Use a system of equations. The sum of three Integers Is 107. The sum of the first and second Integers exceeds the third by 75. The third integer is 74 less than the first. Find the three integers.

Question

Find the 3 integers.

Answer 1

I would like to explain the solution to this problem in a bit unconventional way. Think of X, Y, Z as 3 different people. We can interpret the first equation as: the combined income of X, Y, and Z is $107. The second equation is: X and Y together make $75 more than Z. The third, and last equation can be determined as: Z's income is $74 less than X. Therefore,

EQTN 1: x + y + z = 107

EQTN 2: x + y = z + 75

EQTN 3: z = x - 74

Now the question becomes: can we find the incomes of each person?

First, we will take EQTN 3 and substitute it into EQTN 2. In other words, replace the Z in EQTN 2 with (X-74).

x + y = x -74 + 75

x + y = x + 1

Subtracting X from both sides results in the expression Y = 1.

We will now adjust EQTN 2, replacing Y with it's actual value of 1: x + 1 = z + 75

Subtract 1 from both sides to get: x = z + 74

Take the adjusted EQTN 2 and substitute it into EQTN 1: z + 74 + 1 + z = 107

z + 75 + z = 107

2z + 75 = 107

2z = 32

Divide both sides by 2 to get the expression Z = 16.

Now, solve for X by replacing Y and Z in EQTN 1 with the actual values of Y and Z

x + y + z = 107

x + 1 + 16 = 107

x + 17 = 107

Subtracting 17 from both sides results in X = 90.

X = 90, Y = 1, Z = 16

Answer 2

A + B + C = 107

A + B - C = 75

A - C = 74

First we find B by subtracting the third equation from the second. B = 75 - 74 = 1

Next we add the first equation to the second. C and -C cancel out and get

2A + 2B = 107 + 75 = 182 and after dividing both sides by 2 we get A + B = 91

Since we know that B = 1, A = 91 - 1 = 90

Finally after substituting A = 90 into the third equation we are left with

90 - C = 74, which allows us to calculate C = 90 - 74 = 16

So the first number is 90, the second number is 1, and the third number is 16

Answer 3

Let's "translate" those three sentences into math, using x, y and z for our 3 integers

1. The sum of three integers is 107 becomes: x + y + z = 107
2. The sum of the first and second integers exceeds the third by 75 becomes: x + y -z = 75
3. The third integers is 74 less than the first becomes: z = -74 + x

(Let me know if you need more explanation of turning those sentences into equations!)

Now we have to solve those three equations. Use substitution to plug the last one into the first two, and combine like terms. You should get:

x + y + (-74+x) = 107 -> x + y -74 + x = 107 -> 2x + y = 181

x + y - (-74+x) = 75 -> x + y +74 -x = 75 -> y = 1

Oh look! One of them practically solved itself! y = 1. Then 2x + 1 = 181 gives x=90. Finally z=-74+90 gives z=16

Answer 4

Let's denote our unknown integers by the variables x, y, and z. We are told that the sum of these integers is 107, which gives us the equation

x + y + z = 107.

Next, we are told that the sum of first and second integers exceeds the third integer by 75. This gives us the equation

x + y = z + 75.

Lastly, we are told that the third integer is 74 less than the first, giving us a final third equation

z = x - 74.

All together, we have three linear equations for three unknown variables (which is a good place to be):

x + y + z = 107, (Equation 1)

x + y = z + 75, (Equation 2)

z = x - 74. (Equation 3)

There are lots of ways to solve this linear system. I'll show you one way. Equation 3 tells us z in terms of x. I'll use that equation to substitute x - 74 in for z in Equation 2. When we do this, Equation 2 becomes

x + y = (x - 74) + 75,

which simplifies to

x + y = x + 1.

Subtracting x from both sides, we find

y = 1.

We've determined one of our integers! Now, I'm going to play a similar game with Equation 1. Using Equation 3, I can substitute x - 74 for z in Equation 1. I can also substitute the value 1 in for y because of my result above. When I do this, Equation 1 becomes

x + (1) + (x - 74) = 107.

Simplifying this equation, I get

2x - 73 = 107.

Solving for x, I get

x = 90.

Now that I know the value of x, I can return to Equation 3 to determine z. I get

z = x - 74 = (90) - 74 = 16.

Thus, the integers must be x = 90, y = 1, and z = 16.