Given:
The sum of three integers is 244. The sum of the first and second integers exceeds the third by 48. The third integer is 36 less than the first.
Aim:
We need to find the values of all three integers.
Explanation:
Let x be the first integer.
Let y be the second integer.
Let z be the third interger.
The sum of three integers is 244.
[tex]x+y+z=244[/tex]
The sum of the first and second integers exceeds the third by 48.
[tex]x+y=z+48[/tex]
The third integer is 36 less than the first.
[tex]z=x-36[/tex]
Substitute z=x-36 in the equation x+y=z-48 .
[tex]x+y=x-36+48[/tex]
[tex]x+y=x+12[/tex]
Subtract x from both sides of the equation.
[tex]x+y-x=x+12-x[/tex][tex]y=12[/tex]
Substitute z=x-36 and y=12 in the equation x+y+z=244.
[tex]x+12+x-36=244[/tex]
Add 24 to both sides of the equation.
[tex]2x-24+24=244+24[/tex]
[tex]2x=268[/tex]
Divide both sides by 2.
[tex]\frac{2x}{2}=\frac{268}{2}[/tex][tex]x=134[/tex]
Substitute x=134 in the equation z=x-36
[tex]z=134-36[/tex][tex]z=98[/tex]
We get x=128, y=12 and z =98.
Final answer:
first integer = 128
second integer =`12
third integer = 98.
