The sum of two number is 9 and three time one of them is less then for time the second by one find the largest number
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Answer:
⇒ Let one number be x
, then other number will be
9−
x
According to the question,
⇒
x
1
+
9−x
1
=
2
1
⇒
x(9−x)
9−x+x
=
2
1
⇒
9x−x
2
9
=
2
1
⇒
2(9)=
9x−
x 2
⇒
18−=
9x−
x 2
⇒
x 2
9x+
18=
0
⇒
x 2
6x−
3x+
18=
0
⇒
x(x−
6)−
3(x−
6)=
0
$
⇒
(x−
6)(x−
3)=
0
⇒
x−
6=
0
and
x−
3=
0
∴
x=
6
and
x=
3
⇒ One number is 6 and other number is 3
⇒ The sum of squares of the numbers
=
(6)
2
+
(3)
2
=
36+
9=
45
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