the sum of three intergers is 193.The smaller two are consecutive intergers and the larger two are consecutive even intergers.What are the three intergers?
wouldn't the smaller two be one apart, and the larger two would be two apart?
so
smallest --> x
second ---> x+1
third ---> x+3
then x + x+1 + x+3 = 193
solve for x
x+x+1+x+3=193
3x+4=193
-4 -4
3x=189
x=63
thank you!
3x+4=193 you don't know what x is so you cant go any farther i just got done with this unit
uhmm i think you would subtract 4 from both sides so that would just leave you with 3x
3x+4=193
-4 -4
3x= 189
then you would divide 3 from both sides
which leaves you with 63
x=63 =]
oh and make surewhen you do these promblems you go back and check it
4×4
To find the three integers, let's break down the given information step by step:
1. The sum of three integers is 193:
Let's assume the three integers as x, y, and z.
2. The smaller two are consecutive integers:
This means that x and y are consecutive integers. We can write this as:
x = y - 1 (since y is the larger integer between the smaller two)
3. The larger two are consecutive even integers:
This means that y and z are consecutive even integers. We can write this as:
y = z - 2 (since z is the larger integer between the larger two)
Now, let's put these equations together and solve for the variables:
x + y + z = 193
Substituting the equations for x and y we derived earlier:
(y - 1) + y + z = 193
2y + z - 1 = 193
2y + z = 194
Substituting the equation for y and z:
(2z - 2) + z = 194
3z - 2 = 194
3z = 196
z = 196 / 3
z ≈ 65.33
Since the integers are assumed to be whole numbers, z must be rounded to the nearest whole number:
z = 65
Substituting z back into the equation for y:
y = z - 2
y = 65 - 2
y = 63
Substituting y back into the equation for x:
x = y - 1
x = 63 - 1
x = 62
Therefore, the three integers are 62, 63, and 65.