I am thinking of whole numbers. when I add them, their sum is 123. When I subtract the three lesser number from greater number, their difference is 45?
what are my numbers?
i am having trouble going step by step with this.
i added 61+62=123. i think this is wrong
please help me
thanx
I think you have a typo
If by "When I subtract the three lesser number from greater number" you meant
"When I subtract the lesser number from greater number" then the question is easy
x + y = 123
x - y = 45
add them
2x = 168
x = 84
sub that back in the first equation,
y = 39
the two numbers are 84 and 39
check: 84+39 = 123
84-39 = 45
Yeah!
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Please answer this question
Find the interquartile range (IQR) of the data
98 132 295 68 176 89 287 198 94 68 240 94 369 94 155
To solve this problem step by step, let's denote the three numbers as x, y, and z, with x being the greatest number.
1. First, we know that the sum of all three numbers is 123. So, our first equation becomes:
x + y + z = 123
2. We also know that when we subtract the three lesser numbers from the greater number, we get a difference of 45. So, our second equation is:
x - (y + z) = 45
To solve these equations simultaneously, we can use the method of substitution or elimination. Let's use the substitution method:
From equation 2, we can rearrange it to get:
x = 45 + y + z
Now, substitute this expression for x in equation 1:
45 + y + z + y + z = 123
Combine like terms:
2y + 2z = 123 - 45
2y + 2z = 78
Divide both sides by 2:
y + z = 39
Now, we have two equations:
y + z = 39 (Equation 3)
2y + 2z = 78 (Equation 4)
Next, we can solve equation 3 for one variable in terms of the other:
y = 39 - z
Substitute this expression for y in equation 4:
2(39 - z) + 2z = 78
Distribute the 2:
78 - 2z + 2z = 78
Combine like terms:
78 = 78
This equation is true. It means that any value for z satisfies the equation. So, z can be any number. Let's choose z = 0 for simplicity.
Now, substitute z = 0 back into equation 3:
y + 0 = 39
y = 39
Finally, substitute z = 0 and y =39 into equation 1 to find x:
x + 39 + 0 = 123
x = 123 - 39
x = 84
Therefore, the three numbers are x = 84, y = 39, and z = 0.