which is true, confirming the values of x and y are solutions for the first equation.
the second equation is:
x + y = 19
substituting for x and y, we get:
31/2 + 7/2 = 19 which becomes 38/2 = 19 which becomes 19 = 19, which is true, confirming that the values of x and y are solutions for the second equation as well.
since the values of x and y are solutions for both equations simultaneously, then they are good.
the first number is 7/2
the second number is 31/2
the sum of the first and second number is equal to 19 (7/2 + 31/2 = 38/2 = 19).
the second number is equal to 3 times the first number plus 5 (31/2 = 21/2 + 5 = 21/2 + 10/2 = 31/2).
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