The sum of three numbers is 11. .

Question

The sum of three numbers is

11

The sum of twice the first​ number,

3

times the second​ number, and

4

times the third number is

30.

The difference between

5

times the first number and the second number is

21.

Find the three numbers.

Answer 1

x = 5, y = 4, z = 3

You have 3 unknown numbers x, y, and z in a system of equations

From the information given

Equation 1

x + y + z = 11

Equation 2

2x + 3y + 4z = 30

Equation 3

5x - y = 21

Well there is already one equation available in two unknowns

5x - y = 21

If you eliminate z from the other two equations you can work with two unknowns and solve for the other variables

If we multiply the first equation by negative 4 we have

-4x - 4y - 4z = -44

If we add this to the second equation we will eliminate z

2x + 3y + 4z = 30

-4x - 4y - 4z = -44

We get

-2x -y = -14 another equation in x and y

We can work with the equation above and equation 3 to solve for x

5x - y = 21

-2x - y = -14

Mulitiply Equation 3 by negative 1

-5x + y = -21

Combine it with -2x - y = -14 to eliminate y

-5x + y = -21

-2x - y = -14

-7x = -35

Divide both sides of the equation by -7 to give x

x = 5

Substitute this value for x in the original equation 3 to solve for y

5x - y = 21

5(5) -y = 21

25 - y = 21

Subtract 25 from both sides of the equation

-y = 21 - 25

-y = -4

Divide both sides of the equation by negative 1 to solve for y

y = 4

To find z we can substitute the values for x and y in equation 1

x + y + z = 11

5 + 4 + z = 11

Combine like terms

9 + z = 11

Subtract 9 from both sides of the equation to give z

z = 11 - 9

z = 2

You can check the values in all the equations

Equation 1

x + y + z = 11

5 + 4 + 2 = 11

11 = 11

Equation 2

2(5) + 3(4) + 4(2) = 30

10 + 12 + 8 = 30

30 = 30

Equation 3

5x - y = 21

5(5) -4 = 21

25 - 4 = 21

21 = 21

I hope you find this useful and of course there other ways to approach the problem.

Answer 2

Basically, what we will be doing is solving a system of three equations: Let's define the following:

x: first number

y: second number

z: third number

Here are the equations we get from the three sentences:

(1) 1st sentence: x + y + z = 11

(2) 2nd sentence: 2x + 3y + 4z = 30

(3) 3rd sentence: 5x - y = 21

Let's take the 1st equation and solve for x:

(4) x = 11 - y - z

If we substitute this back into the other two equations, we get:

(5) 2(11 - y - z) + 3y + 4z = 30

and

(6) 5(11 - y - z) - y = 21

If we do the distributive property on both equations, then get both equations to where the variables are on the left and constants on the right, we will have:

(7) y + 2z = 8

(8) -6y - 5z = - 34

We can use the elimination method by multiplying the top equation by 7 to give us:

(9) 6y + 12z = 48

(10) -6y - 5z = - 34

Now if we add equations (9) and (10), the y-terms will cancel out, giving us:

7z = 14

Thus,

z = 2.

If we plug z = 2 back into equation (7), we get:

y + 2(2) = 8

y + 4 = 8

y = 4

Finally, we can plug z = 2 and y = 4 back into equation (4) to give us

x = 11 - (4) - (2)

x = 5

Thus, our solutions are

x = 5

y = 4

z = 2