Answer :
The three integers are 5,17 and -10
Step-by-step explanation:
Let x, y and z be three integers then
[tex]x+y+z=12 \ \ \Eqn\ 1[/tex]
'The sum of the first and second integers exceeds the third by 32'
[tex]x+y=z+32\ \ \ \ Eqn\ 2[/tex]
The third integer is 15 less than the first.
[tex]z=x-15\ \ \ \ Eqn\ 3[/tex]
From Eqn 2:
[tex]x+y=z + 32\\x+y-z=32[/tex]
Subtracting Eqn 1 from eqn 2
[tex]x+y-z-(x+y+z)=32-12\\x+y-z-x-y-z=20\\-2z=20[/tex]
Dividing both sides by -2
[tex]\frac{-2z}{-2}=\frac{20}{-2}\\z=-10[/tex]
Putting z=-10 in Eqn 3
[tex]z=x-15\\-10=x-15\\x=15-10\\x=5[/tex]
Putting x=5 and z=-10 in Eqn 1
[tex]x+y+z=12\\5+y-10=12\\y-5=12\\y=12+5\\y=17\\[/tex]
Hence,
The three integers are 5,17 and -10
Keywords: Linear Equations, Variables
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Answer:
First = 5, second = 17, third = -10
Step-by-step explanation:
First integer = x
The third integer is 15 less than the first, so
Third integer = x - 15
The second integer = y
The sum of the first and second integers exceeds the third by 32, so
The sum of the first and second integers = x + y
The third integer = x - 15
x + y is 32 more than x - 15, hence
x + y - 32 = x - 15
y - 32 = -15
y = 32 - 15
y = 17
The second integer = 17
The sum of three integers is 12, so
x + 17 + x - 15 = 12
2x + 2 = 12
2x = 12 - 2
2x = 10
x = 5
x - 15 = 5 - 15 = -10