The sum of three numbers is 14. the sum of twice the first number, 5 times the second number, and 6 times the third number is 74. the difference between 4 times the first number and the second number is -2. find the three numbers.

Let the three numbers be represented by a, b, c. The problem statement gives you three relationships:

"Elimination" is often used to solve a set of linear equations like this. Here, you can eliminate the c variable by subtracting the second equation from 6 times the first:

... 6(a +b +c) -(2a +5b +6c) = 6(14) -(74)

... 4a +b = 10 . . . . . simplify

Now, you can add the last equation to eliminate b.

... (4a +b) +(4a -b) = (10) +(-2)

... 8a = 8 . . . . . . simplify

... a = 1

Using the equation 4a+b=10, we can find the value of b:

... 4·1 +b = 10

... b = 6 . . . . . . . subtract 4

Now, we can find c using the first equation

... 1 +6 +c = 14

... c = 7 . . . . . . . . subtract 7

The three numbers are 1, 6, 7.

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Many graphing calculators are able to solve a system of linear equations like this. It can be helpful to get familiar with that process. The attached picture shows the solution on a TI-84 calculator. The numbers in the right column of the output are the solutions for a, b, c. The numbers that are input to the "rref" function are the coefficients in the equations in "augmented matrix" form.

don't know sorry bru

step-by-step explanation:

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.