plese solve any meyhod
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5x + 7y=50 -----------------eq. 1
and,
7x + 5y=4 6------------------- eq.2
after multiplying by 5 in eq. 1,
5 ( 5x + 7y)=50×5
25x + 35y=250----------------------eq. 3
and,after multiplying by 7 in eq.2
7 ( 7x + 5y)=4 6 × 7
4 9x + 35y =322---------------eq.4
subtracting qu. 3 from eq.4,
4 9x + 35y =322
-25x - 35y =-250
-------------------------
24x=72
x= 72/24
x= 3.
After putting the value of x in eq. 1,we get,
5x + 7y=50
5× 3 + 7y=50
15 + 7y=50
7y=50- 15
7y= 35
y= 35/7
y=5.
so the value of x= 3and y=5.
and,
7x + 5y=4 6------------------- eq.2
after multiplying by 5 in eq. 1,
5 ( 5x + 7y)=50×5
25x + 35y=250----------------------eq. 3
and,after multiplying by 7 in eq.2
7 ( 7x + 5y)=4 6 × 7
4 9x + 35y =322---------------eq.4
subtracting qu. 3 from eq.4,
4 9x + 35y =322
-25x - 35y =-250
-------------------------
24x=72
x= 72/24
x= 3.
After putting the value of x in eq. 1,we get,
5x + 7y=50
5× 3 + 7y=50
15 + 7y=50
7y=50- 15
7y= 35
y= 35/7
y=5.
so the value of x= 3and y=5.
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