### The sum if the digit of two digit number is 13 while the sum of...

The sum if the digit of two digit number is 13 while the sum of their square is 89. Find the bigger number
(Pls show workings)?

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Let the tens place digit be x and the
units place digits be y
so by the condition given in the
question
x+y=13. ----1
x^2+y^2=89
by squaring both the sides of equation
1 we get
x^2+2xy+y^2=169
89+2xy=169
2xy=169-89
xy=40
so
(x-y)^2=x^2-2xy +y^2
=89-80
=9
(x-y)^2=9
by taking square root of both the sides
we get
x-y=3. ---3
by adding equation 1 and 3 we get
2x=16
x=8
y=5
so the number is 85
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Let the tens place digit be x and the units place digits be y

so by the condition given in the question
x+y=13. ----1

x^2+y^2=89

by squaring both the sides of equation 1 we get

x^2+2xy+y^2=169

89+2xy=169

2xy=169-89

xy=40

so

(x-y)^2=x^2-2xy +y^2

=89-80

=9

(x-y)^2=9

by taking square root of both the sides we get

x-y=3. ---3

by adding equation 1 and 3 we get

2x=16

x=8

y=5

so the bigger number is 8
Follow
Let the small no. be x and the big no be y
x+y=13...(1)
x^2+y^2=89...(2)
x=13-y
sub x in equ (2)
(13-y) (13-y) + y^2=89
169-13y-13y+y^2+y^2=89
2y^2-26y=-80
y^2-13y=-40
y^2-13y+40=0
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• Eddymontana: Factorize y= 8
sub y in equ 1
x=13-8 =5
1 month ago
Let the two digit be x and y respectively
x+y=13
x^2+y^2=89
frm eqn 1 y= 13_x
subtituting
x^2+(13_x)(13_×)=89
x^2+169_13x_13x+x^2=89
x^2+x^2_13x_13x+169=89
2x^2_26x+80=0
simplifying
x^2_13x+40=0
using _b+or _(b^2_4ac)/2a
=_ _ +13 square of + or _ (169_4×1×40)/2×1
=13 + or _ (9)/2
13+3/2=0 or 13_3/2=0
x=16/2=0 or 10/2 =0
x=4 or x=5
for x=4 ,4+y=13 y=13_4 y=11
for x=5, 5+y=13,y=13_5=8 y=8
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