x/2+y/5=5
2/x+5/y=5/6
2006-06-13 04:40:24 · 6 answers · asked by xavier h 1 in Science & Mathematics ➔ Mathematics
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2006-06-13 04:47:01 · update #1
x=65/18
y=115/18
2006-06-13 04:50:18 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
x=6 y=10
2006-06-13 11:45:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋
OK, first of all, get rid of the fractions by multiplying by a common denominator in each equation, then get one variable alone, and plug it into the other equation, then solve it for the other one. When I put in x= (50-2y)/5 from the 1st equation, I get y=10 or 15 from the 2nd one. Now putting y back into the 1st one, you should get that x=6 or 4.
2006-06-13 11:57:27 · answer #3 · answered by smartee 4 · 0⤊ 0⤋
x/2 + y/5 = 5; multiply both sides of equation by 10
5x + 2y = 50
2y = 50 - 5x
y = 25 - 5x/2
2/x + 5/(25 - 5x/2) = 5/6; multiply both sides of equation by x(25 - 5x/2)
(50 - 5x) +5x = (125x - 25x^2/2)/6
50 = 125x/6 - 25x^2/12; multiply both sides of equation by 12
600 = 250x - 25x^2; divide both sides by 25
24 = 10x - x^2
x^2 - 10x + 24 = 0
(x - 4)(x - 6) = 0
x = 4, 6
y = 25 - 5x/2
y = 25 - 5(4)/2 = 25 - 20/2 = 25 - 10 = 15 OR
y = 25 - 5(6)/2 = 25 - 30/2 = 25 - 15 = 10
Solution sets: (4, 15) and (6, 10)
2006-06-13 12:08:43 · answer #4 · answered by jimbob 6 · 0⤊ 0⤋
x = 2*(5 - y/5)
sub x in eq (2)
1/((25-y)/5) + 5/y = 5/6
(5/(25-y)) + (5/y) = 5/6
y +25 - y/(y(25-y)) =1/6
then
25y - y^2 =6*25
y^2-25y+150 =0
y= -(-25)+/-sqrt(25^2 -4*1*150)/2
and then use the calculator
\and sub y in eq 1 to get x
2006-06-13 11:54:30 · answer #5 · answered by allamiro 1 · 0⤊ 0⤋
5x+2y=50 or x=10-2/5y
12y=30x=5xy substituting 12y +30(10-2/5y=5(10-2/5y)y
12y+300-12y=50y-2y^2 or 2y^2-50y=300=0
Factoring 2(y-15)(y-10)=0
y=15 or 10
x/2+15/5=5 or x/2+10/5=5
x=4 or x=6
Solution, (4,15) and (6,10)
2006-06-13 12:11:58 · answer #6 · answered by ecra 2 · 0⤊ 0⤋